'C
'I
Chapter 1 Arithmetic Sequences i"
/
I
Questions : 1.1-1.40
Concepts/ideas
• Number sequences arise from physical contexts • Arithmetic sequences as a special class of nuniber sequences • The difference of two trms in an arithmetic sequence is prop6rtional to the differ encein their positions ..
.
.
..
Q.1.1 Using rnatchsticks, a child makes an equilateral triangle, then .a square, then a regular pentagon and so on, increasing the number of sides by 1 at every stage. Write down the measure of a singl.e angle of each figure in succesion as a sequence. Write down the sum of the measures of angles of each figure in succession as a sequence. Which of the above is an arithmetic sequence? Why? • (d) In this arithmetic sequence, is the dierece of any two t;erms 900? Why? Score: 4, Time : 6 minutes
Concepts/ideas
• Number sequences arise from physical contexts f
Arithmetic sequences as. a special class of number sequences
• In an arithnietic sequence if the common differnce, one term and its position are • ., known, then the term in any other position can be computed • In an arithmetic sequence, if the first term and a. term in a specified position are known, then the sum up to that terni can be computed.
Q.1.2 Anu made an equilateral triangle, then a square, then a regular pentagon and so on, using sticks of the same length for the sides. She also formed the sequence of perimeters of these figures. Is Anu's sequence an arithmetic sequence? Justify The 15' term of this sequence is 85. What is the length of a stick Anu used? What is the total length of the sticks used to make the first 15 polygons?
Score: 5, Time
:
10 minutes
Concepts/Ideas
The general form of terms of an arithmetic sequence . Determination of the individual terms from the general term Determining the position of specified terms from the general form Q.1.3 Consider the arithmetic sequence, 184, 178, 172,1 166, (nth term) of the sequence.,.. (a) Write down the ,agebraic form (b) How many positive terms does this sequence have? (c) Which is the largest nëatiVe number in this sequence?
Score
:
5, Time 8minutes
Concepts/Ideas • Arithmetic sequence are formed by scaling and translating the sequence of natural numbers Finding the general form of an arithmetic sequence • Recognizing terms of an arithmetic sequence
.
• Finding the speciality of the sums of terms, of an arithmetic sequence Q.1.4 Riaz formed a sequence by multiplying the sequence 1, 2, 3, ... of . . natural numbers by 6. and adding 2. Prove that this is an authmetic sequence . . Write the algebraic form of this sequence. (c.Is 250 a.term of this sequence? Why? (d) The alternate terths of this sequence fonn two arithmetic sequences. 'What is the difference, of the sums of the first 20 terms of these sequences?
Score 5, Time 9 minutes
Concepts/Ideas ..
U
• Finding the speciality of the teirns of an authmetic sequence from its algebraic form • Forming an arithmetic sequence frOm its alehraic form • Finding the number of ternis between two specified terms, using the fart that in an arithmetic sequence, the term-difference is proportional to the position-difference
0.5
The n th term ofan arithmetic sequence is 8n + 3 What is the remainder got when the terms of this sequence are divided by 8? How many terms of this sequence lie between 100 and 550?
Score: 4, Time: 6 minutes
Concepts/Ideas .
. ...
..
.
U•.
•
:.
• Arithmetic sequences arising. frOrn physical coitexts . • Forming an arithtm1tié seqiIehce frin its aigebthiC fóirn • Finding the term in a specified position using the algebraic fprrn Q4.6An object falling down from a height travels 9.8n-4.9 metres during the th second. • (a) Write down tie distance travelled by the object during eery second as a sequence (b) During which second does it fall 63.7 metres?
Score : 2, Time: 4 minutes
.•.
Concepts/Ideas
•
U.
.
.
• Forming arithmetic sequences of specified common difference The difference of two trms in an arithmetic sequence is proportional to the differencein their positions. . ..
Q 1 7 Write clown an am ithmetic sequence with common difference 7 In this sequence, is the difference between any two teths equal to 175? JUstifr your answer. . Score : 2, Time.: 3 minutes
concepts/Ideas
.
. ...
• Speciality of the terms in an arithmnetic sequence
ti:ItiiitI • Recognizing terms of an arithmetic sequence Q.1.8 An arithmetic sequence with common differnce 2 has first term 10. Write down the sequence of perfect squares from this sequence. Is this sequence of squares an arithmetic sequence? Justify
Scôe: 3, time: 5 minutes
Concepts/Ideas • For every sequence, there is a rule determining its terms • The rule of formation of a sequence cannot be determined by inspecting a few terms Q.1.9 Find a rule of forming a sequence which gives the first three terms as 2, 3, 5. Write down the next two terms according to this rule. ...
Score 2, Time 4 minutes
Concepts/Ideas . • Algebraic form of an arithmetic sequëiice • Comparison of arithmetic sequences on the basis of their algebraic forms . Q 1 10 Two anthmetic sequences are given belov' .1 8, 15,22, .. Sequence I Sequence II: 21, 27, 33, 39, . ..
. .
. .
.
Write down the algebraic forms of each sequence . Are any two terms at the same position in these sequence equal? If so, what is this number?
Score: 5,Thne: lOminutes
Concepts/Ideas . .
• Recognizing and comparing arithmetic sequences • Comparing arithmetic sequences using their algebraic forms
Q.1.11 (a) Write two arithmetic sequences of natural numbers without any common terms (b) Prove that if two different arithmetic sequences have the same number at thsame position, there is only one such number.
Score: 4, Time 7minutes
Concepts/Ideas
• Sequences arising from physical contexts • Proportionality of two measures become a rule for forming an arithmetic sequence Q.1.12 Aluminium rods of lengths in the arithmetic sequence 10cm, 20 cm, 30cm and so on are cut out, all with cross-sectional area 0.5 square centimetre. Write down the sequnce of their volumes. Is it an arithmetic progressions? Why? The density of aluminium is 2.7gm/cc. Write down the sequence of the weights of these rods. Is it an arithmletic sequence? Why? Score: 4, Time: 8 minutes
Concepts/Ideas
• Arithmetic sequences can be formed by scaling and translating other arithmetic sequences • Comparison of arithmetic sequences using this idea • Comparison of sequences using their algebraic forms Q.1.13 (a) Write down an arithmetic sequence of common difference 4 (h) Write down the sequence got by multiplying the terms of this sequence by 3 and adding 2. What is the relation between the commOn difference of this sequence and the common difference of the original sequence? Prove that if the terms of an arithmetic sequence are all multiplied by the same number and the same number added to these, the resulting sequence also is an arithmetic sequence.
Score: 5, Time: lOminutes
Concepts/Ideas
• The relation between the terms and their positions in an arithmetic sequence • The difference of two terms in an arithmetic sequence is proportional to the difference in their positions
-
Q.1,14
.
..
.
.. ..
.. .
I
Write down an arithmetic sequence and then write down its tèrins at position 1. 4 1 7 and so on, starting with the first and skipping three every time. Do these numbers form an arithmetic sequence? a, b, c,... denote the positions of the terms of an arithmetic sequencC and x,, Xb, ' are the respective terms in these positions. if a, b, c, are in arithmetic séquece, piove that x,, x, x, ...are also in arithmetic sequence. ....
...
Score :5, Time: 8 minutes
Concepts/Ideas
.
. ...•
• If the term at a specified position of an arithmetic sequence and its common differeflc( are known, then any other term can he complyted • The difference of two terms in anarit'hmetic sequence is roportiona1.tb.the difference in their positions and the contmon difference is the constant of proportionality Q.1.15 The 8' term Of an arithmeti.c sequence is 53 and its 15th term is 102. What is the common difference of the sequence? What is the 25 " term of the sequence,? Score: 3, Time: 5 minutes
Concepts/Ideas
..
.
.
.,.
........
...
I.
• Common difference of an ari.ththetic sequenc e ........ .
,:.
.......
=.
.
..
..; . -
• Algebraic form of an arithmetic sequence ..
Q.1.16 Three numbers of the form x - 4, (a) What is the number x?
.
..
ax - 2, 4
2 are in arithmetic seunce .. .. (b). If this arithmetic sequence is continued, what is the nth term? What is the difference of its (ii - 5)thi term and the (n + 5)th term? ScoEe: 5, Time: 10 minutes
Concepts/Ideas
. . ....
.
• Speciality of the terms of an arithmetic, sequence • Finding the speciality of the terms using its algebraic form
11
Q.1.17 Write down the arithmetic sequence with first term and common difference Is any of its terms a whole number? Justif y
Score 4, Time: 8 minutes
Concepts/Ideas
• Various 'opera.tins on numbers result in sequences All properties of a sequence are encoded in its algebraic form • The difference of two terms in an arithmetic sequence is proportional to the difference in their positions
Q 1 18 A sequence is got by taking n = 1,2,3,
in the algebrai expression (n+2)' - (n - 2). Write down the terms of the sequence. Write down the simplified algebra
form of the terms of this sequence. At which, position 'does the number 144 occur in this. sequence? . .. . . . . .
Score.: 4, Time : 8 minutes
Concepts/Ideas
•, If terms of an arithmetic sequence are chosen by skipping the same number of terms, the resulting sequence also is an arithmetic sequence. • The algebraid form of an arithmetic sequence determines all its properties
Q.1.19 Consider the arithmetic sequence starting with
21
and common (lifference
Write down as a sequence, the whole numbers in this sequence Write down the positions of the whole number terms in a sequence (c.) What is the relation between the common differences of these two sequences with the common difference of the original sequence?
Score: 5, Time: 11 minutes
Concepts/Ideas
.
• Finding relations between the terms of an arithmetic sequence • Finding pairs of terms of the same ratio in an arithmetic sequence
Q.1'.20 I'he' 51h and the 8th 'terms of an arithmetic sequence are in the ratio 1: 7 (a), What is the ratio between the 6 t,11 and 15h1 terms? (b) The 7th term of this sequence is 10. What is the 22"< term?
Score: 4, Time 7 minutes
Concepts/Ideas • The sum of consecutive natural numbers, starting with 1 Q.1.21 15 points are marked on a circle. How many chords can be drawn by joining these points in pairs?
Score: 3, Time: 5 minutes
Concepts/Ideas • Finding the sum of consecutive natural numbers, starting with 1 • Finding the sum of consecutive odd numbers, starting with 1 Q.1.22 Consider the pattern below: 22 _3=1 32 _63 42 _1O_6 .52 _15.lO 62 _ 21 = 15 Prove that if the sum of the first n natural numbers is subtracted from the sum of the first n odd numbers, we gt the sum of the first (n - 1) natural numbers.
Scöre:5,Tiine: 9minutes
Concepts/Ideas • Sequences arising out of physical contexts • Sum of consecutive terms of an arithmetic sequence Q.1.23 A sphere moving along a straight line travels 4n + 3 metres during the m second. Write down the sequence of the distances travelled by the sphere, during every second . .. ... . .. What is the total distance travelled during the first 15 seconds?
Score: 3, Time 5 minutes
Concepts/Ideas • The sum of consecutive odd numbers
Q.L24 What is the sum of the fIrst 10 odd numbers? How much more than the sum of the first 10 odd numbers is the sum of the next 10 odd numbers? . Score: 5, Time: 8 minutes
Concepts/Ideas • Sum of consecutive natural numbers. Algebraic form of this sum
Q.1.25 Consider the natural mimbers written in the pattern below: 1 23 456 7 8 9
10
Write the next two lines of this pattern Write the sequence of the last numbers of the rows and find its algebraic form Write down the sequence of the first numbers of the rows and find its algebraic form Score: 5, Time : lOminutes
Concepts/Ideas • Finding the terms of an arithmetic sequexice from its algebraic form • Finding relations between sums of terms of arithmetic sequences
Q.1.26 The nth terms of two arithmetic sequences are 671 + 2. and 6n - 2. What is the difference of the sums of the first 15 terms of these? Score: 3, Time: 5 minutes
Concepts/Ideas • Finding the algebraic form of the sum of terms of an arithmetic sequence
-
The specialities of the algebraic form of the sum H
Q.L27
.
. Write clown the algebraic expression for tlie sum of the flrt in tennis of the aritunetic sequence with first term f and commOn difference d, as a second degree po1ynoni.a1 in n The algebraic form of the sum of the first a terms of an arithmetic sequence is found as 3m2 + 2n ± 1. Is it correct'? Justify your answer. Score: 4, Time: 6rninutes
Concepts/Ideas
• Forming an arithmetic sequence froiñ its algebraic form • Fu1dmg the algebi aic form of the ',uni of consecutive t erins of an an it hmct ic sequence • Finding peculiarities of the sequence from this algebraic expression term of an arithmetic; sequence is Sri —4. Prove that any suril of consecutive Q1.28 The terms of.this.sequence, starting from the first, is a perfect squate.. Score : 3, Time 6 minutes
Concepts/Ideas • Use of the sum of teims of an autlimetic 'equene in piactmcal context Q.1.29 The cost of digging a. e1l is 1000 rupees for the first metre and it increase by 50 rupees for every additional metre. How much would be the cost of digging a well 16 metres deep? How munch more than the first 8 metres is the. cost for the next 8 metres'? Score : 4, Time : 7 minutes
Concepts/Ideas
.
• Sum of conseutivC natural numbers .
. . .
. .. .
..
. .
• Finding the sum of consecutive terms :ofan arithmetic sequenceusing. sums of consecutive natural numbers
Q.1.30 Find the sum of all natural iiumbers from 1 to 20. The term of an arithmetic sequence is 4n + 3. Find the sum of its first 20 terms Score: 4, Time 7 minutes
Concepts/Ideas • The difference of two terms in an arithmeticsequence is proportional to the difference in their positions and the common difference is the constant of proportionality • Relation between sum .of terms and the middle term of an arithmetic sequence Q.1.31 The sum of 15 terms of an arithmetic sequence is 495 and the 5th term is 21 What is the 8th term of the sequence? What is the common difference? Find the sum of the first n terms of the sequence Score : 5, Time : 10 minutes
Concepts/Ideas • The sum of consecutive terms of an arithmetic sequence Algebraic form of such a sum Q.1.32 Prove that the difference of the sum of the first n terms and the sum of the iiext n terms of an arithmetic sequence with common difference d is n2 d Score: 5, Time: 10 minutes
Concepts/Ideas • Sum of consecutive terms of an arithmetic sequence • Relation between such a sum and the middle term Forming an arithmetic sequence using two terms and their posi.ions Q.1.33 All students in a class drew polygons of 9 sides with angles in arithmetic sequence Anu says one angle of all these polygons would be the same. Do you agree with her? •Justify Anup says the smallest angle in his polygon is 116°. What is the largest angle in his polygon?
Score : 5, Time: 8 minutes
Concepts/Ideas • Sum of consecutive terms of an arithmetic sequence Q.1.34 If 22 x 26 x 2 1 x
x 21 = (0 . 125)_ 24 , what is x?
Score: 4, Time: 8 minutes
Concepts/Ideas
• Sum of consecutive terms of an arithmetic sequence • Finding all terms of an arithmetic sequence from two terms • Forming an arithmetic sequence Q.1.35 The sum of the first 20 terms of an arithmetic sequence with common difference 6 is 1300 (a) Write down the sequence (b.) Find the sum of the first n terms of this sequence
Score 5, Te.:. 9 minutes
Concepts/Ideas
• Finding the first term and the cornrxion difference of Art arithmetic sequence Q.1.36 Consider the arithmetic sequence —150, —145, —140,
What is the next term? What is the common difference? Is 0, a term of this sequence?
Score: 3, Time: 3 minutes
Concepts/Ideas . Forming an arithmetic sequence from a physical problem and findirig a specified term of this sequence
•
•
•
.,..•..•
Q.1.37 Consider a sequence of equilateral triangles of sides 1 cm. 2 cm, 3cm and Write down the perimeters of these triangles as a sequence What is the common difference of this sequence? (c) What is the 21 term of this sequence?
SO
On.
Score.: 3, Time: 3 minutes
Concepts/Ideas Sum of specified number of terms of an arithmetic sequence Q.1.38 Find the sum of the first ten multiples of 10 Score: 2, Time: 3 minutes
Concepts/Ideas • Sequences consisting of a sin.1e number repeated are arithmetic sequences 0.1.39Writ&dówn therdmainders got by dividing consecutive odd numbers by 2, as a sequence. What is the algebraic form of this sequence? Score : 2, Time: 3 minutes
Concepts/Ideas • Computing the sum of consecutive terms of an arithmetic sequence, using sums of consecutive natural numbers Q.1.40 Using the fact that the sum of the first hundred natural numbers is 5050, compute the sum of the terms of the arithmetic sequence 3, 6, 9 1 ... from 3 to 300 Score: 2, Time : 3 minutes
.•
-
IQuestions: 2.1-2.43 Concepts/Ideas,...: • The angle gotby joining...two .ends...of a diameter of a. circle..to a. point, on the circle. is 90°; joining ends of a diameter to a point within the circle gives an angle greater -than 90° and to a point outside the circle gives an angle less than 90°. - 's O
ositeangles of -a
np1thnta
Q.2.1 In the quadri1teral ABCD, we have LA = 1000. LB = 110°, LC = 600 \i th respect to the'cucle drawn v itli dialneteE AC, 'what are the postion of the poinfs B 'and D? \. - 1 Can a ..ircle.be'rawn through the four vertices of this ' a.dtiléial?' Justify. 4.
'Score 3, Time 5 minutes
Concepts/Ide'as' • The angle got by joining the ends of a diameter of a circle to a point outside the circle is less than 90 0 .
Oppoiteangies: of,a cyclic
..
- .. ......
Q.2.2 Prove that if two vertices of a quadrilateral are both outside the circle drawn with a diagonal".s . diaméter. then the quadrilateral is not cyclic. Score: 3, Time: 5rninutes
-j
Concepts/Ideas
• The angle got by joining two ends of a diameter of a circle to a point on the circle is 90°; joining ends of a diameter, to a point within the circle gives an angle greater than 900 and to a point outside the circle gives, an angle less than 90°. •
• Opposite angles Of a cyclic quadrilateral are supplementary. Q.2.3 In the quadrilateral ABCD, we have LA = 80°,. LB = 1200 , LC = 100° (a.) Does the circle drawn with any of the diagonals of this quadrilateral as diameter pass through all the four vertices? Justify. (b) Can a circle be drawn passing through the four vertices of this quadrilateral? ,Justify. Score: 4, Time: 8 minutes
Concepts/Ideas
• The angle made by an are of a circle at any point on the opposite arc is half the central angle of the arc. B Q.24 In the figure, 'G.is' the centre of thefl:.. circle. (a) How. much is LOCA? b) Calculate .LAOC .. ( c.) Compute the three angles of '..
. .
(
40-
Score: 4, Time: 6 minutes.
Concepts/Ideas
.
.
• The angle made by an are of a 'circle at any .point on the opposite arc is half the . . central, angle of the arc. . . A. Q.2.51n ihe figure, 0 isthé center of the -. . circle' ., ..... Provethat LABC + LOCA 90 0 Taking the measure of LOCB as x°, calculate LEOC and LA
-V
Score: 4, Time: 6 minutes
I,
Concepts/Ideas
• Angles in the same segment of a circle are equal • Angle in a semicircle is a right angle Q.2.6 In the triangles below with the given dimensions, LA = LP also. R 601
B What is the circumradius of
P
ABC? Score: 4, Time: 6 minutes
Concepts/Ideas
• Properties of the parts Of intersecting chords of a circle • Angle in a semicircle is a right angle
Q.2.7 In the figure, 0 is the centre of the circle and AB is a diameter What is the length of PC? What is the length of PB? Calculate the length of OQ
w
A
2.I
Score: 5, Time: 10 minutes
Concepts/Ideas
• The angle mdë by ati are Of a circle at any point on the opposite are is half the central angle of the arc 1#J Q.2.8 In the figure, () is the centre of the circle What is the central angle of the arc ABc? Calculate LABC and LADC How much is LOAB + LOCB? D
Score: 4, Time: 5 minutes
Concepts/Ideas • The angle made by an arc - of a circle at any point on the opposite are is half the central angle of the are Q.2.9 Ammu draw an angle of 75° by first drawing an angle of 150° and t.hen drawing its bisector. But Appu drew such an angle using ruler and compass, without drawing an angle bisector. Draw such an angle by Appu's method Score: 4, Time .7minutes
Concepts/Ideas
• The angle made by an arc of a circle at any point, central angle of the arc
on the opposite, are is half the
Q.2.10 Draw. an angle of 11 0 using ruler and compass, without drawing any angle bisector Store
Concepts/Ideas
.
..
:
4, Time: 6minütes
,.. .
• The ahgle made by an are of a. circle at any point on the opposite arc is half the central angle of'the are Q.2.11
. . Draw a circle of radius 4.5 centimeters Draw a triangle of angles 30 0 , 70°, 80°, with its vertices on the circle Mea.sui e the lengths of the sides of the triangle and write them down Score: 4, The: 7minutes
Concepts/ideas
1
• Angles in thesame segment of a circle are equal • Opposite angles of a. cyélic quadrilateral are supplementary
.
...........\/'
Q.2.12 In the figure, A, B, C, D are points on the circle. How much is LACB? How much is LBCD? How much is LBAD? Calculate all the angle of the quadrilateral ABCD Score: 4, Time : 6 minutes
Concepts/Ideas Angles in the same segment of a circle are equal
up
Q.2.13 In the figure, A, B, C, D are points on the circle. Write down two pairs of equal angles in this.
Score: 2, Time: 2ininutes
Concepts/Ideas • Opposite angles of a cyclic quadrilateral are supplementary • If a circle is drawn through three vertices of a quadrilateral and if the fourth vertex is inside the circle, then the sum of the angles at that vertex and its opposite vertex is greater than 1800 Q.2.14 In the figure, ABCD is a cyclic quadrilateral and P is a point on its side AD. To prove that LABC + LAPC is greater than 180°, a student wrote the folIowiiig steps. Give the justification for each StC1) 180 0
ER
LABC + LADC = LAPC = .PDC + LPCD LABC + LAPC> 1800
Score: 2, Time: 2 minutes
Concepts/Ideas
. Recognizing cyclic quadrilaterals -Q.2.15 Find.out the cyclic quadrilaterals among the following classes of quadrilaterals: Rectangles
Squares
• Parallelograms
Rhombuses • flapeziums • Isosceles trapeziums :
Score : 2, Time 3 minutes
(FQ)
ITht
a
FHi1
Concepts/Ideas Properties of cyclic quadrilaterals •
• Locating a vertex of a non-cyclic quadrilateral with respect to the ëircle passing through the other three vertices
Q.2.16 In the quadrilateral ABCD, LA = 750, LB = 110°, LC = 90 0 (a) Where would be the vertex D with respect to the circle through the vertices A l B and C? Justify (b). Where would be the vertex C with respect to the circle through the vertices A l B and D? Justify Score : 3, Time: 6 minutes
Concepts/Ideas
• Properties of cyclic quadrilaterals
Q.2.17 In the figure, P and Q are the points of intersection of the circles and AR, CD are lines through. P and Q. with the points A. B, C, Don the circles. Calculate LAPQ and LBPQ How much is .LPQC? Is the quadrilateral ABCD cyclic? Jistify Score: 4, Time: 6 minutes
Concepts/Ideas
• Properties of cyclic quadrilaterals
Q.2.18 In the figure, P and Q, are the points of intersection of the circles and AB, CD are lines through P and Q, • with the points A, B, C, D on the circles. Which are the two cyclic quadrilaterals in the figure? Prove that if ABCD is a cyclic quadrilateral, then LA=LD Score: 4, Time : 6minutes
Concepts/Ideas
• Properties of cyclic quadrilaterals Q.2.19 In the figure, P, Q R, S are the points of intersection of the various circles and A, B, C, D the points on the circles such that P, R lie on AB and Q, S lie on CD
Which are the two cyclic quadrilaterals in the figure? Taking LA = x° , find LPQD and PQS in terms of L . Score: 4, Time: 8minutes
Concepts/Ideas .
.
.
• The vertices of a regular polygon divide the circumscribing circles into equal parts .. The angle made by an are of a circle at any point on the opposite arc is half the central angle of the arc • The sum of opposite angles of a cyclic quadrilateral are supplementary
Q2.20 In the figure, the vertices of the regular hexagon ABCDEF and the point P are on the circle. (a.) What is the central angle of the arc APB? What is the central angle of the opposite are of APB How much is LAPB?
A
Score: 3, Time: 5 minutes
-
Concepts/Ideas
.
• The angle made by an arc of a circle at any point on the ojposite are is half the central angle of the arc
Q.2.21 In the figure, the points A, B, P are on the circle centered at 0 and OA = AB (a) How much is LAOB? (h) What, is the centra.1 angle of the opposite arc of APR.. (c) How much is ZAPB?
.
B Score: 3, Time 5 minutes
Concepts/Ideas
• Angle in.a semicircleis a right aaugle • Angles in opposite segments of a::circic are supplémentaiy. . Q.2.22 .,
•
.
. in this figure, if a circle is drawn with AB as diameter, which of the points C, D, EwOuIdbe:iit? Whichof.them would be inside the circle? Which of them would be outside? Can a circle be di awn passm.g through •: A,B,('andE?Justify .• If a circle isdrawn with AB as a, chord and passing through D, which of the points C and E would be on it?
P
a
1100
A
B
..
..
.. •.,.- .
'o .
•
•. .
B
Score 5,Tmie 8mmutes
I1YIr'1HI I Concepts/Ideas
The angle made by an are of a circle at any point on the opposite arc is half the central angle of the are I,
Q.2.23 In the flgtire, the central angle of the are AXB is 1100 and the central angle of the arc CYD is 30° How much is LCAD? How much is LACB? Compute the angles of AAPC x Score: 4, Time: 6 minutes
Concepts/Ideas
Opposite angle in a cyclic quadrilateral are supplementary
Q.2.24 In the figure, A, B, C, D are points on the circle. Prove that AABC and AADC are congruent Prove that AC is a diameter of the circle Compute the area of the circle
C
A
Score: 5, Time.: 10 minutes
Goncepts/Ideas
.....
.... .
• The angle made by an are of, a circle at any point on the opposite arc is half the central angle of the arc .. Q.2.25 In the figure on the right below, the length of the are BDC is 18 centimeters What is the central angle of the arc
BDC?, What fraction of the circumference of the circle is the arc BDC? What is the circumference of the circle? What is the length of the arc QSR?
A
.
:.
.
. P
S
R .
B
c
Score: 5, Time 8 minutes
Concepts/Ideas
• Product relation between parts of chords intersecting within a circle Q.2.26 In the figure, A, B, C. D are points on the circle Write down two pairs of equal angles in the figure • (b) Are the triangles PAC and PB.D similar? Why? (c) Write two other qi.iotients of lengths in the figure which are equal to
A
PA TD
jJS .JVA
UtULL S Lt±t!SSS IJt WV _1L USS
ucts PA x PB and PC x PD?
-
C
Score: 4, Time: 6 minutes
Concepts/Ideas
• Product relation between parts of chords intersecting outside a circle Q.2.27 In the figure, A, B, C, D are points on the circle. The steps leading to the conclusion PA x B PB = PCx PD are givenbelow Write the justific ion for each step
A
P
(a)LPBC_—LPDA
LPAD and L\PBC are similar
D
(c)= Score: 4, Time: 6minutes
Concepts/Ideas
• Product relation - between parts.of chords intersecting inside a circle Q.228 In the figure j the chord AB ,bi-.. sects the chord CD Also, AB = 13cm and PB = 4cm What is the length of CD? If -a chord MN is drawn through',P with PM = 3cm. what would be the length of
D
•
A
•
P
B
C
Score: 4, Time: 6 minutes
........
./\........
Concepts/Ideas • Product relation between parts of chords intcrseting inside a circle Q.2.29 In the figure, the chord RS cuts the chord PQ in the ratio 3 2 and PQ cuts US in the ratio 8 11 at A. Also PQ 20cm What fraction of PQ is PA? Find the lengths of PA and AQ
What is the length of RS?
..
R
Score: 4, Thne: 6 minutes
Concepts/Ideas • Product relation between parts of chords intersecting outside a circle
Q.2.30 In the figure, A, B, C,D are points on the circle with PA = 4cm, AB 5cm, PC = 3cm. What is the length of CD? .
Score: 4, Time: 6 minutes
Concepts/Ideas
0
Product relation between partsof chords intersecting outside a circle . Q.2.31 From a point P outside : a circle, aline is dra*n, cutting the circle at C and D, with PC 9cm and CD 3cm. The shbttest distance from P to the circle is equa.l to the radius of the circle.. How long is the radius of the circle? as
1
Score: 3, Time: 6 minutes
rArrrr1ftfl
ii ri
Concepts/Ideas
-
.
• Product relation between parts of chords intersecting outside aircle Q.2.32 In the .fire A, B, C, D are the points where the circle cuts the tnangle. Also, PA = PC. Prove that the sides PQ and PR of the triangle are equidistant from the centre of the circle.
A B
D
R Score: 4, Time 6 minutes
Concepts/ideas Angles in the same segment are equal Angle in a semicircle, is a right angle
Q.2.33 The lengths of the sides of a triangle are a, b, c and its circurnradius is R. Prove that the area of the triangle is abc
4R
Concepts/ideas
'
. Construction of a square equal in area to a given polygon ..Q.2.34 Draw an equilateral triangle of sides 6 centimetre each. Construct a square of area equal to that of the triangle. Measure the length of a side. of the square and write it down.
Score: 5, Time lOminutes
Concepts/Ideas • Construction of a square equal in area to a given polygon
_
. .:.".:.. ...
iL
Q 2 35 Draw a square of area equal to that of the rectangle with dimensions as shown measure the length of a side of the square and write it clown. 6cm
Score 5, Time lOmmutes
Concepts/Id'a
.. .,
.
• Relation bétwëëñ ëëhtfal angles of an arc and its opposite arc
Q.2.36 Thr
s,ABCD, CDE are marked on a circle T. .ecentral angle of ABC is 60°. What is the central angle of its opposite arc? The central angle of BCD is 260° ks than the central angle of its opposite
arc. Find the central angles of these arcs The central angles of are CDE and its opposite are are in the ratio 1: 8. Find theseangles ';•:'.
.....
Concepts/clea
.
.-
.:,
.
..
.. ..
• Relation bctwceii angles in a cu dc Q.2.37 In the .figur. 0 is the center of the circle and CF is a diameter. The other points are all on the circle. Based on the figure, write the cpt underlying the following conclusions
D
A
(a)LA=LC0E (b)LALG
ii
(C) £1i=9U''
..
LA+LD=1°'
..
Score : 5, Time 10 minutes
Concpts/Ideas Product
. ..r:
.
910n parts of chords intersecting inside a circle
C
Q.2.38ln the figure, a semicircle is drawn on AB asdianieter Wha± is the length of PC? Explain, how we can draw a line of length /in thefigure.. .. .
... ..y
. ..
'
.'
p:
B
Score 4, Time : 7 minutes
Concepts/Ideas • Properties of a cyclic quadrilateral Q.2.39 In the quadrilateral ABCD, we have LA = 1000, LB = 70°, LC 50° The circle through A, B, D does not pass through C. Why? Without clmnging LA, by how i:iuch should LC be increased, so that this circle passes through C?
Score: 2 7 Time: 4minutes
Concepts/Ideas
Relations between central angle of an arc, angle in the opposite arc-, angle semicircle
ml. a
Q.2.40 Draw a circle of radius 3 centimeters and mark an arc of central angle 1000 Using ruler and pencil only, draw angles of measures 500, 130° and 90° in this circle Write down the mathematical principles used to draw-these
Score: 5, Time: 8 minutes
Concepts/Ideas • Product relation between parts ofadiameterand. a perpendicular chord
Q.2.41 in the figure, B and C are points on the line AD and semicircles are drawn with diameters AD and BD What is the length of cP? What is the length of QP?
A
B
C2cmD 6cm. • ............ 10cm
.........................................
Score: 4, Time.:. 8-minutes
Concepts/Ideas • Relations between are length, circumference 9f circle and central angle Q.2.42 A. piece of iron wire is bent to form an angle of 300 If the corner of this is placed at the centre of a circle, what fraction of the circle would he cut between the the two arms of the wire? If the corner is placed on the circumference of a circle, wha.t part of the circumference would be cut off? Score 3, Time : 3 minutes
Concepts/Ideas • Drawing a si.ecified geometric figure of area equal to that of a given figure Q.2.43
. Draw an equilateral triangle of side 8 centimeters Draw a rectangle equal in area to this triangle A rectangle is to be drawn with area equal to that of an equilateral triangle of side 20 centimeters and one side of length 20 centimeters. What should be the 1eigth of the other side of the rectangle? Score: 5, Time: lOminutes
r
1•
chapter 3 Second Degree Equations Questions: 3.1-3.22] Concepts/ideas
. Formulation and solution of second degree equations Q.3.1 When all sides ofaçuare Nvere reduced by 6 centimetres, the area becane 900 square ceitirnetres. (a) Take the length of a side of the original square as x centimetres and formulate an eq.iation. (h)' What i the legthof a side of the original square? What is the length of a side of the reduced square? Score : 3, Time : 5minutes
Concepts/Ideas
. Formulation and solution of second degree equations Q.3.. The hypotenuse of a right angled triangle is three more than twice the length of a hbrter side. The third side is one less than the hypotonuise. Take the length of the second side as x and write the lengths of the hypotemse and the third side in terms of x Compute the lengths of all three sides. Score: 4, Time: 7 rninutes
Concepts/Ideas
FoñnulàtiOn and sohition of second degree equations
concepts/Ideas • Solution of geometric problems through the solution of second degree equatiOns
Q.3.7 The perimeter of a rectangle is 44 centimetres and its area is 117 square centitnetres. (a) If the length of one side of the rectangle is taken as 11 + x, what is the length of the other side? (h) Compute the lengths of the sides of the rectangle
Score
Concepts/Ideas
;
3, Time
:
5 minutes
.
Solution of practical problems through the solution of second degree equations
Q.3.8 To complete a job, one company asked for five more clays than another company. When both of them did the job together, it was completed in six days. How many (lays would eaeh of tlieixi have, needed to do the job alone?
Score: 5, Time: 10 minutes
Concepts/Ideas to
Application of second degree equations in other brandies of mathematics
Q.3.9 Can the sum of the first few terms of the arithmetic sequence 8, 14, 20, . . . , be 280? Justify your answer
Score: 3, Time: 5 minutes
Concepts/Ideas • Using the method of solution of second degree equation in practical situations
Q.3.10 To get a resistance of 4S1 in an electric circuit, anelectrican connects two resistors in parallel. The resistance of one is 6l more than the other. Calculate the resis tance of each. The effective resistance R of two resistances .R1 and R2 connected in parallel is giveil by the equation -- = +
R R 1 R2 Score: 4, Time; 8iniuutes
Concepts/Ideas • Using the inel;hod of solution of secouçl degree e(ltlatiOfl in practical situations
Concepts/Ideas Solution of geometric probkxps through the solution of second degree equations
Q.3.7 The perimeter of a rectangle is 44 centimetres and its area is 117 square centitnetres.
-
(a) If the length of one side of the rectangle is taken as 11 + x, what is the length of the other side? (l.) Compute the lengths of the sides of the rectangle
Score : 3, Time: 5 minutes
Concepts/Ideas Solution of practical problems through the solution of second degree equations
Q.3.8 To complete a job, one company asked for five more days than another company. When both of them did the job together, it was completed in six days. How many iays would each of them have needed to do the JO1) alone?
Score: 5, Time: 10 minutes
Concepts/Ideas Application of second degree equations in other branches of mathematics
Q.3.9 Can the sum of the first few terms of the arithmetic sequence 8, 14, 20, :.., be 280? J .istify your answer
Score : 3, Time: 5 minutes
Concepts/Ideas • Using the method of solution of second degree equation in practical situations
Q.3.10 To get a resistance of 4Q in an electric circuit, anelectrican connects two resistors in parallel. The resistance of one is M more than the other. Calculate the resistance of each. The effective resistance R of two resistances R1 and R2 connected -=-+in parallel is givexi by the equation R R 1 R2
Score : 4, Time 8minutes
Concepts/Ideas • Using the xnel;hocl of solution of second degree equation in practical situations
I-
Q.3.11 An object thrown vertically upwards with a speed of 29.4 rn/s. The :lleight of the object in metres after t seconds is given by the expression 29.4t - 4.9t 2 How high would the object be after 6 seconds? How much time would it take for the object to reach a height of 44.1 metres? Score : 4, Time : 8 minutes
Concepts/Ideas • Understanding the nature of the solutions of a second degree equation, using the - discriminant Q.3.12 The common difference of an arithmetic sequence is 2 (a) Can the product of any two consecutive terms of this sequence be —5? Why? (h) Can the product of any two consecutive terms be —1? Why?
5 Score: 3, Time: 5 minutes
Concepts/Ideas • Solution of a problem by formulating a second degree equation • Determining the nature of the solution using the discriminant Q.3.13 The sum of a fiumber and its reciprocal is12. What is the number? Prove that the sum of a natural number and its reciprocal would not be
30
5 Score: 5, Time ilminutes
Concepts/Ideas
• Relations between second degree polynomials and equations Q 3 14 Can we get 0 from the polynomial x 2 - 3 i + 7 by taking 'r as some numnbei 7 Can we get 1fith the polynomial x 2 - 5x - 8 by taking x as some number? 5 Score 4, Time 8 minutes
Concepts/Ideas
. Relations between second degree polynomials and equations
Q.3.15 (a). Prove that whatever number be taken as x, we will not get a negative number from the polynomial p(x) = - 4x + 4. What number should be taken as x to get p(z) = 0? Can we get a negative number or 0 from the polynomial x 2 - Gx + 10 by taking x as some number? 5 Score : 4, Time : 6 minutes
Concepts/Ideas Solutions of problems involving second degree equations Q.3.16 In copying a second degree equation from the blackboard, a student made a. mistake and wrote the coefficient of x as 7, instead of —7. He found the solutions as —4 and —3 What was the correct equation? What are its solutions?
5 Score: 4, Time : .8 minutes
Concepts/Ideas . Solution of practical problems by formulating second degree equations Q.3.17 Fom the four corners of a rectangular sheet, small squares of the same size were cut off and the tabs folded upwards to make a box of length 21 centimetres and volume 2499 cubic centimetres. The width of the box is 10 centimetres more than the height. . Taking the height of the box as x, what is its width? Calculate the. height and width of the box Calculate the length and width of the original sheet 5 Score : 5, Time : 10 minutes
Concepts/Ideas • Geometrical visualization of the method of completing squares • Solution of problems by formulating second degree equations
Q.3.18 The figure shows a square and two rectangles joined together (a) To make this a square, what is the smallest geometrical figure to be joined to this? (b). What is the area of the figure so joined? (c) What is the length of a side of the new square so forme.I?
5 Score : 4, Time 7 minutes
Concepts/Ideas.
• Solving p.rrctical problems by formulating second degree equatior.is
.
Q.3.19 A rectangular plot of land has one side 4 metres longer than the other and its area is 32() square metres. Taking the length of the shorter side is x, what is the length of the lotiger side? Compute the lengths of the sides of this plot . . 5.
Score : 4, Time .6 minutes
Concepts/Ideas.; ..
• Relation between the terms in the same position of two arithmetic sequences with the same common difference ... • Solutions of problems by formulating second degree equations Q.3.20 Two arithmetic sequences have the game cominoli difference. One of them starts with 5, while the other starts with 8. What is the difference between their 111h terms? The product of the 11th terms is 2160 Find these numbers
Score 5, Time 8 minutes
Concepts/Ideas
.
.
.
.
. ...
.
..
s. Solving problems involving second degree equations by completion of SquareS
Q.3.21 When two consecutive even numbers are multiplied and 1 added to the product, weget225 Taking the smaller number as x, what is the other number? .., Write down a second degree equation in x. based on the given facts Find out the numbers . ... ..
Score : 5, Time: 7minutes
Concepts/Ideas
Solution of practical problems by formulating second degree equations
Q.3.22 The sum of the numbers giving the perimeter and area of a square is 192 (a) Taking the length of a side of the 'square as x, what is the area and perimeter? (h) Write down a second degree equation expressing the facts given (c) Solve the equation to find the length of a side of the square
Score: 4, Time: 6minutes
Chapter 4 Trigonometry Questions: 41-4.41 Concepts/Ideas -. In triangles with the same angles, side oppOsite equal angles are proportioiial Q.4.1 Are the triangle shown below similar? Why? 1(1 em
7x0 12cm
/
Calculate the lengths of the third skies of the triangles Score: 3, Time: 5 minutes
Concepts/Ideas • In triangles with the same angles, side opposite equal angles are proportional
Q.4.2 A ladder rests against an upright wall. A meter-scale is also placed against the wall, parallel to the ladder. The foot of the ladder is 6 metres away from the foot of the wall, while the foot of the scale is 75 centimeters away from the foot of the wall. Draw a rough sketch showing these details What is the length of the ladder? Score: 4, Time: 7 minutes
4:.;...
11
Concepts/Ideas A triangle with angles 30°. 600 and 900 has lengths of sides in the ratio 1:
: 2
Q.4.3 A circle is drawn with diameter AB and a point P is chosen on it with LABP = 60° and it is found that BP = 6cm Draw a. rough sketch showmg these details What is the len gth of A.P? What is area of the circle? =
Score
:
3 Time
5 minutes
Concepts/Ideas AtiiaiigIe \ith arigks 30°. 60° and 90° has lengths of sides in the ratio 1: NF3 2 Q.4.4 From the top of an upright electric-pole on a level ground, a. wire is stretched to the ground, making an angle 30 0 with the ground. The lower end of the wire is 12 metres away from the foot of the pole. Draw a rough. sketch showing these details Calculate the height of the pole (v'
1.73)
Score
Concepts/Ideas
..
:
3, Time: 6 minutes
.
• A triangle with angles 45°. 45° and 90° has lengths of sides in the ratio 1
1:
Q.4.5 The hypotenuse of all isosceles right angled triangle is 10 centimetres Calculate the lengths of its perpendicular sides What is the area of time tiiangle?
Score
Concepts/Ideas
.
:
2, Time
:
2 minutes
.
• A triangle with angles 450, 45° and 90° has lengths of sides in the ratio 1: 1 :
Q.4.6 A right angled triangle with another angle equal to 45° has one of the perpendicular sides 8 centimetxes long. How long is the hypotenuse of this triangle? What is the distance from the midpoint of the hypotenuse to the right angled vertex?
Score: 3, Time: 6 minutes
Concepts/Ideas A triangle with angles 45°, 45° and 90° has lengths of sides in the ratio 1: 1:
Q.4.7 A triangular card-board sheet has two sides of length 20 centimetres and 16 centimetres and an angle of 135° between them. Draw a. rough sketch and mark the given measures What is the perpendicular distance to the 20 centimetre long side from the opposite vertex? What is the area of the triangle?
Score: 4, Time: 9 minutes
Concepts/Ideas A triangle with angles .30° 60° and 90° has lengths of sides in the ratio 1:
V's: 2
Q.4.8 Find the area of the triangle shown 15cm
Score: 3, Time: 6 minutes •1
Concepts/Ideas • A triangle with angles 30°, 60° and 90° has lengths of sides in the ratio 1: v": 2
Q.4.9 The figure shows a sheet of paper cut in the shape of an isosceles trapezium What is the distance betweeri its parallel sides? VThat is the area of the sheet?
60° 10cm
Score: 4, Time: 10 minutes
Concepts/Ideas A triangle with angles 30°. 60° and 90° has lengths of sides in the ratio 1:
:2
Q.4.10 In the figure, a triangle and its circumcircle are shown. What is the radius of the circle?
Score: 4, Time:: 10 minutes
Concepts/Ideas • A t;riangle with angles 45°, 45° and 90° has lengths of sides in the ratio 1: 1: • A triangle with angles 30°, 60° and 90° has lengths of sides in the ratio 1:
2
C
Q.4.11 What is the length of A.B in the figure? What is the area of ABCD?
U
What is the radius of the circiirncircle of ABCD?
A Score: 5,. Time
:10m1fl1.ttes
Concepts/Ideas • A triangle with angles 45°, 450 and 90° has lengths of sides in the ratio 1 1
10cm
120 °
Q.4.12 What is the area of the rectangle shown?
Score: 4, Time: 9mhiutes
Concepts/Ideas
• A triangle with angles 300, 60° and 90° has lengths of sides in the ratio 1. :.
:2
• A triangle with angles 45°, 450 and 90° has lengths of sides in the ra.ti( 1: 1:
C
Q.4.13
B (a) Fiiidthe.lengths QA, OB, OC in the figure
(b)FindAB:BC
A
OI2cm..
P
Score: 5, Thne: liminutes
Concepts/Ideas
• A triangle with angles 30°, 60° and 90° has lengths of sides in the ratio 1: '/ :2 • A triangle with angles 45°, 45° and 90° has lengths of sides in the ratio 1: 1.:
Q4.14 Calculatethe lengths of the diagonals of the kite shown
Score : 4, Time: 8 minutes
Concepts/Ideas
• A triangle with angles 30°, 60° and 90° has lengths of sides in the ra.tio 1: v': 2
Q.4.15 The diagonal of a rectangle is 12 centimetres long and it makes all angle of 600 with one of the sides Draw a rough sketch and determine the lengths of the sides of the rectangle What is the area of the rectangle? Score: 4, Time: 6 minutes
Concepts/Ideas
• sine and cosine of an angle are numbers to measure the angle
A
Q.4.16 In the figure, what is the measure of LA? What are the lengths of AC and BC?
Score : 2, Time 4 minutes
Concepts/Ideas
• Sine and cosine of an angle are mmabers to measure the angle
S
R
Q.4.17 In the figure. PQRS is a square If the length of PQ is x, what is the length of SQ? Calculate cos 45°:
Q Score : 2, Time : 3 minutes
Concepts/Ideas • Inc and cosine of an angle are numbers to measure the angle •
A
Q.4.18 In the figure. if LA = x°, then how, much is sin c° and cos x° ? Amiiu says sin(90 -. x)° = cos 'v 0 What is your opir.uori? Justify II
6 cm
B
C
Score: 4, Time : 5 minutes
Concepts/Ideas Solving geometric problems using trigonometry Q.4.19 The sides of a rhombus are 7 centimetres long and one of its angles is 40°. Calculate its area (sin 40° = 0.643, cos40° = 0.766, tan 40° = 0.839) Score: 4, Time: 8 minutes 1
Concepts/Ideas Solving geometric problems using trigonometry A
Q.4.20 In the figure, BD is a diameter of the circle and]3G = a; also the radius of the circle is R (a.) Which angle in the figure is equal to LA?
(h) Viji says a = 2R. What is your sin A opinion? Justify Score: 4, Time: 8 minutes T
Concepts/Ideas • Solving geometric problems using trigonorrietry
..
.......................................
...................... ..
.
.
.
Q.4.21 In the figure, 0 is the centre of the circle, LAOB = 90 1 AB = 6crn, LBOC = 300 . What is the area of the triangle OAB? Calculate the area of the quadrilateral
B
OABC A
Score: 5, Time: 8 minutes
Concepts/Ideas
• Solving geometric problems using trigonometry ,ij
Q.4.22 In the figure, 0 is the centre of the circle, AC = 8cm, LA = 40° Wha.t is the area of 0BC? What is the area of 0AB? sin80° = 0.985, cos80° = 0.174, tan80° = 5.671 sin 40° = 0.643, cos40° = 0.766, tan 40° = 0.84
C
A
Score: 4, Time: 8 minutes
Concepts/ideas
Solving geometric, problems using trigonometry
In the figure, 0 is the centre of the circle, 9 .4.23 LAOB = 90 0 , LBOC = 400, AB = 5/ (a) What are the lengths of QA and OC? (h) What is the length. of BC? sin 40° = 0.643, cos40° = 0.766, tan 40 = 0.839
C
Score: 4, Time: 8 minutes
Concepts/Ideas • Solving geometric problems using trigonometry Q.4.24 In the figure, LBAC = 90°, AD = 6cm, CD=9cm,LACD=
(a.) What is tanx? How mw.th is L.BAD? What is the length of BD?
-
-
tN
Score 3, Time: 5 minutes
Concepts/Ideas • Solving geometric problems using trigonometry
Q.4.25 In the figure, AQB is an are of a circle centred at 0. Also, LAOB = 120° 1 LAOQ = 60 0 7 PQ = 3cm What is the radius of the circle?
I
Q
Score : 3, Time: 5 minutes
Concepts/Ideas • Solving practical problems using trigonometry Q.4.26 A boy stahding on the ground sees the top of a tree at an angle of elevation of 40°. After walking 20 metros towards the tree, he sees it at an angle of elevation of 80°. Draw a rough sketch and compute the height of the tree sin 40° = 0643, cos40° = 0.766, tan 40° = 0840 sin 80° = 0,985, cos4O° = 0.174, tan 40° = 5.671 Score: 4, Time: 8 minutes
Concepts/Ideas • Solving practical. problems using trigonometry
Q.4.27 Hari standing on the bank of a river, sees the top of a tower. 10 metres away fromthe opposit.e bank, at an angle of elevation of 20° .John standing on this bank, on the line joining the feet of the tower and Han, seés it at an angle of elevation of 40°. Draw a rough sketch and calculate the width of-the river - sin 40° = 0.643. cos40° = 0.766, tan 40 = 084 --
-
flij
- - -
-
Score : 4, Time 8 minutes
Concepts/Ideas • Solving practical problems using trigonometry
Q.4.28 Two buildings of different heights stand 16 metres apant. From the fo. of the taJler building the top of the shorter one is Seen at an angle of elevatioiibf5° and from the foot of the shorter building., the top -.f. tle a0er. biilcfing i. seen - at, an angle of elevation of 70° - -- (a) Draw a rough sketch (h) W hat is the hc ight of the shorter bmldmg 7 ( ) What is the height of the taller hmlding ,in 70° = 0.94., 70° = 0 312 tan 706 = 2,747
till
Score 4 1 Time 8 minutes
Concepts/Ideas - 'Slving pat1cl problerns using trigonometry
Q.4.29 A boy standing atop a ligl'ithoi.ie on the seashore sees a- ship at sea at'ai'i angl5 of depression of 20°. A man standing 100 metres away from the light ho!4s-.sees. boy at an angle of elevation of 45°. (a) Draw a rough sketth ba$ed on - tllese:1 pieces of information-(h) What is the height of the light house? (c) How fam fio the shore is the ship? tan 20° = 0.342 -
Concepts/Ideas, --
• Geometric problems involving sine and cosine
- --
- Score-: 4,Tiiné-:: 9minutes
A
Q.4.30 In the figure., LA = 40°, BC = 3cm Calculate the circumradius of LABC sin 40° = 0.64, cos400= 0.766, tan 40° = 0.84 Haritha says that if one angle of a triangle is 30 0 1 then its circumradius is equal to the side opposite this angle. What is your opinion? lus-. tif'y Score 4, Time 7 minutes
Concepts/Ideas
. Geometric problems involving sine and cosine A
Q.4.31 In the figure, C) is the circurncentre of AABC. Also, BC = 8cm, LB= 70°, LC = 40° Find the diameter of the circumncircle Find the length of AB sin 40° = 0.64, cos4O° = 0.766, tan4O° = 0.84 sin 70° = 0.94, cos70° = 0.34, tan40° = 2.75
VA
C
Score 5, Time 7 minutes
Concepts/ideas
.. Geometric problems involving sine and cosine Q.4.32 In APQR, we have PQ - 8cm and PS perpendicular to QR. Also, LQ = 20° , andLR=50°. . Draw a rough sketch. Find the length of QS . . Find the length of RS . sin 20° =0.3420, cos20° = 0.9397, tan 20° 0.640 Score 5, Time 8 minutes
Concepts/Ideas
.
.
...
Trigonometric measures to compute heights and distances
....
..
Q.4.33. A man standing on the deck of a ship, 5 metres above the sea level, sees the top of a light house at an angle of elevation of 700 and its foot at an angle of depression of 30°. (a) Draw a rough sketch (h) How far is the light house from the ship? (c) Calculate the height of the-light house Score :5, Time: 8minutes
Concepts/Ideas • Problem involving ratios of sides of triangle with angle 30°, 60°, 900 Q.4.34 The picture shows a still model made for the Math Festival, where ABCD is a rectangle with AD = 15cm and LBEC = 300 Find the length of the rectangle
D
E
C
Find the lengths of all sides of AAEB A
B Score: 5, Time: 7minutes
Concepts/Ideas Trigonometric measures to compute heights and distances
Q.4.35 When the sun is at an angle of elevation 45°, the length of the shadow of a tree is 6 metres. Calculate the height of the tree Calculate the length of the shadow when the sun at an elevation of 30° What is the difference in the lengths of the shadows?
Score: 5, Time : 8minutes
Concepts/Ideas Trigonometric measures to compute heights and distances
Q.4.36 A man on top of a tree ses a car, 30rnetrs away from the foot of the tree at an
•
angle of depression 30° (a) How hih from the ground is he sitting? (b) He climbs down a bit and sees the car at an angle of depression of 18°. How high is he now? sin 18° = 0.3090, cos 18° = 0.9511, tan 18° = 0.3249 sin72° = 0.9511, cos72° = 0.3090, tan 72° = 3.0777 Score:. 5, Time: Sminütes
Concepts/Ideas Solving geometric problems using sine and cosine In the figure, PQ = 10cm and PS is perpendici.ilar to RS. Also, LPQR = 1200 and LPRQ = 400 • •(a) How much is LPQS? Find the length of QS • (c) Find the length. of PR sin 40° = 0.6428, cos40° = 0.7660, tan 40° = 0.8391
Q.4.37
S
Q
Score : 4, Time: 7 minutes
Concepts/Ideas • Solving practical problems using tangent Q.4.38 A regular pentagon PQRST is drawn within a circle centred at 0 and OA is drawn perpendicular to PQ. The length of PQ is 10 centimetres. (a) How much is LOPA? (h) What is the length of PA? What is the length.of OA?
sin 54° = 0.809, cos54° = 0.588, tan 54° = 1,376 Score: 5, Time 8minutès
Concepts/Ideas
.
. Trigonometric measures to compute heights and distances Q;4.39 There are two buildings on either side of a 20 metre high towem', the feet of all
three on a line. A 1.6 metre tall man, standing atop the tower, sees the feet of the buildings at angles of depression 20° and 30°. (a) Draw a rough sketch (h) Calculate the distance between the buildings sin 200. = 0.34, cos 20° = 0.94, tan 20° = 0.36 sin 30° = 0.87, cos30° = 0.50. taii30° = 0.58 • . •
• Score .4, Time 6 minutes
Concepts/Ideas
• Solving geometric problems using sine and cosine Q.4.40 In the figure, the circumcentre of the isosceles triangle ABC is 0 and its circumradius is 6 centimetres. The midpoint of BC is D and the the central angle of the are APB is 1400 Calculate the length of the chord
A
P
AB
Calculate the perpendicular distance from A to BC sin 20° = 0.34. cos20° = 0.94, tan 20° = 0.36, tan 70° = 2.75 Score: 4, Time: 7 minutes
Concepts/Ideas
Solving geometric problems using sine and cosine .
1)
C
Q.4.41 In the figure, ABCD is a rectangle and PB = 7cm. Also, LAPD = 80° and LPBD=40°
Wh.at is the height of the rectangle? What is the length of the rectangle? sin 80° = 0.98, cos80° = 0.17, tan 80° = 5.67 sin40° = 0.64, cos40° = .0.77, tan 40° = 0.84
400
A
P
B
Score : 4, Time: 6 minutes
Chapter. 5 Solids Questions: 5.1-5.22
Concepts/Ideas Square pyramid
Q.5.1 To make a square pyramid of base 20 centimetres and slant height 24 centimetres, Bobby cut out a square and four isosceles triangles. What are the lengths of the sides of one such triangle? •.
..
.
. .
Score : 3, Time 5 minutes
•
Concepts/Ideas
..
Square pyramid
.•..
. . .
Q.5.2 The figure shows some of the measures of a lateral face of the square pyramid Meenu made. How, long are the base edge and the slant height of the pyramid?
Score : 2, Time : 3 minutes
Concepts/Ideas . Square pyramid
-
.
.
.....
.:T
..........••
. .••
Q.5.3 The figures show that the square and one of the triangles Rabia cut out to make a square pyramid
H 4.2 an
_ 42cm
Can she make a square pyramid with these? Explain the reason Score : 3, Time 5 minutes
Concepts/Ideas
• Surface area of a square pyramid Q.5.4 All edges of a square pyramid are of length 10 centimetres. Find its lateral surface area and surface area. Score
Concepts/Ideas
.
:
3, Time: 5minutes.
..
• Surface area of a square pyramid Q.5.5 Find the area of colored paper needed to cover a square pyramid of base edge 18 centimetres and lateral edge. 41 centimetres. Score: 4, Time 5minutes
concepts/Ideas.. . . • Volume of a square pyramid Q..5.6 A square pyramid is of base edge 24 centimetres and slant height 20 centimetres. Calculate its volume. . Score: 3, Time: 5 minutes
li Concepts/Ideas . Volume of a square pyramid Q.5.7 Ramya made a, square pyramid of base edge 10 centimetres and height 6 centimetres and Gopu made a square pyramid of base edge 5 centimetres and height 24 centimetres. Compute and compare their volumes. Score: 3, Time: 4 minutes
Concepts/Ideas • Cone Q.5.8 What are the radius and central angle of the sector needed to make a. radius 9 centimetres and height 12 centimetres?
COflC
of base
Score: 3, Time : 5 minutes
Concepts/Ideas . . .. .
.
• Volume of a square pyramid Q.5.9 A circular meta.l sheet of radius 12 centimetres is cut into 6 equal sectors and bent into cones. Calculate the slant height and base radius of one such cone. Score : 3, Time: 5 minutes
Concepts/Ideas
..
• Surface area of a cone ,Q.5.10 A sector of angle 120° is bent to form a cone. What is the ratio of the radius to the slant height of the cone? The curved surface area of such a cone is 1087t square centimetres. 'What are its slant height and base radius? Score: 4, Time: 6minutes
Concepts/Ideas • Concept : Surface area of a cone
Q.5.11 A circle of radius 15 centimetres is cut into two sectors and rolled into cones. What is the sum of their perimeters of their bases? What is the sum of the areas of their curve.I surfaces?
Score: 3, Time: 5niinutes
Concepts/Idea . Surface area of a cone
Q.5.12 The base radius of a wooden cone is 30 centimetres and its height is 40 centimetres. What is its slant. height? What would be the cost of painting 10 such cones, at the rate of 50 rupees per square metre? Score
5, Time
8 nunutes
Concepts/Ideas . Volume of a cone
Q.5.13 The base perimeter of a cone is lGrr centimetres and its slant.. height is 1.7centimetres. What is its height? Calculate its vohme.
Score: 3 1 Time 5 minutes
Concepts/Ideas • \To luine of a. COlIC
Q.5.14 A conica.l vessel of base radius 9 centimetres is used to fill a cylindrical can of base radius 12 centimetres and height 15 centimetres. The can was filled by 4 cones full ..• What is the height of the conical vessel? . .
Score: 4, Time: 7 minutes
Concepts/Ideas • Volume of a cone Q.5.15 Neena. and Sini made cones using sectors of radius 10 centimetres. Neena's sector had central angle 216° and Sini's sector had central angle .288°. Compute the volumes of both cones. Wijat is the ratio of their volumes?
Score: 5, Time : 10 minutes
Concepts/Ideas Volume of a sphere Q.5.16 The edges of a cube are 24 centimetres long. Find the volume of the largest sphere that can be cut from it.
Score
:
2, Time: 4 minutes
Concepts/Ideas Surface area of a sphere Q.5.17 What is the radius of a sphere of surface area 1447rsquare centimetres? What is the surface area of a sphere of half the radius of this sphere?
Score: 3, Time: 4 minutes
Concepts/Ideas
Volume of a hemisphere Q.5.18 TIie radii of two hemispheres are in the ratio 1 : 3. The volume of the smaller one is 50 cubic centinietres What is the volume of the larger one?
Score: 2, Time: 3 minutes
Concepts/Ideas s Comparison of volumes Q.5.19 A metal sphere of radius 6 centimetres is melt and recast into a cone of base radius 6 centimetres. What is the height of the cone?
Score : 3, Time: 5 minutes
Concepts/Ideas • Comparison of volumes
59
Q.5.20 A solid is made by joining a hemisphere and
--
cone of the same radius, as in the figure. The radiis of the hemisphere is 9 centimetres and the tctal height of the s1icl is 21 centimetres. What is the volume of the solid? ....
Score: 2, Time: 3minutes
Concepts/ideas
..
. ...
• \Tjflj'f L hemisphere
.
.
.
.
Q.5.21 The diameter of a hemispherical vessel is 60 centimetres. How many litres of water
can it
tmr/
-.
Concepts/ideas
.
.
..
.
Score.: 2, Time :. 4mhiutes
.
.. Surface area of a sphere and hemisphere Q.5.22 Two hemispheres of the same radius are joined to form a sphere...The, surface.
area of each hemisphere is 120 square centimetres. What is the surface area of the sphere? . . .. .. Score: 2, Time: .3mi!lut.es..
...___
Chapter .6 Coordinates
LQuestions: 6. 1-6 .ii Concepts/Ideas
• Concept Finding the coordinates of a point Q.6.1 In a coordinate system with 2 centimetres as the unit, of length, the coordinates of two points A, B are (2,4) and (6,8). With the same axes, if the unit of length is changed to 1 centimetre, what would be the, coordinates of these points? If the coordinates of a point C is (3,6) with respect to the unit 1 centimetre, what were its coordinates when the unit was 2 centimetres? Score: 3, Time: 5 minutes
Concepts/Ideas
.
.
Finding the coordinates of a point
Q.6.2 A circle is drawn with centre at (-1,0) and radius 5 units in a coordinate system. What are the coor4inates of the points at which it cuts the x-axis? And the points where. it cuts the y-axis?.
Score: 4, Time: 8minutes
Concepts/Ideas
• Finding the coordihates of a point
Q.63 A rectangle.'is .drawn with sides parallel to the coordinate axes and two of its vertices are (2,4) and (8,12). Draw the rectangle without drawing the coordinate axes and mark these vertices properly. Find the coordinates of the other two vertices. Also, find its length and breadth. What is the length of its diagonal? Score: 4, Time: 7 minutes
Concepts/Ideas
Finding the coordinates of a point Q.&4 In the equilateral triangle ABC, the coordinates of A are (-4,0) and the coordinates of 13 are (6,0). The perpendicular from C to AB meets it at P. Draw coordinate axes and a rough sketch of the triangle. What are the coordinates of P? What is the length of CP? What are the coordinates of C?
Score: 4, Time: 7 minutes
Concepts/Ideas
• Finding thecoordinates of a point V Q.6.5 In the figure, the semicircle with An as diameter passes through the point P. What are the lengths of OA
andOP? What are the coordinates of B?
Score: 3, Time 5 minutes
Concepts/Ideas
• Finding the coordinates of a point Y C Q.6.6 In the figure. OA = 6 units, LAOP = 45°,
LBOP=.15°. Whit are the coordinates of P? If a circle is drawn with 0 as centre and 013 as radius, wha;t would be the coordinates of the points where it cuts the x-axis?
o
6
A
Score: 3, Time: 6xninutes
Concepts/Ideas • Finding the coordinates of a. point;
Q.6.7. In A ABC, drawn above the x-axis. with. LA K 90 the coordinates of A are (2,0) • and those of-B are (8,0). The J.eigth.. pf.AC is .5 units and the area of the triangle is 12 square units. Draw a rough sketch of coordinate axes and the triangle. Find the coordinates of C . ... .. .
..
Score: 4, Thne: 8 minutes
Concepts/Ideas
.
• Finding the coordinates of a point
.. •. ...
Q.6.8 In .PQR, drawn above the z-axis with LP > 90°, the coordinates of P are (-4 7 3), those of Qarc (6, 3). The length of PR is 13 units and the area of the triangle is 60 squaT i unit', Draw a rough sk t( 11 of the triangh Find the coordinate', of R
Score: 5, Time : ilminutes
Concepts/Ideas .Findrngthe • coordintes
of a point Y E
Q.6.9 In the figure, ABCDEF is a regular hexagon with each side 6 units. Find the coordinates
of A, D, E
•, A
What is the radius of the circle whic.h, touch.s all ',ix sidcs of this hexagon?
.At
Concepts/Ideas • Coordinates of a. point on lines paraihal to axes
59
..
U
Q.6.10 From the points given below, find the pair which are on a line parallelto the x-axis and the pair which are on a line parallel to the y-axis: A(4. 3), B(3, 5). C(-6, 3), D(3, —2), E(5, 4)
ScOre: 2, Time: 2 minutes
Concepts/Ideas Distance between points on lines parallel to axes
Q.6.11 Wha.t is the distance between the points (-3, 2) and (4,2)? Find the coordinates of the points on the line joining them, at a distance 5 units from (4, 2) Score
2, Time
:
3 minutes
:
Concepts/Ideas Distance between points on. lines parallel to axes
Q.6.12 The coordinates of a point on a line parallel to the y-axis are (5,2). What is the distance between this line and the y-axis? Find the coordinates of the point where this line meets the x-axis. What is the distance between these two points?
Score: 3, Time: 4 minutes
Concepts/Ideas Distance between points on lines parallel to axes line joining (-2,5), (4,5) and the line joining (3,6), (3, —3) intersect at a Q.6.13 The point. What are the coordinates of this point?. What are the coordinates of the points on the first, line which are 6 units from this point of intersection?
Score
:
3, Time
•:
5 minutes
Concepts/Ideas . Distance between points on lines parallel to axes
Q.6.14 The pdint (-4,3) is on a line parallel to the x-axis and the point (6, —3) is on a line parallel to the y-axis. Draw a rough sketch based on these facts. What are the coordmates of the point of intersection of these lines? Which of the given points is farther from this point of intersection?
Score
:
4, Time
:
6 mnutes
Chapter 7
•
Mathematics of Chance Questions: 7.1-7.11
I Concepts/Ideas
. Probability as a number Q.7.1 There are 24 beads in a box, some white and some black. The probability of drawing a white bead from it is How many white beads are there in the box? How many black? How many black beads should be removed to make the PrObabilitY of drawing a white bead ? Score : 3, Time: 5 minutes
Concepts/Ideas
s Probability as a nutnber li
Q7.2 In the picture, P, Q, B, S are the midpoints of the sides of square ABCD If the length of a side of the larger square is a, what is the length of a side of the smaller square? If a point is marked oil the figure ithout looking, what is the probability of it to be within the small square?
R
C
N
S
A
P
B
Score: 3, Time: 5 minutes
Concepts/Ideas • Probability as a number Q.7.3 In a box there are 6 black beads and 8. white beads and in another box, there are 8 black beads and 6 white heads. From which box is it more probable to draw a black bead? If all the beads are put together in a box, which coloured bead is more probable to be drawn? What is the mimber denoting the probability of getting a white bead froixi this combined collection?
Score : 3, Time 4 minutes
Concepts/Ideas • Probability as a number Q.7.4 One says a three digit number. What is the probability of all digits beihg the same?
Score 3, Time 5 minutes
Concepts/Ideas • Counting techniques to compute probability Q.7.5 Prime nurners less than 20 are written in paper slips and put in a box. All natural numbers upto 10 are written in another set of paper slips and put in a second box. How many slips are there in the fiist box? How many slips in the second box have prime numbers on them? If one slip is drawn froin each box, what is the probability of both being primes?
Score
4, Time 6 minutes
Concepts/Ideas • Counting fechniques to compute probability Q.7.6 Two:bokes contán paper slips with numbers written on theii. One box contains 10 even numbers and 15 odd numbers; the other has 20 even and 30 odd. If one slip is drawn from each box, What is the probability of both being odd? What is the probability of at least one of them being odd?
Score : 4, Time: 6 minutes
Concepts/Ideas
Counting techniques to cOrriJ)ute probability Q.7.7 In Class 10A there are 20 boys and 15 girls and in Class lOB, there are 15 boys and 15 girls. One student from each class is to he selected for participation in the Math Fair. What is the probability of both being hoys? What is the probability of both being girls? What is the probability of one being a boy and the other agirl?
Score : 5, Time : 10 minutes
Concepts/Ideas
• Probability as number Q.7.8 In the rectangles.showu AE = 10.and EF = 4. Points are marked in it and it is found that the probability of a, point being marked within the rectangle ABCD. is.. What is the area of the rectangle ABCD? What is the probability of the point being marked within BEFC?
D C
F
A
E
B
Score 3, Time 4mmutes
Concepts/Ideas
. Probability as number Q.7.9A box contains 15 beads, some black and some white. The probability of drawing a black bead is known to be How many black beads are there in the box? If one black bead is removed, what would he the probability of drawing a black bead from the remaining beads?
Score: 3, Time: 5 minutes
i* nBankf1. ts/Ideas
robability as number one orange from each, the probability of Liatis the probability of at least one being
Two boXes contain oranges. In tal both being ripe is computed to be 18 unripe?
Score : 2, Time : 2 minutes
Concepts/Ideas
• Probability as miiniber Q.7.11 In selecting a. two digit number upto 50, what is the probability of the digit in the ten's place to he larger than the digit in the one's place? what, is the probability of the digit in the ten's place to be smaller than the digit in the one's pla.
Score: 3, Time: 5 minutes
68
Chapter 8 Tangents Questions: 8.1-8.32
Concepts/!deas • Tangent is perpendicular to the radius through the point of contact
Q.8.1 In the fijure, AB is the tangent at B of the circle centred at 0 How, mubh is•i()BA? How, much is LAOB?
B A Score : 2, Time: 2 minutes
Concepts/Ideas Line perpendicular to the radius through a point on the circle is a tangent
Q.8.2 In the figure, the semicircle with diameter OP intersects the circle centred at 0 at Q. Prove that PQ is a tangent to the circle centred at 0
Score 2, Time : 2 minutes
Concepts/Ideas
Line perpendicular to the radius through a point on the circle is a. tangent
Q.8.3 In the figure, 0 is the centre of the circle. Draw this figure with the given specifications
Score.: 2, Time 4 minutes
Concepts/Ideas
• Line perpendicular to the radius through a point on the circle is a tangent
Q.8.4 In the figure, 0 is the centre of the circle, 300. Draw this figBC = 6cm. LABC ure according to specifications anddraw the tangent at A.
B
6
A
C
))
Score: 3, Time: 2 minutes
Concepts/Ideas
. .. .
.
.
Tangent is perpendicular to the radius through the point of contact Q.8.5 In the figure, 0 is the centre of the circle and GA is the diarricter of the semicircle with GB = 1 cm and AB = 3 cm. The tangent to the circle at B cuts the semicircle atC (a) What is the length of BC'? (h) Find the angles of 0BC Score : 4, Time : 6 minutes
...
Concepts/Ideas Tangent is perpendicular to the radius through the point of cont act.
Q.8.6 Draw a cirle of radius 3 centimetres and draw the chord AB of length 4 centimetres in it. Draw the tangents at A and B
Score 4, Time; 6 minutes
Concepts/Ideas . Tangent is perpendicular to the radius through the point of contact Q.8.7 Draw a• circle of radius 3.5 centimetres and draw two non-perpendicular diameters. Draw the tangents at the ends of these diameters. What is time specialty of the quadrilateral formed by these tangents?
Score 5, Time : 8iuinutes
Concepts/Ideas
.
• Two tangents can be drawn from an external point
Q.8.8 Draw a circle of radius 3.2 centimetres and mark a point P, 8 centimetres froth its centre. Draw tangents from P to the circle and write (iowim the measurements of their lengths
Score 4, Time : 6 minutes
Concepts/Ideas Tangent is perpendicular to the radius through the point of contact . .
Q.8'.9 Draw two circles of radii 3 centimetres and 6 centimetres with the same centre. From a point on the larger circle, draw tangents to the smaller circle. Write dmvn the measurements of these tangents. .
Score 4, Time 6 minutes
Concepts/Ideas • Lengths of tangents from an external point are equal
71
-
Q.8.10 In the figure, 0 is the centre of the circle and AB is the tan-
gent at B on the circle, with OA = l3centirnetres and OB = 5 centimetres What is the Iength of AB? What is the length of the second tangent from A to the circle?
A
Score : 3, Time: 5 minutes
Concepts/Ideas
• Lengths of tangents from an external point are equal
Q.8.1I In the figure, 0 is the centre of the circle and the tangents at P and Q meet at R. Find the angles of APQR
Score: 3, Time: 4mhiutes
Concepts/Ideas
.
• The central angle of the smaller aic determined by two points on the circle and the angle between. the tangents at these pomts are supplementary
Q.8.12 In f he figtire, 0 is the centre of the circle and P, Q, R are points on it. Find the angles of the triangle formed by the tangents at P, Q, R.
R
P
Score: 3, Time: 4 minutes
Concepts/Ideas • The central angle of the smaller arc determined by two points on the circle and the angle between the tangents at these points, are supplemeiitar
Q.8.13 In the figure, 0 is the centre of the circle and AB, AC are tangents from A, with LBAC = 80°. Find LBOC and LBPC
Score: 3 Time: 4 minutes
Concepts/Ideas • The central angle of the smaller are determined by two points on the circle and the angle hetween the tangents at these points are supplementary Q.8.14 Draw a circle of radius 3.5 centimetres and draw an equilateral triangle with all three sides touching the circle. \Vhat is the circumradius of this triangle? Score 4, Time : 6 minutes
Concepts/Ideas • The central angle of the smaller arc determined by two points on the circle and the angle between the tangents at these points are supplementary Q.8.15 Draw a circle of radius 3 centimetres and draw a triangle with angles 60° and 70°, and all three sides touching the circle. Score: 4, Time: 6 minutes
Concepts/Ideas
• The central angle of the smaller are determined by two points on the circle and the angle between the tangents at these points are supplementary
Q.8.16 Draw a circle of radius 2 centimetres and draw a regular hexagon with all sides touclnng the circle.
Score : 4, Time: 6 minutes
Concepts/Ideas • Each angle between a chord and the tangent at one end is equal to the angle in the segment on the other side
P
Q.8.17 In the figure, PQ is a diameter of the circle and LAQR = 450 •
S
(a) Find L.PRQ, LPSQ, LQPR •. (b) Given that QR = 3cm, what is the radius of the circle?
a
A.
..
.
Q
B
Score : 4, Time :6 minutes
...
Concepts/Ideas • Each angle between a chord and the tangent at one end is equal to the angle in the segrneht on the other side Q.8.18 In the figure, ABC is a triangle with LA = 1.00° and AB = AC. Its incircie touches the sides at Q
P,Q,R How much are LB and LC? Calculate the angles of B
P
Score : 4, Time: 6 minutes
Concepts/Ideas • Each angle between a chord and the tangent at pne end is equal to the angle in the segment on the other side
Q.8.19 In the figure, the circumcircle of the regular pentagon ABCDE is drawn and also its tangent. PQ at the point A. How much is LPAE? -
P
A
Q
Score : 3, Time : 4rninutes
Concepts/Ideas
Relation, between tangent from an external point and portions of line cutting the circle at two points In
Q.8.20 In the figure, 0 is the centre, of the circle and f'Q is a tangent., with PQ = 12 cm and PR = 18cm. How much is PS? What is the radius of the circle?
R
I,
Score: 3, Time: 5 minutes
Concepts/Ideas
• Relation between tangent from an external point and portions of line cutting the circle at two points B
Q.8.21 In the figure. BC is a diameter of the circle and PC is the tangent at C. Also, A is a point on the circle with AC = 5 cm and LACP = 45° How much are LP and LB? Find the lengths of PC, BC, PB
.
.
X A
C
1'
Score: 5, Time: 7 minutes
Concepts/Ideas • Relation between tangent from an external point; ari4 : portions of line cutting the circle at two points Q.8.22 In the figure, the tangent from M to the circle touches it at N and two other lines through M cut the circle at A, B and C, D.
A
-
> 13 M
Alsc, MN =.12.m, MB = 9cm and MD = 8cm.
Which are products in the fig iire are equal to MA x MB? Calculate the lengths of AB and. CD Score 4, Time : 6 minutes
Concepts/Ideas • Relation between tangent from an external point and portions of line cutting the circle it two points . .. ... Q.8.23 in the circle the tangent from P to the circle toi.iches it at C and another line from P cuts it. at A and B. Also, i'A = 16cm, AB = 9cm, LCPB =
.
B
.. P
600
What is the length of PC'? Calculate the area of PCB Score 4, Time: 6 minutes
Concepts/Ideas • Relation between tangent from an external point and portions of line cutting the circle at two points ..
Q.8.24 In the figure, 0 is the centre of the circle and the tangent from B touches the circle at A. Also BD = 16cm and
.
-
D
/ o (
What is the length of AB? What is the radius of the circle?
A
..
B
Score: 4, Time: 6 minutes
Concepts/Ideas • The angle bisectors of a triangle meet at a. point
Q.8.25 In the figure, AD and CF are bisectors of the angles A and C. Is BE the bisector of LB? Why? B
.D
C
Score : 2, Time : 4minutés
Concepts/Ideas • Angle bisectors of a triangle pass through its incentre A Q.8.26 In the figure, the sides of zABC are tangents to the circle. Also, AB = 4cm, AC = 6cm, BC = 8cm. What are the lengths of BD and CD? B
D
C
Score : 3, Time : 5 minutes
Concepts/Ideas Incircie of a triangle
Q.8.27 Draw APQR with PQ = 6cm, PR = 6cm and LP = 65°. Draw its inciicle. Measure the inradius and write it down
Score : 4, Time: 8 minutes
Concepts/Ideas • Incircle of a triangle
77
ni
Q.8.28 In the figure. BC = 35cm, AC = 37cm and
A
LB = 90
How much is AB? Calculate the inradius of the triargie C
B
Score : 4, Time : 7 minutes
Concepts/Ideas
Incircie of
it
triangle
Q.8.29 Draw a triangle of sides 6 centimetres, 8 centimetres and 10 (:entilnetres and draw its inceiitre. Measure the mradius and write it down Score: 4, Time: 8 minutes
Concepts/Ideas
• Tangents drawn from an external point are of equal length
Q.8.30 In the figure, AB. BC, CD. DA are the tangents to tlie circle at P, Q, R, S. Prove that the perimeter of the qua.cirilateral ABCD is 2(AP + BQ + CR + DS)
C
A
PB
Score : 4, Time : 6 minutes
Concepts/Ideas
• The bisector of the angle formed by two tangents to a circle I)S through the centre of the circle Q.8.31 Draw quadrilateral ABCD with AB = 7cm, BC = 4.5 cm, AD = 6cm, CD = 5cm,LA=70° Draw the circle touclung the sides An, BC, AD. Is the side CD also a tangent to the circle? Score: 5, Time : 10 minutes
Concepts/ideas r
General properties of tangent to a circle
Q.8.32 In the figure, circles centred it A and B touch at C, with AC = 9cm and BC = 4 cm. The line PQ touches these circles at P andQ How rmich are angle APQ and LBQP? Calculate the length of PQ
Score: 5, Time: 10 minutes
79
Chapter 9 Polynomials Questions: 9.1-9.20
Concepts/Ideas
• Factors of a polynomial Q.9.1 Find the remainder on dividing the polynomial x 3 - 3w2 - x - 3 by .x - 2. What number added to x 3 - 3w2 - x - 3 gives a polynornia.I for which x - 2 is a factor? Score : 4, Time: 7 minutes
Concepts/Ideas
• Factors of a polynomial Q.9.2 Check whether x + 2 and x + 3 are factors of x 2 + 5x + 6 Score: 2, Time: 4 minutes
Concepts/Ideas
• Factors of a. polynomial Q.9.3 Find the remainder on dividing ax2 + bx + c by x - 1. If x - 1 is a factor of this polynomial, what is the relation between a, b, c? Score: 3, Time: 5 minutes
Concepts/Ideas
• Factors of a polynomial Q.9.4 Show that if b = a+c, then x+ 1 is a factor of ax2 + bx+c. Write down a polynomial
for which x + 1 is a factor. Score
:
3, Time
:
5 minutes
Concepts/Ideas
• Factors of a polynomial Q.9.5 Prove t.hata' —1 a.ndx+1 arefactorsof3x 3 -2x2 -3x+2. If3x 3 -2x2 -3x+2r = (3.2
- 1)(ax + b), what are a. and b9 Score: 5, Time 8 minutes
Concepts/Ideas
• F'aetors of a polynomial Q.9.6 x - 1 and x + I are factors of ax 3 + bx2 + cx + d
Prove that a = —c and b = —d \Vrite down a polynomial for which x - 1 and x + 1 are factors Score
:
5, Time
:
8 minutes
Concepts/Ideas
. Factorization of polynomials Q.9.7 Write x 2 - 7x - 60 as the product of two first degree polynomiais Score 3, Time: 5 minutes
Concepts/Ideas
• Factorization of polynomials Q.9.8 Prove that in the polynomial p(x) = x 2 + 3x + k
(a) if k = --4 1 then it has factors (h) if k = 4, then it has no factors Score : 4, Time : 8 minutes
Si
Concepts/Ideas
• Factorization of polynomials
Q.9.9 Write x 2 x - 1 as the product of two first degree polynomials Score: 4, Time: 6 minutes
Concepts/Ideas
• Factorization of polynomials
Q.9.10 If x - 1. and x - 2 are factors of x 3 - 6x2 - ax + h, what are a and b? Score : 5, Time 8 minutes
Concepts/Ideas
Factorizatiori of polvnormals Q.9.11 Find the remainders on dividing x 3.+ 6x2 + liz —6 by x + 1 and x + 2. If x + 1 and x + 2 are factors of x + 6x 2 + liz - 6 + k 7 then what is k?
Score
:
5, Time
:
8 minutes
Concepts/Ideas
• Factorization of polynomials Q.9.12 Which number added to the polynomial 2x 2 - 3x - 1 gives a polynomial with x - 1 as a factor?
Score : 3, Time: 5 minutes
Conepts/Ideas
• Factorization of polynomials
Q9.13 Write dowr a second degree polynomial with the coefficient of x as 1.. Check whether x - 1 is a factor of this polynomial
Score : 3, Time: 5 minutes
qBank Concepts/Ideas
• Factorization of polynomials Q.9.14 The solutions of x2 + ax + b = 0 are -3 and 5
Write x2 + ax + b as the product of two first degree polynomials What are a and b? Score : 3,
Time : 5 minutes
Concepts/Ideas
• Factors of polynomials Q.9.15
Find the remainder on dividing p(x) = x 2 - 7x + 5 by x - 2 Find the remainder on dividing q(x) = x2 - 5x + 7 by x - 2 Find the remainder on dividing p(x) + q(x) by x - 2 Score :
4, Time: 6 minutes
Concepts/Ideas
Factors of polynomials Q.9.16 The remainder on dividing the polynorma.l p(x) by x - a is k and the reniainder on diving the polynomial q(x) by x - a is —k. Prove that x - a is a factor of P(X) + q(x) Score:
4, Time: 6 minutes
Concepts/Ideas
• Factors of polynomials Q.9.17 If x - 1 is to be a factor of x 3
-
kx 2 - x + 2, what should be k?
Score : 2, Time: 4 minutes
RKE
Concepts/Ideas
• Factors of polynomials Q.9.18 Check whether 2x + 3 is a factor of 2x 3 + 3x2 + 4x +7. Write down a third degree polynomial for which 2x + 3 is a factor Score: 3, Time: 4 minutes
Concepts/Ideas
• Factors of polynomials Q.9.19 Check whether x 1 is a. factor of ax3 + bx2 .-- ax - b. Write down a polynomial for which T. - 1 is a factor. Score: 3, Time: 4 minutes
Concepts/Ideas
• Factors of polynormals Q.9.20 Which first degree polynomial added to 5? + 3x 2 gives a polynomial for which - 1 is a factor? Score : 5, Time: 8 minutes
U
Chapter 10 Geometry and Algebra Questions: 10.1-10.24
Concepts/Ideas -
-
• Distance between two points
-
Q.10.1 A circle is drawn with centre at the origin and radius 10 units. Classify the points below as tliose within., on and oirl;side this circle-
(-4 1 12). (8 1 -6) 1 (8,2), (10,0)
Score : 3, Time: 6 minutes
Concepts/Ideas
-
-
• Distance between two points
Q.10.2 A line drawn from the origin cuts the circle centred at C at A(3, 4) and B((3, 8). The tangent from the origin to this circle touches it at P. Draw a rough sketch and calculate the length of the tangent OP Score: 4,-Time: rminutes
Concepts/Ideas • Distance between two points
-
Q.10.3 The vertices of a triangle are 0(0,0). A(6. 0). B(2, 3). Draw a rough sketdi of time • axes and the triangle. Calculate the perimeter of the triangle Score : 4, Time : 6 minutes
-
-
.
-
•
II
Concepts/Ideas • Distance between two points
Q.10.4 In the figure. OABC is a parallelograrn where the coordinates of C are (2 1 6) and the coordinates of B are (10,6). What are the coordinates of A? Find the ieigths of BC and OC
Score : 3, Time : 4 minutes
Concepts/Ideas • Distance between two points
Q.10.5 The coordinates of Some points are given below: A(2, 4), 13(2, 6), C(5, 4). D(5. 9), E(8, 4). F(8, 12) Calculate the lengths of AB, CD, EF and show that they are in arithmetic s& quence.
Score 3, Time Sniinutes
Concepts/Ideas • Distance between two points
Q.10.6 A circle is drawn with its centre on the x-axis and radius 5 units and it passes
through the point (4.3). Taking the x-coordinat,e of the centre as a, what are the c:oordjnates of the Centre in terms of a? Find the value of a and hence the coordinates of the centre. Score 5, Time 12 minutes
Concepts/Ideas • Distance between two points
Q.10.7 A circle drawn with its centre on the x-axis passes through the points (-5. 12) and (12, —5). Taking the x-coordiriate of the centre as p, find the value of p and hence the coordinates of the centre of the circle.
Score: 5, Time: 10 minutes
Concepts/Ideas • Distance between two points Q.10.8 Draw the rough sketch of a circle write down the coordinates of its
through the points (4,0). (-3,2) and
Score: 5, Time : 12 minutes
Concepts/Ideas • Distance between two points Q.10.9 A is a point on the y-axis, equidis sketch. If the u-coordinate of A is p, Calculate the value of p and hence t
I from (3,5) and (-2.6). Draw a rough at are the coordinates of A, in terms of p? coordinates of A.
Score: 5, Time: 10 minutes
Concepts/Ideas • Slope of a line Q.1O.10 What is the slope of the line joi Is the point (8,12) on this line?
(3,2) and (5,6)?
Score : 3, Time : 5 minutes
Concepts/Ideas • Slope of a line Q.10.11 A line of slope passes through • (8,9)? Find the coordinates of the
point (4, 5). Does this line pass through nt where this line meets the s-axis.
Score: 4, Time: 7 minutes
Concepts/Ideas • Slope of a. line Q.io.12 Prove that the line through the pointS (-2,5), (3,8) and the line through the points (5 1 —2), (8,3) are not parallel. Write clown the equation of a line parallel to one of t;hese lines Score : 4, Time: 9 minutes
Concepts/Ideas • Slope of a line Q1O.13 (a.) What is the slope of the hne through the points (2,5) and (-3, —5)? (b) Write down the coordinates of a point on the line parallel to this and passing through (4, 6) Score : 4, Time : 8 minutes
Concepts/Ideas • Slope of a line Q.10.14 What is the point of intersection of the lin.e through (2, 6) with slope 1 and the line through (6,2) with slope Score : 4, Time : 7 minutes
Concepts/Ideas • Slope of a line Q.10.15 The coordinates of the points A, B, C are (-2,-1), (1,5) and (3,9). Find the slopes of AB and BC. Prove that we cannot draw a triangle with A, B, C as vertices. Score: 3, Time: 7 minutes
88
Concepts/Ideas
• Slope of a line y
Q.10.16 In the hgure, the centre of the circle is the origin and its radius is 1 unit. The points A, B are on the circle with LAOP = 30° and LAOB = 150 . Find the coordinates of A and B. Find the relation between the slopes of the lines OA, OB and the tan measures of angles they make with the xaxis
x
Score: 5, Time : 10 minutes
Concepts/Ideas
• Slope of a line Q.10d7 Tangents are drakvii at the endpoints A and B of the diameter of a circle. Two points on the tangent at A are (4,5) and (12, 10) and one point on the tangent:. at B is (8, 5). Finch the coordinates of another point on the tangent at B
Score: 4, Time: 7 minutes
Concepts/Ideas
Slope of a line
Q.10.18 Without drawing axes, draw a rough sketch showing the points A(2 4), B(8, 4), C(10, 12) 1 D(412) marking them with their coordinates. Prove that ABC.D is a parallelogram
Score : 5, Time : 9 minutes
Concepts/Ideas
• Equation of a line
.
11111
..I..IJI.I
Q.10.19 What is the slope of the line passing through the points (5. 2) and (8,6)? Find the equation of this straight line and find the coordinates of another point on it
Store: 4, Time: 8 minutes
Concepts/Ideas
Equation of a line
1:
Q.10.20 Find the coordinates ófany two points on the line 3x - 6y + 10 = 0 and find the slope of this line Score: 3, Time : 6 minutes
Concepts/Ideas • Equation of a line
Q.1O.21 What is the slope of the line 4x + 2y - 9 = What is the equation of the line with the same slope, passing through (4; 7)?
Score 5, Time : 10 minutes
Concepts/Ideas
• Equation of a. line
Qd0.22 A line is drawn through the points (0,2) and (2,4) What is the slope of this line? Find the coordinates of another point on this line Prore the the y coordinate of any point on this line is 2 more than the xcoordinate.
Score: 5, Time : 10 minutes
Concepts/Ideas • Equation of a line
IIIf
flflI
Q.10.23 What is the slope of the line joining (2. 5) and (3, 7)? Find the equation of this line. Prove that if (x, y) is on this line, so is (x ± 1.. y 4- 2). Score : 4, Time
9 minutes
Concepts/Ideas
• Equation of a line -
-
Q.10.24 (a) Find the point of intersection of the lines 2x - 3y + 7 = 0 and 3x ± 2y - 9 = 0 (b) Find the equation of the line of slope through this point Score 5, Time : 10 minutes
Al
Chapter 11
Statistics Questions: 11.1-11.7
Concepts/Ideas
. Mean from a frequency table Q.11.1 The table below shows the classification of students in Class 1.0 according to their weights: WEIGHT (K(.:)
.
•
30-35 35-40 40-45
NUMBER OF CHILDRE:N 3
•
•
-
8 12
45-50
9
50-55
6
55-60
.
2
Find the meaii weight Li
Score: 4, Time: 6 minutes
Concepts/Ideas
Mean from a frequency table
I
Q.11.2 The table shows the classification of the days of November according to the minimum temperature at Munnar TEMPERATURE (°C)
NUMBER OF DAYS
0-5
2
510
3
1.0-15
7
15-20
10
20-25 -
6
.
25-30
2
Find the mean minimum temperature for this month
Score: 4, Time: 6 minutes
Concepts/Ideas
• Median from a frequency table
Q.11.3 The table below shows the classification of students in Class 10 according to their heights: HEICET (CM)
.
NUMBER OF CHILDREN
135-140
3
140-145 145-150 150--155
14 .
12
155-160
6
160-165
2
Find the median height
Score: 5, Time: 8 minutes
Concepts/Ideas
• Median from a frequency table
-<93
Q.11.4 The table shows the classification of families in a locality, according to their monthly electricity bills E1,EcrlucJrY
NUMBER OF
BILL
FAMILIES
(Rs) 5....... 150
4
150...250
5
.
250-350
7
350....450
18
450-550
3,2
550-650
20
650-750
8
750-850
4
850-950
2
Find the median amount paid in electricity bills Score: 5, Time: 8 minutes
Concepts/Ideas . Median from a frequency table Q.11.5 For nak.ing cones. children cut out several sectors of circles. The taile shows the number of sectors made, classified according to central angles 'NUMBER OF
CENTRAL ANGLE
SECTORS
0-45
1
45-90
3
90135
10
135-180
12
180-225 225_L270 270-315 315-360
11
,
8
,
'
3 2
Find the median angle Score: 5, Time: 8 minutes
Concepts/Ideas
• Computation of mean Q.11.6 The table shows the number of children getting different scores in a test. NUMBER OF
ScoRE
CHILDREN
5
1
6
3
7
10
8
12
9
9
10
.
Calculate the mean score Score 3, Time 4 minutes
Concepts/Ideas
.
.
Computation of mean
.
..
Q.11.7 The table shows the number of workers earning different daily wages in a locality DAILY WAGES
NUMBER OF
(Rs)
WORKERS
-
.
..
300
1.
325
4
400
6
425
6
450
4
475
2
500
2
Find the mean daily wages Score : 3, Time: 3 minutes
.95
1. Arithmetic Seciuences 1.1 Sequence of the angles :60°, 90°, 108°, 1200. 1 180, 360,:540, .... 1 The sum of the angles is in arithmetic sequence with common difference 180 0 1 Yes. 900 = 180 x 5 900 is a multiple of the common difference 180. 1
Thtal4 1.2 The perimeters are in arithmetic sequence, for as we move from one polygon to the next, the number of sticks increases by one. This makes an increase in the perimeter. The common difference of this arithmetic sequence is the length of a stick. 15' term = the length of 17 sticks 1 That is, 17 x length of àstick = 85 1 Length of a stick =5 1
x1 =x15 -14d = 85-14x5 =15 = 15+85
1
=100 Total length of the sticks =
15
x100
= 750cm.
1 ThtaI5
1.3
nthterm,
x
= dn+(f—à) = 6n 4- (184 —6) =190-6,z
1 1
If dterm in zero, then
190-6n = 0 ii
= 190
1• 31.3
There are 31 positive terms in the sequence The greatest negative term = =190-6x32 =2
1
1
Totals
1.4 Sequence: 8, 14, 20,26. It is an arithmetic sequence with common difference 6. .•. . .•. natural number= ii x = dn ±(f- a) or x=6n+2 (multiply by 6 and add 2) = 6n + 2
1
.
1
6n+2 = 250 n = !, not a natural number
250 is not a term of the sequence 1 (d)Sequencé:8,20;32,44,.;. ......................... Sequence: 14, 26, 38, 50, ... Difference : 6, 6, 6, 6,... . . The difference of the sums of 20 terms =20 x 6 .2 =120 . Totals
1.5
..
Since the i/' term is 8n + 3, the terms.leave.a remainder 3 on division by 8 1 1
The first term in the sequence, between 100 and 500 = 107 The last term = 547 The number of terms = 56
1 Thtal4
1.6 The sequence of the distances covered by the falling object in each second 9.8x1_49 = 4.9 ............................... =.1 .....-. n=.2, 9,8x2-4.9= 14.7. .......... 11=3, 98x3-49=245 Sequence49, 147,245, 98,z-49 = 637 = 1.7
686
1
=7 seconds
...................,Thtal 2 . 1 (a) To write an arithmetic sequence with comij1pn difference 7 175=7x25 175 is a multiple of the common difference 7 . 1 Therefore, the difference between two terms is 175 Thtal2
1.8 Sequence: 10, 12, 14, 16, ... 1 10 and even numbers greater than 10 are terms of this arithmetic sequence. Therefore, all even squares greater than 10 will be terms of the sequence. 1 Perfect squares in the sequence : 16, 36, 64, 100, ... 1 The sequence has no common difference. Therefore it is not an arithmetic sequence 1 Ibtal3
1.9 To explain the law of formation of the sequence To write two consecutive terms 1.10
1 1 Total2
..
.... Sequence 1: x. = dn±(f-a)
.
1 Sequence 2: .
x.=6n+15 ........................ The n' term is a common term . 7n-6 = 6n± 15 7n-6n= 15+6 . . . .. n=21
1.
........ The 2lstterm isacommon term Common term . . x21 = 7x21 —6 or 6x21+15 .......................... 1.47. - ... 126 + 15 =141 . 141 ... . ..
1
...
1
-.
1
1.11 For writing two arithmetic sequences with no common term Eg 2, 12, 22, , , 3, 13, 23, 1 If an + c, bn + dare two arithmetic sequences with common differences a and b, thena*b . . ... . Ilthereisacommontenn,an±c=bn +d .. 1 Ifc=a thenan=bnanda=bareimOssible . Ifc#d,then an+c=bn+d an- bn=d- c
........,...
.
d-c
... fl=.__-_..:... a-b
.
. .
.. .
. .
.1 .
ii has only one value. There wIll be only biie common term: . .
.
.
. TotnJ4
1.12 Sequence of the volumes : 5, 10, 15, 20, ... 1 ;• This is an arithmetic sequence with common difference 5. 1 If the sequence of volumes is multiplied by density, we get the sequence of weights. 1 Sequence of weights: 13.5, 27,40.5,54, ... 1 This is an arithmetic sequence with common difference 1.3.5 .1 lbtaI4 .
1.13
1 Writes an arithmetic sequence with common difference 4. Writes another sequence by multiplying each term of the first sequence by 3 and adding2 . . 1 Obs&vs that the common difference is 12. This is 34 times the common difference of the first sequence 1 th The term of an arithmetic sequence, x = an + b. If the terms are multiplied by kand radded, kx+r=/(an+b)+r =kan+/th+r
This is the ,jth term of an arithmetic sequence kr+ ris another arithmetic sequence
1
.
with common difference k
.
. .
1.14
Writes any.arithmetic sequence Write the terms in positions 1,4, 7, 10, This is an arithmetic, sequence . Sincea, b, c, ... are in arithmetic.sequence. b—a=c—b X — X =(b—a)d x — x(— b)d
.
.
..
1
.
1
.
I
.
.
xb —x =xc —xb a . .rc xx, ... is an arithmetic sequence. ...
..
1 Ths
,
.1.15
1 ThtalS
.
.
.
x15 —x8 = 7d
. . 7d= 102-53=49. d=7 x = x15 +10d = 102+10x7
=172
.
.
.
.
. .
.
..
..
. '.
.
.
1 1
.
1 Thtal3
1.16 Since x+ 4, 3x- 2, 4x- 2,... are in arithmetic sequence, (3x-2)- .(x+4) (4x-2).-.(3x-2). ... 3x-2-x--4= 4x-2-3x+2 . . 2x-6=x x6 . . Sequeiice:1O,16,22, ....
1
I
x=a+f-d
=6n+4 I 5)th There is a distance.of 10 common differences from the (nterm to the 5)1terim (n± . = lOd Difference between (ii— 5)th term and (n+ 5)th term ..
•.
10x6 .=.60
i :.
1.17
..
Sequence: The
4 11 14 17
.....
...
rrn of the sequence,
..
•..
•..
.
1
.
.
..
.
...
.
..
....
1
.
..
1
..
.....................1................. =n3n+5
.
2
.
Since 3n+ 5 is not a multipleôf3,itis n ota nai ltiple of 6. Therefr an integer.
,..
3n±5
is not
.
Thtal4
) The it" term of the sequence... x = d+f-d = ...•..
3 4
11 3 4. ,. 4
3n+8 4
If the taken=4, 8,12,16,... '
1
. . ...
".
.. .
'•'
1
3x4+8 ...=
x8 = =
=
3x8+8
=8
'•
...
3x1.2•+8' '=11
3 11 3 444
-n+---
3n+8 4
The sequence of integral terms: 5, 8, 11, 14, The sequence of position of terms : 4, 8, 12, 16, ... The common difference of the first sequence =3 The common difference of the secondse4uence= 4 The common difference ofthe given sequence = Totals
1.20 (a)
(b)
'
1:7 =A x8 =7A 3,d = "6k .d .2k = x5 +d=k+2k=3k .x15 ......5 ±iOd=.k-'20k . 21k .............. x615 .3fr:Zk=1 .,..
X 5 :X8 ..=
x7 x22 x7 =10,x22
17 7x10=70
1
1
1 1 Total 4
121 The number of chords is the sum of the natural numbers from 1 to 14 The number of chords =
2
14x 15 2
lO5
1 Thtal3
1.22
Sequence of odd numbers 1, 3,5,7, x = 2n-1 x1+x a = 1+2n-1 =2n
1
The sum of a terms = (+ x) = x 2,z= ,i
1
The sum of first a natural numbers
-
=
2+n n(n+1) 2 = n2
1
(n-1)(n--1+1) 2
The sum of first (a - 1) natural numbers -
- n(n-1)n2—n 2 2 n
2_ . fl 2 +fl_fl 2 .2
1
—fl
1
.
•.
.:
1.23
Writes the sequence using the n term 7,11, 15., 19, The total distance travelled is the sum of 15 terms of this sequence. X15 4x15+3=63irietres ...
Sum -=
(x+x)
-
1 ..... .
15 15
--.(7+63) = 525 metres 1.24
.. .
.
1 ThM3
..,-
Sequenceofoddnumbers 1, 3, 5., 7, ... ThelOthoddnumber= 2x10-1=19 x1+x10-1+19_.20 ...The sumof 10 terms =
10
x20= 100
1 -1
The 201 odd number = 2 x 20 - 1 = 39 x,+x20 = 1+39=40 The sum of 20 terms =
20
x 40 = 400
1
The sum of the nextiO odd numbers = 400 - 100 = 300 1 The difference between the sum of the first 10 odd numbers and the sum of the next 10 odd numbers =300- 100 = 200 1 TotalS
1.25
For writing the next two lines Observes that 1, 3, 6, 10, ... are continued sums of natural numbers,
1 1
Algebraic form
1
= 2
1, 2,4, 7, ... are 1 more than continued sums of natural numbers
1
2 + 1 = n2-n+2 Algebraic form = ( n -1)n
1
2
Total S
1.26 Sequence 1: 8, 14, 20, 26..... Sequence2 4,10,16,22,... . The difference between terms in the same postion is 4 The difference between stimsof 15 terms = 15 x 4 =60
1 1 1 Ibtal3
1.27 xi =1
x =dn-i-f—d +x f + dn +f— d = dn + 2f— d Sum = n -(x +x) 2 ' =
(dn+2f—a) d
1 = 2 +(,f)n There will not be any constant term in the algebraic expression for the sum of n terms of an arithmetic sequence. TherefoEe, 3n 2 + 2n + 1 cannot be the sum. 1 lbtal4
...
128 Sequence 4, 12, 20, 28, ...
1
The sum of ii terms
x)
=
( +
=
(4+8n-4)
1
=41 2 =(2n)2
1 ThtaI3
•
1.29 Note that the wages of digging to a depth of each metre are in arithmetic sequence 1000, 1250, 1500,... I 16thterm =1000+250x15 = 4750 1
• •
Wages for digging 16 metres
= =
•
8 1h term = 1000 + 250 x 7
=
Wages for digging the first 8 metres
=
(1000 + 4750) 46000 2750
1
(1000, + 2750)
=
15O00 Wages for digging the next 8 metres =• 46000 - 15000 = 31000 Extra amount needed = 31000 - 15000 =
•
1
1
16000
TotalS
1.30. 1+2+3...+20=
20x21
1
=210
The sum of 20 consecutive terms of the arithmetic sequence• with = 4(1--2+3-i-...+20)+3+3+3...+3 '4n+3' =
4x210+3x20 900
flth
term 2
20terms 1 Ibtal4
131 5 1, term = x8_
.
1
495 -j--
1
X8
d
...,
Sum of 15 terms 15
= 33 33-21
X5
1
3
Firstterin,x1 = - 7d=33-7 x4=5 . n thterm , x =dn+f—d=4n+i The sum of n terms =
1
( + x)
=
(5+4n+1)
=2n2 +3n
1 Thtal5
1.32 The sum of the first n terms = - d
.
.
. d . .. -- )n
1
n2 + (f--
(2n)2 + (f - ) 2n
The sum of the first 2nterms =
= 2dn2 + (f-
)2n
1
The sum of n terms following the first n terms = 2dn2 +(f-)2n _[ 3d
= -- n2 +(f-
d
fl2 + (/_)n ]
)n
2
The difference between the sum of the first ii terms and the sum of the next ii terms = Wd
1 ThtalS
1.33 (a) Sum of the angles= 7 x 180 Middle term (5 1 term) 7x180 = 140 The fifth angle of all polygon will be 1400.
1
. . .
1 1
I
Anu got it right. InAnup's polygon, 116 x5 = 140 d
X5
-
X
-
4
4
-
6°
1
9th angle, x9 = 4d = 1404x6=1640
1 Totals
1.34 (0.125)_24
1000 ('i n
272 The sumof,iterms of 2, 6, 10,...n..= 2n2. . . 2,z=72 n=6. . x4 2-f5 The 614 term of 2, 6, 10, ... = x=22. .., • 2 2+6+
10 +
..
1
.
1
.
.,.
1 Thtal4 .............
1.35 Thesumof2Oterms = 1300 1300 x1 +x2o=-i.-
=1301
130
... x1 +x1 +19d
....... 2x = 13019x6=16 •.
..;
....
.
.
1 1
x=8
Sequence: 8, 14, 20. 26, = dn+f-d
,..=6+.2.......................... .1.
The sum of n terms = =
Ix,± xl
.
.
[8±6n±2]
1 Total 5
136 Common difference - 145 - - 150 = Nextterm=140+5=135 0 is a term of this sequence
I 1 1 IbtaL 3
1.37 Sequence 3, 6, . Common difference•= 3 . 2Pt term = 3 + (20 x 3) = 63
I
1.38
1 1 Thtal3
. The first multiple of 10= 10 The10"r ultipleof 10= 100 ThësuthófthesèlOtejrns
1 10
(10+100)
•.
= 5x110 =550 1.39
1 Thtal2
, .. Sequence.:. 1,1, 1 ........................... Algebraic form
1.40
.,
.
2n+1 2
.
1
.. . .
. . :•.. 3+6+9+..+300 = 3(1+2+3+...+I100) = 3x5050 .
. .:.
..1 1 Thtal2
2. Circles 2.1 1 • (a) LD=90° 1 Dis a point on the circle, with diameter A C Bwill be inside the circle. (b) L4+LC=160° ABCDis not a cyclic quadrilateral .. Bus not a point on the circle. We cannot draw a circle through the four vertices of 1 the quadrilateral. 'lbtal3 2.2
A circle is drawn with diameterACof quadrilateral ABCD. If h Dare outside the circle, 1 LB<90°,LD<90° 1 LB+LD< 1800 1 ABCDis not a cyclic quadrilateral lbtal3 2.3 LD=360-300=60° Since LA = 80°, LC= 100° , the circle with diameter LID will not pass through the . 2 points Band D and vice versa. • • 1 LA+LC=180° .• ABCDis a cyclic quadrilateral 1 A circle can be drawn through the vertices of the quadrilateral TotaI4 2.4 LOCA=40° L4OC=100° . L4BC=50° LOAB=20°. L4BC=50°, LBAC=60°,L4CB=70°
1 1 I
. •
1
•
Thtal4
A
2.5 Answer key - Model 1
C
• -
IfwetakeLOBC=LOCA=jLOBA=z LBOC=180-2x ZBAC=90—x . . . )'+z=90—x x +y + z = 900, x + y= 90° ZABC + LOCA = 900
1 1 1 Thtal 3
• •
Answer key - Model 2
Draw dmmeterAD.Join BD ... • LABC+LOCA = LA.sc,+LOAC LABC+ LDBC . =LABD • .... =90°
• ...
1 1
1 Total 3
A
Al
,....
6.•
12cm.
R 5cm. Q
Move the point A along the circle to obtain the diameterA'B Being angles in the same segment,
1
•
-.
•
LA =LA' BC=QR LA'=LP LA'BC = LPRQ M'CB LPQR A'B=PR •A 'B A'B
=
V52 +122
=
13cm.
1 1 1 Thtal 4
2.7 QC= 15 cm.; PQ= 9cm.; PC= j152._92 = 12cm.
1
PC= 12cm.;BC=13cm.;PB= ..j132 122 =5cm.
1
PAxPB=PO PA x 5 = 122 = 144
1
144
R4==28.8cm.
1
ABFA+PB=28.8+5=33.8cm. 33.8 +16.9cm.
0QOB-QB=16.9-14=2.9cm.
1 Thta15
2.8
Thecentralangleofarc ABC=360- 110 = 2500
1
LABC=x1O0=55°• •
1
LADC = x250 = 125°
•
LOAI? + LOCh = LOBA + ZOBC = LABC 55°
1 •.. •
••
1 lbtaI4
2.9
For drawing angle of bU" For drawing angle of 1500 For drawing angle of 750
1 2 1 Total 4
2.10 Draw a circle and construct an angle of 900 at the centre Drawanangleof45°oftheminorarc With the vertex of 45°angle as centre, draw a circle and construct an angle of
I 1
22'inthesmallerarc.
1
With the vertex of 22 angle as centre, draw a circle and construct an angle of 11°inthesmallerarc.
1 Total 4
2.11 Draws a circle of radius 4.5 cm. Divides it into arcs of central angles 60°and 140° Draws a triangle by joining the end points of the arcs For measuring the length
1 1 1 1 Total4
2.12
ZACB = 500 LBCD = 105° LBAD=75° LACD=55°, LARD =55° ZABC=85° LADC=95° LADJI = 50°,
1 1 1 1 Thtal4
2.13 LABC LADC are angles in arc ABC LBAD, LBCDareanglesinarcBAD If BC, AD intersect atM, LAMB =LCMD LBMD = LAJWC
I 1
lbtal2
2.14 The sum of the oppoiste angles of a cyclic
quadrilateral is 1800.
1 1 2 Thtal 4
An exterior angle of a triangle is the sum of the interior opposite angles LAPC> LFDC -1
2.15 Those that are cyclic quadrilaterals • square • isosceles trapezium rectangle
2 ThtaI 2
Wei
LB=1l0°,LD=85° LB+ LD= 195 0 > 1800 Dis within the circle LA = 750 LC= 900
1
1
LA + LC= 165 0 < 180° Cis outside the circle
1 IbtaI3
2.17 JoinPQin the figure APQD, PBCQ are cyclic quadrilaterals LA = 110°, LFQD=.80°, LPQC= 100°, LB= 80° LD= 800, LAPQ= 1000, LBPQ— 80 0 , LC= 1000 LA+LC=100+100=200° ABCDis not a cyclic quadrilateral
JoinPQiñthefigure ABCDis a cyclic quadrilateral IfLA=x, then LFQD=180—x IfLD=y,thenLAPQ=180—y . • LPQC=x • PQCBis a cyclic quadrilateral LB= 180— x If ABCD is a cyclic quadrilateral LD+LB= 180
1 1 1 .• 1 lbtal4
1 1
•
•
• •
1 1
y+180—x= 180 )' —x=,0 y=x 2.19
•
•
i
-; JoinPQ,RS APQD, PRSQ, RBCS are cyclic quadrilaterals Z-A =x, LPQD=180—x zPQS=x LPRS=180—x ZBRS=x LBCS=180—x LA+LC=x+180—x 180
.
Y
1
.
I
ABCDis a cyclic quadrilateral
1
:
Thtal4
2.20 ThecentralangleofarcAPR
=
1
=60° The angle subtended by arc APR in the minor arc •. =30° LAPB=180-30 =1500 •. .
1 1 Thtat3
2.21
•
•JoinOR .................................. LAOB=60° If Dis a point on the minor arc, then •.• LADB=30° /4PB = l50°
1
.
.
•
•
.
•
•
.
:•..•.
•
1 1
..................... .2.22
.
•' (a) An angle in a semicircle is a right angle Cwill lie on the circle LD=110°>90°,Dwfflliewithinthecircle ...' LE=70°<90°, Ewill lie outside the circle
•...
.
.•.
•
,
•• ..
1 1,
•
Hi I
LC+LE=90+70=160 #180 0 Cand Eare not points on the same circle ZD+LE=110+.70=180 0 Ewill be a point on the circle through D
i liii
1 1 1 Total 5
2.23 The central angle of arc CYD = 300 LCAD=15° The central angle of arc AXB = 1100 ZACB =550 LC'PA=55— 15 =40° ZACP=125°
.
.
I 1 1 1 Total 4
2.24 JoinAC AD=AB CD=CB L\ADC. MBC.. ZD =LB
1 1
LD+LB=180° LD=LB=90°
1
AC= .il02 +6 2 = ,1136
1
cm. 2 = ,J Area of the circle=. 34,r sq.cm Radius =
1 Total 5
2.25 The central angle of arc ADC= 60° The length of arc ADCis
=
of the circumference of the circle
Circumference of the circle = 6 x 18 = 108 cm. Thecentral angle of arc RSQ=36° The length of arc RSQis
= 10 of the circumference of the circle
Length of arc RSQ = 108 x 10 =10.8cm.
1 1 1 1 1 TotalS
2.26 LPAC=LPDB LPCA—LPBD.
1
Since the angles are equal APAC—APDB .cJ
1
-
PA_PC_AC PDPBBD
1
PAxPB=J'CxFD
1 Thtal4
2.27 1 1 1
Angles in the same segment are equal Triangles are similar if their angles are equal The ratio of corresponding sides of similar iriangles will be equal
Total 3
2.28
1
Since PC= PD, PDI= FAXPB 9x4
=36 PD= 6cm. CD= 12cm.
1
.
PMxJW=PAXPB 3xPN=9x4
.
1
I'N=l2cm. eWjV=l5cm. 'lbtal4
2.29 FQ=2Ocm. AF:AQ= 3:2
.
AJ'= 20x=12cm. AQ=8cm AS:AR=8:3 If AS = 8x.thenAR3x .... ARxAS=4PXAQ.
.,
.
12x8 .3xx8x ..24=96
1
-:
1 .. .
AR=3x=6cm. -AS = 8x,= 16cm 22cm. -
.
Thtal4
2.30 PA = P11= PC= PCxPD =
4cm. 4+5=9cm. 3cm. PAxPB
1
2 =12cm.
PD=
CD= PD—PC=9cm.
1 Thtal 2
2.31 10
is 1
if ris the radius of the circle, PA = ,AB= 2,, PAxPB= 9x12 r2=36 r= .6cm.
I 1 'l'btal 3
2.32 If chords Ah CD cut at P outside the circle PAxPB=PCxPD . PA=PC:.PE=PD PA+PB=PC+CD AB=CD .
..
ABand CD are chords of equal length . . Therefore the distances from the centre of the circle to the chords are the same. ...
1
1 1 Tot9J4
2.33
A
\I:1 1
111111111111111117
' 2
Move A along the circle A' to obtain a diameter BA' LA = LA' Draw CD = Ii, perpendicular to AB M'CB:. MDC
1
b Ii a.2R hc= 1
hc
Area of 1\ABC
abc
1
abc
1
abc
TotalS
2.34 Draws an equilateral triangle of side 6 cm. Draws a square equal in area to that of the equilateral triangle Measures a side of the square
I 3 1
i
Totals
2.35 5
Draws a square equal in area to that of the rectangle and measures a side.
Totals
2.36.
The cenfral angle of the opposite arc Of arc ABC= 360 60 = 3000 x-i-x+26O = 360° x= 500 50 The central angle of arc BCD Theceniralangleoftheoppositearc = 360-50 310° ..
..
1
.....
1 1
x+8x= 360 x=40° The central angle of arc CDE = 400 The centraiangle of the opposite arc 360 - 40 3200
1 1 Totals
L5/
The central angle of an arc is twice the central angle of the opposite arc Angles in the same segment are equal An angle in a semicircle is a right angle Angles in opposite arcs are supplementary
1 1 I 1 Total 4
2.38 PAxPI? = PG 4x2 = PG Pc= sJi
1
,j2 =8
Area of the square = 1 Mark a point 0, 5 cm away from A. Draw a line perpendicular to A/I, from Q. The distance between Qand the point of intersection of the perpendicular and the semicircle wifibe ..j cm. Ibta2
2.39 Since LA, LCare not supplementary, adding the measure 30° of LC
i 1 Total 2
2.40 For correct figtre For writing the concept used
2 3 Totals
2.41
•
•
C8xCD=CP2 4x2 = CF2
1
CP= CAxCD =CQ2 8x2=C
1
CQ=4 •
QF=4—j
1 1 lbtal4
2.42 1
30 = 360x1 If the folded corner is placed at the centre, the length of the arc is
12
of the 1 I
circumference of the circle
Total 3
2.43 1 2
For drawing a similar triangle. For drawing a rectangle Heightofthetriangle= 2x20= 10,j j The breadth of the rectangle in half the height Breadth =
=5 ,j cm.
1 TotalS
3. Second Degree Equations 3.1 A side of the first square (x6)2 (x-6) x x A side of the first square A side of the second square
x 900 ±30 36or-24 —24 is inadmissible 36cm. 36 - 6 = 30cm.
= = = = = = =
1
Total 3
3.2 base = x diagonal=2x-i-3 thirdside = 2x+2 (2x + 3)2 = xI + (2x + 2)2 —4x=5
—4x+4 =9 (x-2)2 = 9 x = 5or-1 x = — 1 is inadmissible base =5 diagonal = 2x5+3=13 thirdside = 2x5+2=12
1 1
1
1 Total 4
3.3 For a rough figire
1 PA x PB 9x5 . 45 245: thscnnmmnt:
The length of chord CD cannOt be 8 cm
PCxPD x(8—x) 8x - 0 (_8)2 _4 xl x45 64-180 —1.16
1
1
1 Total3
Thepriceoflkgofapple =x -5 The price of 1 kg of mango = Total weight of apples
=
Total weight of mangoes
=
600 600 x-5 x 600x- 600x+ 3000 4 - 20x 22 -5x-750
kg. •
600 kg.
= = -
I 1
=
1
4-20x 3000 0 5±.J(-5)'-4xlx-750 2x1
x =3Oor-25 x = -25 is inadmissible The price of 1kg ófapple = 30 The price of 1 kg of mango = 25 600 = 24kg. The weight of mangoes =
I 1 1 'IbIal 6
3.5 Next number = 45 - x • (45 + x) (45 -4 = 2009 45.2 _22 = 2009 • x2 = 2025 - 2009 x2 =16 x =±4 Ifx= 4, the numbers are 45 +4=49and45-4=41. If x = -4, the numbers are 45 + -4 = 41 and 45 - 4 = 49.
I 1
1
1 Ibtal 4
3.6 •
If the number be x, then the second numberis x+ 8 x(x+8) = 105 j2 +8x = 105 The smallest number to be added to make it a perfect square = 16 +8x+16 =121 (x+4)2 = 121 11 or-li x+4 x = 7'or-15 The numbers are 7, 15 or -15, -7 •
.
.
...
..
........
1 1 1
1 Total .4
3.7 Length Breadth
11+x 11—x
ui
Area=(11 +x)(11 —x) 121 -
117 4 2
22
x Length=11+2=l3cm. Breadth = 11 — 2 = 9 cm. 3.8
Jfxbe the number of days the first.company takes to finish the work, the second company takes x + 5 days. 1 1
The part of the work done by the first compnay in a day =
The part of the work done by the second company in a day = Thatis,.±-4-
=
.
..
..
.
1
6(x+5)+6x = x(x+5) 6x-i-30+6x = +5x t-7x-3O.. = 0.. . —7x=3Q.. (.7'2
49
= 30+--
(7•\2
169
.
1313 ---or 2. X = 10 'Fhé first company takes 10 days and the second 15 days 7
1
1
=
.S
1 ..
.3.9
.
..
S
The sum ofii terms of the arithmetic sequence 8, 14,20 3,±5n= 280... 3,+5n28O =
280 ......
..
i
b2 -4ac = 5-4x3x-280
= . 385 .... Even though b2 - 4ac= >— 0, this is not a perfect square. So ,zcannotbeanaturalnumber. . . . . That is 280 cannot be the sum of consecutive terms of the sequence 8, 14,20, ... 1 ..... .. flg3 ..
.
3.10 If x and x +6 be the resistances x+64 4(+6)+4x = x(x+6) +6x 8x+24 = —2x=24 —2x+1 = 24+1 (x-1)2 = 25 x-1 =5 x=6 The resistances are 6, 122
1 1
1
Total4 3.11 29.4t— 492 =. 44.1 Dividing by 4.9 1
= = = = = =
0 0 (t3)2 3 seconds t 6 seconds If, t 29.4 x 6 - 4.9 x 6 2 Distance 176.4 - 176.4 =0 The object reaches the grOund after 6 seconds
1 1
Thtal 3.12
.
. .
x(x+2) = -5 +2x+5 = 0 22 -4xix5 = —16 Dicriñinant Theproductcaniiotbe-5. . x(x+2)=-1 22 +2X+1 = 0. . Discriniinant = 22 _ 4 x 1 x 1 = 0 The product can be —1
1
1
..
Totid 3.13
. Number = x, reciprocal =
.,
.... 1
25
.
..
1
StIO 122 -25x+12 = 0 b2-4ac = (-25) 2 —4x 12x 12 =. 625-576=49
1
25±.i 24. 4 3 x=orx=
.
.1
1
61 = -
x
30
3012 -61x+30 = 0 b2 -4ac = (-61)2 -4x30x30 3761 - 3600 = 121
1
61±11
.x..6O. 61+1i 60 61-11 60
In either case, xis not a natural number
3.14
1
...... . .
..
.
Total
..
..
.
P.(x) = 0 = 0 1 Discriminant=(-3) 2 x4xlx.7=9-28=-19 Whatever be x, P (x) cannot be zero. , ..5x-8 = 1 . —5x-9 = 0 Discriminant = (-5) 2 —4x 1 x-9=25±36= 61 1 Here, the values of xcannot be natural numbers as. the discriniinant is •not.a perfect square. Therefore, if x is taken to be some natural, number, the value of the polynomial cannotbel. . 1 .11btal4 3.15 P(x)=+4x+4 = (x+2) 2 Since P (x) = (x+ 2)2, P (x) cannot be negative
1
.. .
.
'.
.
. .
Q(x)=-6x+ 10=-6x+ 9 ± 1 =(x-3) 2 + 1 Here, too, Q(x) is neither zero nor negative.
1
1
'Ibtal4
-r<
3.16
.
1 x+4=O,x+3=0 1
.
(x+4)(x-i'3) = 0 xl +7x+12 = 0 Actual equation ..t2 - 7x+ 12 = 0
.
I
7±..J49-48 2 7-1
7+1
x =
1 Thtal4
x=4or3. 3.17 Height of the box = x Breadth = x+10 2499 x (x + 10) x 21 119. +. lOx lOx+25 =. 144..• .
I
=7orx=-19 x= -19 is inadmissible Height = 7cm. . •. . Breadth = 17 cm. The length of the rectangular sheet of paper =21+14=35cm. Breadth = 17+ 14=31 cm.
1
1 TotalS
3.18 The smallest figure to be adjoined = square . . Its area = 64 sq. cm . ............................ .•
+16x+64 .= 74. (x+8)2 = 784 x+8=28orx+8=-28 . . .. x+ 8= -28 is inadmissible 28cm: Aside of thenew square'. ........... 3.19
..
..
...
Breadth: Length =x4-4 x(x+4) = 320 320 +4x+4= 324
1 1 r1btal
. .
........
1 1
. .
...
..
.,
. . . . ,1
(x+2)2 = 324 •x-i-2=±18 I
x=l6or-20 x= —20 is inadmissible Breadth of the plot = 16im Length=20m.
1 lbtal4
3.20 The difference between the 1 li" terms x(x+3) +3x-2l60 9-4ac=
1 1 1
3 2160 . 0 32 -4xlx-2160 S649
= = = = =
—3±93 2
-3±J 2
: x=45or-48 If the 1 1th term in the first sequence is 45, then, x11 48 is the second sequence If x11 = —48 in the first sequence, then x 11 = -45 in the second sequence
1 1 Ibtal 5
3.21 Thenextoddnumber. =. xx+2)±1 = 225 x + 2x + 1 = 225 (x+ 1)2 =. 225 x+1=. ±15 :. x =14 or x= ±15 x=14 orx=-16
..
. .
1
x=-16isinadmissible . ., Therefore, the even natural numbers are 14, 1.6• Thtal 5 322.
. .Area= zI Perimeter = 4x 192 TI +4x+4 = 196 = 196 x+2. . ±14... x = 12 or x=-16isinadmissible Aside of the square =12
..
.
ibtal 4
4. Trigonometry 4.1 Tiiangesaresirnilar The side opposite toyare in the ratio j
63
1
The conespondig sides are in the ratio 3 : 5 The third side of the first triangle = 15 x =9cm.
1
The third side of the second triangle= 12 x = 20 cm.
1 Thtal 3
4.2 Draws a roiih sketth 91 1
A
6 m.
'75cm 1
AOPQ, EOAB are similar. OA
AB
1
= -x1=8m. .75
Total 3
4.3 Draws figure
41
19
LP=900,A=600 AP:PB:AB=1:,J:2
1
AP =2 -,r3 cm, AB 4 ,j cm.
1
Radius of the circle = 2 cm. Area of the circle =
it
(2,J
sq.cm.
= 12itsq.cm.
1 Thtal 3
cc
A 12m Draws aroughfigure The angles Of AAOB are 300, 600, 900
1
1
OB:OA:AB = OB =
4x1.73 6.72m.
1 ThtaI 3
4.5
Side = 10 12 cm.
1.
10.10 1 Area =xx = 25 sq.cm.
1 Thtal 2
4.6 The distance from the mid point of the hypotenuse to the vertex of the right angle is the 1 circuniradius Circumdiameter = hypotenuse 1 =8icm.. 1 Circumradius = 44cm. Total3
4.7
1
Draws a rough figure Draws L\ADC LCAD = LACD = 45°. AD:CD.:AC = i:i.:J. CDr8 ,J
1 1
1
Area of AABC = x 20x 8 = 80 -vI2 sq.cm.
Total 4
4.8 Draws the perpendicularfromA to BC
12 1500 C
I
The angles of £&DB are 30 0,600 ,900 AD:BD:AB =
1
:2
ADCDAC = 11 ,f AD =6cm
.......
1
TheareaofMBC = . - x 15.x6.. = 45sq.cm. Total4
MO
D
C
-
A
M
10
N
1
B
Draws the figure InADM AM:DM:AD = 1::2 AD=6cm.,AM=3cm. h=DM = 30cm.
1
BN =. 3cm. DC=MN 10-6=4cm. The area of the trapezium
Ii (AB + CD)
= .
1 1
x3.(1O+4)
= 3x7 = 21..J sq.cm.
1. ..Tôta15
4.10
The point B is moved along, the circle to B So that AB'is a diameter. LACB' = 90 0 The angles of A ABC will be 30, 60o, 90°
1
B'C:'AC':AB 1:'0':2'' . AC'r'6,',AB = •40cm.;radius=2'0cm. .
1 1 Total4
.
4.11 The angles of AABD are 45 0, 450,900 AB:AD:BD.=1:1:.j
.
.131 ,..
BD=6cm.,AB==3.Jcm..• x3,Jx3,J
The area of A ABD =
9 sq.cm.
1
The angles of A BCD are 30°, 600, 900 CD:BC:BD = 1:,j:2
I
BD=6cm,CD=3Cm,BC3'J cm. The area of A BCD =
x3x3.i
= 4.5.J sq.cm .
1
Area of 0 ABCD = 9+. 4.5 ..J sq.cm. The circumcentre is the mid point of BD Circurnradius = 3 cm.
1 Ibtal 5
4.12.
A
El
In the figure, LEBD = 300 ABED is an isosceles triangle BC The angles of ABEC are 30 0 .600, 900 CE:BC:BE CB BC ,
•
= 10cm. . . 1: = 5 cm. = 5Jcm..
I 1 1
The areaoftherectàngle = CDX BC =.15x5.J = 75 ..i sq.cm.
1
Thtal 4
Is 4.13
The angles of MOP are 30 0, 600, 900 OAOF:PA OP12,OA The angles of zQBP are 450, 450, 900 OPOB:PB OP=12,OB AB
1:j:2 cm.
1
i:1:.J 12cm.
1
12 -4,j cm. 4(—.1:)cm..
The angles of LOCP are 300 , 600, 900 U
OP:OC:PC OP=2OC • BC
1:
:2
12 5
•
120-12cm. i2(J-1)cm.
ABC
—1) :.1 2.(J —1)
1 1
4,j:12 1 TotalS
D
4.14
A
C .
The angles of AAPC are 450, 45, 900
.
.
.
AC=4Jcm.
.
.
AP=4cm.,PC=4cm.
.
PD=4cm.
.1 . .
CD=8cm. The angles of LBPD are 300, 600, 90 0
. 1
PD =4cm,PB=4cm. AB = AP+PB
1
= 4+40cm. The lengths of the diagonals are 8 cm., 4 + 4 cm.
lbtat4
.
Forarough figure The ratio of thelength, breadth and diagonal 1: ..i3 2 .Breadth = 6cm., Length =6 r3 cm. Area = 6 -.J x 6=36 sq.cm .
..
. -..i.Ti ....................................................................
. 1 1
Thtal 5
4.16 LA=60°
1
BC= 10 ,j cm.,
1
AC=20 cm.
Thtal 2
4.17 SQ=4
1
cos 450 =
=
1
xT2
Thtal2
4.18 sinx=i,cosx=i
1+1
sin(9O—x)==cosx 10
1+1 Thtal4
4.19
s
sin 400 •
ST
1
ST=FSxsin4O°=7xsin4O° Area =
R
1
PQxST
= 7x7sin40°
1
= 7x7x0.6428
1
= 31.4972 sq.cm. Total 4
D
MEH
A
C The angles of AAPC are 45 0 45, 900 ,
AC 442 cm.
.-
AP=4cm.,PC=4cm. PD=4cm.
.
CD=8cm.
• • •
...........
..
.
1
.
1
The angles of ABPD are 300 600 90° ,
=
.
,
,.
PD 4 cm, PB 40 cm. ................................. AP+PB
1
4+4Vcm. The lengths of the diagonals are 8 cm., 4+ 40 cm.
1
= =
••
Total 4 4.15
•
:I
H:.
wz .
For a rough figure The ratio of the length, breadth and diagonal 1: .,J :2 Breadth = 6cm., Length = 6 F3 cm. Area= 6 0x6=36isq.cm. •
., •
.
1 1 1 TotalS
-
I
•
(a) LA=60°
1
(b)
1
BC= 10 ,JT cm., AC= 20 cm.
Thtal2 4.17 1 cos45°=-=
1
Thtal2 4.18 Sifl
XTd , COSXI
1+1
sin(90-x)= =cosx
.
1+1
Thtal4 4.19
ST sm40°= PS
I
ST= PSx sin 40° =7 x sin 40 0
Area =
1 1
FQxST
= 7x7sin40°
1
= 7x7x0.6428
1
= 31.4972 sq.cm .
ibtal 4
4.20 .
LA=LD
1
sinD=,Vijigotitright.
1
LA=LD :.sinA=sinD
1
:.sinA=
1
sinA
Thtal4.
4.21
•
The area of M)AB=
.•
..
(a) OA=OB=OC=6.cm.
1 1
.x OA x OB= 18 sq.cm. •
•
(b) OAf=6sin30°=6x=3cm.
1
..Area of AOBC = x 6 x 3 = 9 sq.cm .
1
The area of quadrilateral OA8C= 18 + 9 = 27 sq.cm.
1
OR AreaofAOBC=
•
=
x 6 x 6x sin 3O0
x 6.x 6x - = 9 sq.cm.
•
. ThtMS
-
• .
........r..,...x
................
4.22
A
BD OB
sin8O° :.BD
(a)
Area of iOBC
C
=
OBxsin8O°
=
4xsin8O°
=
x OCxBD
=
x 4 x 4 sin 8O0
=
2x4x0.9848
=
7.8784 sq.cm .
1 1
OR Areaof&7BC
Areaofz\OAB
=
x 4 x 4 sih8O0
=
8x0.9848
=
7.8784 sq.cm.
=
xOAxBD
=
2x4x0.9848
=
7.8784 sq.cm. ThtaI 4
4.2 C
OA=OC
=
5cm.
1
sin4O°
=
ç
B?
=
5xsin40°
=
5x0.643
1•
.
3.215cm. cos40°
=
OP
=
5xcos40 °
=
5x0.766
=
3.830 cm.
PC
=
5
BC
=
0? I
3.830.= 1.17 /BP2 +PC 2 215)
~
.
..
,.
(1.17)2
3.337cn. ThtaIS
4.24
.
thn x =
..
2
... .
.
LBAD=x C.
BD
2 - 3
BD=4cm:
.
.
....
Thtal3
425
.
If0B=thenQ?X-3 3o0o900rjghtthan9ie.
..
.
2/i
.
• :.0?=0A
I x=3 x=6cm.Thatis,radiUS6Cm.
•
1 Total3
4.26
1 C LACD
= 400
:.AD
=CD=20m.
sin8O°
=
OA
= ADxsin8O°
D
1
= 20x0.9848=19.6960 m.
1
Height of the tree 19.6960 m
Total 4
4.27
B
0 B
A
cos40°
=BC
BD = .7660
ABL= Width of the river
lOim C
I
m.
20 .7660
1
1
26.11 m. Total 4
no
4.28 (a)
1 16m.
A
p
16m.
1
tan 70° = PQ
1
PQ
= 16tan70°
-
I
= 16x2.747. = 44m. 4.29
A
(a) 0
C
lOOm.
B
D
Height of the lighthouse = AB= lOOm.
1
tan2O° = AB BD AB - tan20
-1
100 0.342
1
= 292.4m. Thtal
4.30
A Pj
1
LA =
LD=40°
If BDis a diameter, then Lb'CD=900 3 , BC sin4O == BD = 0.64 = 4.6875 cm.
Radius = 2.34 cm. BD=2xBC ..OB=BC
Haritha's opinion is correct OR 2R=
a sinA
3 = sin4O 3 0.64 R
=2.34 cm. ThtaI4
M , 4.31
A
1
To draw the right angled triangle ABC sin70°
AC
AC==8.51 cm.
1
Move Calong the circle so that BCis a diameter .sin40° AR
=
AB
= BCxsin4O°
1
= 8.51 x 0.6428
1
= 5.47cm.
1 Totals P
4.32 a.
s
b. cos20° QS
= = 8 x 0.9397 = 7.52 cm.
.
.1
R
I
c. sin2O°
PS
= T,PS=8x0.342=2.736 PQ
1. 4
tan50°
=
2.736
PS
1
= 2.296 cm. Total 5
4.33
C
E
D
5m.
1
5m.
A
DE
=
tan70°
=
EC
= =
1 EC DE
1 x2.7475
23.79.m.
1 1 TotalS
4.34 a.
InrightangledtriangleBEc EC
=
iicm.
BE
=
30cm.
1
LDEA
=
60
1
DE
=
AE AE
r
.
15 j5,JCtfl.
=
2XDE
=
15 30 2x=cm.
=
ioJ cm.
1
11111
-
11111
Length of the rectangle =
( isJ + 60
b.
jI) 1
cm.j cm.
=
7
AE
=
ioJcm.
BE
=
30cm.
AR
=
2ojcm.
-
1 Thtal 5
4.35 a. A
0
B
C
• 6m. 6 AB
tan 45°
1
= 6m.
•AB
1
b. A
II
AB BC
1
-
6 0.5774
1
=
10.39m.
•
OR
tan 30° BC
I••'
BC
=
6JflL
10.39m. c Change.in the length of shadow = 10.39-6=4.39 m.
1 'Ibtal 5
4.36
C
a.
30m.
A
B 1
BC=
30
10 x 1.732 =
17.32m.
2 Total 3
4.37 LPQS=60°
I
QS
= 5cm.
1
PS= 5,r3 Cm.
1
C.
sin40 = PS PR PR=
sin 40
= ____ = 13.47 cm.
0.6428
1 Ibtal 4
4.38 Forarough figure
1
LOPA=54°
1
PA
= 5cm.
tan54° = 24
1
1
OA
= 5x1.3764 = 6.90 cm.
1 Total 4
4.39 - E C
B
c
A
tan20°=
. 21.6 0.36
AC
= BC=
=
59.35m. 21.6x 37.4iñt
Distance between the buildings +37.41 .96.76m. Total4
4.40
A
a. In right angled triangleAEO
1
sin7O°
-
AE
AE
=
6x0.9397
OA
1
=
AVB=
5.64 cm. 11.28cm.
.b.&48D
.
sin70°
=
AD AB
AD
=
11.28 xO.9397
=
10.59cm.
V
I
1 Thtal4
4.41 LPDB=40°
1
PD=7 cm.
sin8O°
=
AD
AD.
=
7x0.9848
=
6.89cm
cos80°
=
AP PD
A?
=
7x0.1736
=
1.22cm.
=
AP+PB=1.22+7=8.22cm.
:.AB
1
. .
.
1
. .
.
.
..
1
N
5. Solids 5. t Lateral edge =26 cm. A side of triangle =20 cm. Other sides =26 cm. each
1 1 1 Ibtal 3
5.2
Base edge = 12cm. Slant height = 6 cm.
1 1 ThtaI2
5.3
Slant height =20 cm. Knowing impossibility The slant height must be more than half the base edge
1 1 1
OR Height = =
120 2 _212
, impossible Thtal3
5.4 Slant height =5 ,j cm. Lateral surface area = 100 ..j sq.cm. Surface area = 100 + 100,J sq.cm;
1 1 1
Slant height = 40 cm. Base area = 324 sq.cm. Lateral surface area = 1440 sq.cm. Surface area = 1764 sq.cm
1 1 1 1
5.5
Thtal4 5.6
Height = 16cm.
1
Volume =
1
x 24 x 24 x 16
=3072cu.cm
1. Thtai3
stto u
Bank
IlI
5.7 Volume of the first cylinder = Volume of the second cylinder =
xlO x lOx 6 = 200 cu.cm. x 5 x 5 x 24 = 200 cu.cm .
The volumes are equal
1 1 1 Thtal3
5.8 Slant height= 15cm. Radius of the sector= 15cm. Central angle = 216°
1 1 1 Thtal3
5.9 Central aiigle of the sector = 60° Slant height= 12 cm. Radius =2 cm.
1 1 1 •Thtal3
5.10 Radius r 120 1 Slant height = I = = 3 rl-108 n 3iti2 =l08icorr=ócm. Slant height, / = 18cm.
1 1 1 1 Total4
5.11 For equating the arc length of the sector with the perimeter of the pyraniid Sum of the base perimeters= 30it cm. Area of the sector = Area of the curved surface Sum of the areas of the cCrved surface = 225it sq.cm.
1 1 1 ThtaI3
5..12 Slantheight=SOcm. Area of the base = 9007c sq.cm. Areaof the curved surface= lSOOitsq.cm. Area of 10 pyramids = 2400m x 10 sq.cm. = 7.536 sq.cm. Expense =376.80
1 1 1 1. 1 TotalS
5.13
Radius =8 cm. Height = 15 cm. Volume=mx82 x15=32Oitcu.cm.
. -
1
TotaI3 5.14
•
Volume of the cylinder = It x 122 x 15
1
Iftheheightofthecone is Ii, itsvolume= 1tx92Xh
1
4xItx9 2 xh=Itx122x15
1
h=2Ocm.
1 Ibtal4
5.15 To find the radius of each cone (6cm, 8 cm.) To find the heights of the two pyramids (8 cm, 6 cm.) To find the volume of the two cones (967t cu.cm , 1287r cu.cm.) To find the ratio as 3 :4
2 1
•
1
•
1 'lbtalS
•
5.16
Radius of the sphere
=
1
12cm.
4
Volume = 7c x 121 = 23047t cu.cm .
•
1 Thtal2
5.17 4it,2 =144it Radius =6 cm. Surface area of a sphere of half the radius
=
1 1 1 lbtal3
36it sq.cm.
5.18 1
Ratio of the volumes = 1: 27 Volume of the larger sphere = 1350 cm3.
•
1 Thtal2
5.19
4 Volumeofthesphere=.ytx6 3 fftheheightoftheconeish, its volume Height= 13.5 cm.
1• n x 82 xh
1 1
'Ibtal 3
P. 5.20 Height of the cone= 12 cm.
1
Volume of the hemisphere =7E x
93 =
4861r cu.cm .
1
Volumeofthecone= 1tx92x12 =324m cu.cm .
1
Total volume = 4861t + 3247t = 81Oit cu.cm .
1 Total 4
5.21 Volume of the vessel
it x 30 = 180007t cu.cm .
1
56.521.
.
1 'lbtal2
5.22 31c,= 120 sq.cm. 4iti2 = 160 sq.cm.
..
.
..
.
..
1 1 Thtal2
.....
...
..
.
.
.........
1••.
6. Coordinates .6.1 If the unit is 1 cm,
A(4, 8)
If the unit is 2 cm,
B(12, 16) Q1.5,3)
1 1 1
.
6.2
For a rough figure A( -6, 0), h4, 0) OP=,152_12
fO,
=h
OR42-4
1 ThIal 4
)
6.3 !1I
For marking A and Cm a rough figure Tofindh8,4);L2, 12) Length = 6; Breadth = 8 Diagonal= ,j62 +82 =10
A(2,4)
C(8, 12)
B
.
1 1 1 1
ThtaI 4
6.4
For a rough figure For writing the coordinates of Pas (1,0)
1 I 1
PC=5J
C=(1,5,j)
1 Total 4
6.5 OP=4units OAxOB =OI• 8xOB=42 011=
1 1
=2
h2, 0)
1 Total 3
6.6
Since LOAPhas angles 45°, 45°, 90°, we have AP = 6 P6,6)
1
Since AOABhas angles 300, 60°, 90°, the sides are in the ratio 1 OA =6 :. AB=6,j b6,6,J)
3 2.
Li
OB= I62 + (6,I)2 = 12
The circle cuts the x-axis at (12,0), ( - 12,0)
1 Thtal 3
6.7
V
For a rough figure
1
x
A
ABxh =12
FA
.x6 x/i =12 /z=4units
1
SinceAC=5,PC =4 AP=3 OP =5, PC = 4, Cr5, 4)
1 1
OR '7
x
A
ABxh=12
h=4units AP=3units
OP=lumts The coordinates of Care (-1,4) lbtal4
6.8
y For a rough figure
R
I
1
El
xPQxh=6O 1
x1Oxh=6O
/z=l2units In APMJ?,• RAI= 12, RP = 13 Therefore P11=5 The x- coordinate of R=(4 + 5) = They- coordinate of R= 3 + 12 = 15
1 1 1 1 Totals
estionBkj
A( -6, 0); D(6, 0)
1 1
For drawing the figure and marking the measut OP=3,PE=3
3,3)
1
The inradius, OiW =3 'J
1 'Ibtal 4
6.10 Points on the line parallel to the x-axis (4, 3), (-6, 3) Points on the line parallel to they-axis (3,5); (3, —2)
1 .
1 'lbtal2
6.11 Distance between the points (-3,2); (4, 2) =7 units Points5unitsawayfrom.(4,2)are(-1.,2);(9,2) 6.12
.. . . . . ..... Point of intersection with the x- axis (5,0) Distance between the points =2 units .. . Distance between they-axis and the lihe =5 units
1 .
1 Total2
.
1 1 I Total3
6.13 1
Point of intersection (3,5) Point at a distance of 6 units is (3,5) or (9,5)
1
Ibtal 2v
.6.14
... ........
x 1 1 1
For a rough figure Point of intersection of the lines (6,3) Distance to the point (4,3) = 10 units Distance to the point (6, 3) =6 units The point (4,3) is at a greater distance
1 ThtaI
7.
..
stion ank
7. Mathematics of Chance 7.1 Number of white beads = 24 x = 8
1
Nurnberofblackbeads =24— 8=16 Number of black beads to the removed = 8
1 1
Probability of getting a white bead =16= ThtaI 3 7.2 Side of smaller square=
..
Area of the larger square = d
.
.
.
Area of the smaller square =
.
.
..
.
Probability of marking a point within the smaller square, without loong Total 7.3
. Drawing from the second box Probabilities are the same. Probability of getting a white bead = 14
1 Thtal3
74
Number of three digit numbers with all three digits equal = 9 Total number of three digit numbers = 900 Probabih.of telling a three digit number with all three digits equal =
1 1 =
1 TotaI3
7.5 Number of slips in the first box =8 Number of slips in the second box having prime numbers in them = 4 If one slip is drawn from each box, then the total number of number pairs = 8 x 10=80 Number pairs with both numbers prime = 32 .
1 1
Probability of both being.prime = 32= 2
1
.
1 1
'Ibtal 5
7.6 Total number of number pairs such that one slip is taken from each box = 25 x50= 1250 Number of number pairs in which both are odd numbers = 15 x 30 450 Probability that both are odd numbers
450
=
9
1
1
= 25
Number of number pairs in which both are even numbers =10 x 20=200 Number of number pairs with at least one odd =1250-20Q= 1050 Probability that at least One of them is an odd number =
1050
=
1
I lbtal4
7.7 Number of pairs in which both are boys = 20 x 15= 300
1
300 2
Probabiltiy of both being boys = j. =
Number of pairs in which bOth are girls =15 x 15 = 225 Probability of both being girls
.
= 1050 14
Number of pairs in which one is a boy. and the other a girl =300+225=525
1
.
Probability of one being aboy and the other a girl =
1 525
1050=
1
1
7.8 a): Area of rectangle ABCD= x lOx 4=10 square units.
1
b)
1
Area of rectangle BEFC= 40— 10 = 3 0 square units Probability that the point is marked in BEFC =
=
..
1 ThtaJ3
79. Number of black beads = 15 x
5
1
Number of black beads = 4 Total number of beads = 14
. 1
Probability that drawing a black bead =14=
1
..
. .
7.10 ........................ a) Totalprobability=l Probability of at least one is unripe = I Th2 71.1: Total number of two digit numbers up to 50 =41 a) Number of numbrs with the digit in the tens place to be larger than the digit in the ones place = 11 Probability that the number with tens place digit is larger ..
1
.... .
11 . . 41 b)... Probability that the number with the tens plate digitis smaller than than the digit in the ones place =
the digit in the ones place =
26 41
.. . ..
.. .....
. ..
1
. ..
1
8. Tangents I LOBA=90°
.
.
1 Total2
8.2 LOQP=90° PQisthetangentatQ
8.3
1 1 Thtal
.
. To draw the circle and mark LAO? To draw the tangent
. 1 1
8.4 To draw the circle and mark LABC ..::... TodrawOA . To draw the tangent . 8.5
. BCOBx.A; ....... (or B = 1 x 3) BC = .: :...: OC=2 ... Angle measures are 300,600, 900
1 1 1 lbtal3
... . . .
..• .
.
...
. .
. ......
.
... .
..
1 1 1
'lbtal4 8.6 . . To draw the circle and the chord TodrawtherathibyjoiningA, Bwiththecene To draw the tangent atA .... . ........ To draw the tangent at B
...
.... . . .
1 1 1 'lbtal4
8.7 To draw the circle and diameters To draw the tangents through the end points of a diameter To draw the tangents through the end points of the second diameter To know that it is a rhombus
1 2 . I 1
MIM r.]
To draw the circle and mark the point P The draw the circle with the line joining the point Pand the centre as diameter To draw the tangents To write the length of the tangent as 7.33cm by measuring
1 1 1 1 Thtal4
Me 1 To draw the circles To draw the circle with the line joining the centre of the smaller circle and any point on the larger circle as diameter 1 1 To draw the tangents To write the length of the tangent by measuring 1 'LbtaI4
WE LOBA =900 AB=l2cm.
Length of the second tangent = 12 cm.
1 1 1 'LbtaI3
8.11
To draw the rough figure LR=70° LRPQ=LRQP=55°
1 1 1 Thtal3
8.12 To know that one angle of the triangle and the central angle opposite to that are supplementary One angle 40° (or to write the measure of any angle) To write the measures of the other angles.
1 1 1 Total.3
8.13 LBOC=100°
Central angle of the opposite arc of arc BPC= 260° (or angle in the opposite arc) = 50° LBPC=130°
1 1 1 'Lbtal3
8.14 To draw the circle and a radius To draw three radii making central angles 120° each To draw tangents To draw the triangle
1 1 1 1 TotaI4
To draw the circle and a radius To draw two radii making central angles 140°and 1200 To draw the tangents To draw the triangle -
1 1 I 1 Total
8.16 1 I 1 1 lbtal4
To draw the circle and a radius To draw radii making central angles 600 each To draw the tangents To draw the regular hexagon
1 1
To know that LPRQ, LPSQ are of measure 45° each To know that LQPR= 450 PQ=3jcm.
Rathus= =
3 2
3
.
jcm.
1
cm.
Thtal4 MH .
LB=LC=40° LBPR = 70°,LPQR =70° LPRQ=70° ZQFR = 40°
1 1 1
'ibtal 4 8.19 To draw AD and to know that LAED
108°
.
.::LALE=36° ZPAE = 360
. .. .1.
.•,
.
.
1 1 .
..
lbtal3
8.20
122 = 18xPS PS =8cm. Radius =3 cm.
1 1 Tht.al 3
LBCP = 900 LBAC = 900 LB =45°,LP =45° PC=5j BC=5J
PB=10
1
- -
1 1 I 1
ThtalS
8.22 M4 x MB = MCx MD MA xMB=MM
1
MC=l8cm. A14=16cm. CD = 10 cm. AB=7cm.
I 1 1 'lbtal4
8.23 PG=16x25 PC=2Ocm. Perpendicular from Cto PB = 10 ,J cm. Area of triangle =125 ,J sq.crn.
1 1 1 1 Thtal4
8.24 Aff AB=2Ocm. LBAC=90
Radius = 7.5 cm.
1 1 1 1 Thtal4
8.25 Not a bisector Angle bisectors of the triangle intersect at a point
1 1 Thtal2
8.26 The circle intersects ACand ABat Eand Frespectively. BD=x,CD=8—x CE=8—x,AE=x-2 AF=x-2
x+x-2=4 x=3cm.
.1
1 lbtaI3
8.27 To draw the triangle To draw the bisectors of two angles To draw the radius To draw the circle andto write the radius by measuring
1 1 1 1
Total4
8.28 Third side = 12 cm. To draw the radü from the centre to the sides of the right triangle. Toknow that if the radius is .ç then 12—r+35 - r = 37 r = 5 cm.
1 1 1
1 Total4
8.29 To construct the triangle To draw the angle bisectors To draw the radius ........................ To draw the circle and write the radius by measuring . H
8.30 AP—AS BP = BQ, CQ = CR, DR = DS
....
.
•:.
. ..
1 1 1 1 Thtal4
...
.•
1
Perimeter.=AB+BC+CD+AD =AP+PB+BQ+QC+CR+DR+Ds+As =2(AP+BQ+CR+D5)
1
.8.31 To.contruct the quadriiaterai . To draw the bisectors of LA and LB To draw the circle To know that CDis not a tangent to the circle
2 1 1 Total5
832 .
................
z4DQ= 900,
QP. 900
BRis drawn perpendicular to A)? In the right triangle, BR =12. cm.. To know that PQBftisarectangle To know that PQ= 12 cm.
.
p
1 1 1 I 1 Total S
9. Polvnomi Ix)=23 -322 —x-3 Remainder 23-3x22-2--3 = 8-12-2-3
. .
1
.
For (x— 2) to be a factor, 192) must be equal to zero
1
If 9 is added to 11x), then .(x— 2) is a factor of Fx). = (_2)2 + 5 x —2 + 6 =0 x+2jsafactor . Remainder I1-2)
.
.
,.... .
=9-15+6 =0 .(x+3)isafactor
... Total 2
9.3 i(l)=ax12 +bxl+c =a+b+c
.. ..
.
.
=0
. . 1
i 1
Th ....
...
If b=a+c, then =b—b
;.
.
If (x— 1) is a factor, then P1)=O
94
.
1
. .
..
.
3
...
.
.
1
If b= a+ c,then (x+ 1)isafactorofax2 +bx+c Writes a polynomial having (x-i- 1) as a factor, (with b = a + c) Eg:2+5x+3 ....
1 Thtal 3
9.5 x)=3-2-3x+2 /1)=3-2-3+2 =0 x — lis a factor.
1 1 1
=0, x+lisafactor (-1)(tzr+b)= ax+b.-ar--b• a =3,b=-2 9.6
1 1 Total 5
...
a+c=b+d F(1)=a+b+c+d=0 =a+c+a+c=0
1 1 1
.
a=-c b+d=0
1 1
The polynomial with factors(x+ 1) and (x— 1) = 2.+ 3— 2x —3 'Ibtal5 97.. .
. - 7±V49-4xlx--60
.
2 7±J 2
1
= l2or-5 - 7x —60= (x— 1 2)(x+ 5) 9.8
.
..
I&+3x+k=0,then ,.discriminant =32 -4xlxk =9-4k . (a) Ifk=-4,then 9-4k=9--4x-4 =9+16 =25>0 i9x) has factors
. . :
.
1 Total 3
. .
.
1 ..
. .
.
. .
1 1
(b) Ifk=4,then —4k=9-4x4 916 =-7<0 P(x) has no factor.
.. .
9
....
. ..
.. .
1 .
-. 9.9 ff—x-1=0,then +i±Ji 2
i±-J
=
i2—x-1=
.
2 X
(i+.i)
(i_)
2
2
.
. .
.
..,
. (x— 1) is a factor :./1) = 0 1 3 -6x1—a+b=O b5—(l) .. (x— 2) is a factor ..I2) = 0 23 -6x22 -2a+b=0 —2a+ b = 16. —(2) Solving equations (1).and (2), a=-1l,b=--6
2
............
9.10
911
I
.
..
. .
.
.
..
.
.
..
. ...
. ..
. .
.
1: 1.
.
..
.
..
.
.
1
1
...
Remainder, when divided by (x+ 1) = i9-1) _1 + 6116 . . . . . =-12 Remainder, when divided (x+ 2) = P-2) . =(-2)3 +6x(-2)2 +11x-2-6 . =-8+24-22-6 = . . .. =-12 12 must be added to make i9-1) = 0 and P-2)= 0. . ..k=12 .:.
....
..
1. 1 1 .
.
...
.
.
1
1 Total
a
it 9.12 Fx)=2-3x-1 . Suppose that (x— 1) is a factor of Px) + k. . . 1x)+k=2-3x-1+k ..2x1 2 -3x1-1+k=0 2-3-1+4=......... . .. . —2+krrO 1
OR Jx)=2i2 -3x-1 flhl)=2 - 3 - 1 =-2 IfJ1)+k=O,thenk=2 9.13
. lbtaI3
....... . Writes the po1omia1 Checks whether A 1) =0 Justifies whether it is a factor or not.
.
•.
.
.•.
1 1
9.14
+ ax + b = (x+ 3)(x— 5)
.
e +ax+b —(x+3)(x-5) =-2x-15 a=-2,b=--15 9.15
i 1 1
.
lbtal3
. 112)=4-14+5=-5 . . . . . . .. . . 2)=4-10+7=1 . ... . . . The remainder obtained on dividing Jx) + Q(x) by (x— 2) = J2) + =-5+1=--4
. 2)
. .
1 1 1 1
Thtal4 9.16
. .. The remainder obtained on dividing Jx) by (x— a) = Ja) = k The remainder obtained on dividing x) by (x— a) . = a) = —k The remainder obtained on dividing. Px) + x) by (x— a) .=Ja) + a) =k—k =0 (x— a)isafactorofJx) +Qx). ..... .
1 1
1
1 ¶btal4
I-
9.17
Since (x– 1) is a factor of i9x), i91) 0 ie 1 3 –kx1 2 --1+2=0
1
1–k-1+2=O -k+2=0;k= 2
1 'Ibtal 2
9.18 x)=2+3 +4x+7 1 27 12 . = -27 + -i-+7 = 1; (2x + 3) is not a factor
.
The polynomial hav.ing (2x+ 3) as a factor is 2 + 3 + 4x + 6 9.19
1 1 Total3
.
. i9x) =ax3 +b–ax–b . . H.. .P1) =axl 3 +bxl 2 –axl–b =a+b–a–b=O (x- 1) is a factor of P (x) . By giving different values to a and b, writes the polynomial with (x– 1) as a factor.
1 1
lbtal3 9.20
Considerx)=5+3 + ax-i-b Ifs–i isafactorofJx),thenJ1)=Oand i9-1)=O
.
1
i91)=5x1+3x1+axl+bO 1 F(-1)=5x-1+3x1+ax-1-i-b=O –a+b=2—(2) Solves equations (1) and (2) b=-3, a=-5 Polynomial to be added = –5x-- 3 .
..
. ..
.
1
1
10. Geometry and Algebra. 10.1
-Distance from the origin to (-4, 12) > 10 = 4(l 4)2 +12 2 = .. (4, 12) is a point outside the circle. Distance from the origin to (8, -6)
I
= 182 +C6)2 = 10 (8, -6) is a point on the circle. Distance from the origin to (8,2).
1
=82 +22
1
(8,2) is a point inside the circle (10,0) is a point on the circle
Total3 10.2
to
I 1
To draw a rough sketch as shown above OA=.j3242=,j5=5 :
6±8 2 = rJ00 OAxOB= OP
IOB =
10.
1 1
... ...... ..
.5x.10=OP
1 lbtal4
OF==units.
...•.:•-•. -.
.......
.. .-.
10.3
x
I
To draw the rough sketch as shown above OA=6
1
OB = AB = V(6_22+(3_0)2
.
Jj=5units Perimeter 6 + 5 -- ..Jj = 11 + ,Jj units.
.•.
.
1
lbtal4
10.4 S1opeofOC.T ABC(x, 0), Then
.
3zs1opeofAB 6-0
=3
.
C (2,6)
30-3x1 :.x1 =8 :.Ais(8,0) ............ BC=0A8 . o OC= ,jT = .Jj5 units (0,0)
B (10,6)
1
.......
1•
6
.
10.5 CD.,J0
25
.
.
/ / A
1
..
zr5
..AB,CD,DEarejnAF Since 2,5, 8 have a common difference, they are in arithmetic sequence.
. .
1 1
10.6 Centre (a, 0) Radius= jT)2+(T)2 _.5 89 16+d-8a+925 - 8a = 0 a(a-8)=0 a=0,8
I (10,6) 1
1
1 1
Co-ordinates of the Centre are (0,0) or (8,0)
Thtal 5
10.7 1
To draw the rough fIgure
x
x
1
To write the centre as (p. o)
1
012) 2
• •
(p+5)2 ± 122 (p 12)2 52 p2 + 10p+25+ 144=p2.24p+ 144+25 J/+ 10/1+24p 169-169 34p= 0 . pO Co-ordinates of the centre are (0,0)
Pd 1
I Thtal 5
10.8
Point of interse(tjofl with thex- axis is (4, To draw the rough figure
1
I
,j(x +3)2 +( y
2) 2 = \/(x4)+(y0)
1
+6x+9+y4y+4=-8x+16+y2 6x+8x-4y= 16 7 13 14x-4y=3 All the (x, y) values satisfying the equation 14x— 4y=3 can be the céntreof the circle passing through the points (4,0) and (-3,2) 1 Ifwe putx= 1 in the equation 14x-4y= 3, then 4y=11,y='3' .. Centre of the circle is (1,
IY4
)
1 Total 5
10.9
The number pair representing A is (0,p) To draw the rough figure J(O- 2)2 +(p-6)2 = j(O_3)2 +(p-5)2 ,J4+p 2 _12 p +36
1 1 I
= ..J9+p 2 .10p+25
p— 12p+40_?— 10p+34 p2 -12p+lOp= 34-40 2p=6 p=3 Co-ordinates of A are (0, 3) p2
1
Total 5
Slope of the lme=
6-2
---= 4
=2
Slope of the line joini g the points (5,6) and (8, 12) =
1 ==2 1
Since the slopes of these lines are the same, the point (8, 12) lies on the line. 1 Total 3
10.11 Slope of the linejoining the points (4,5) and (8,9) =
9-5
= 1
- Since the slope is not equal to ,the line doesn't pass through the point (8,9) 1 If the line intersects the x- axis at (x, 0), then Slope, -= 0-5 2
1
5x3=2x-8 -15 + 8 = 2x 2x=7,x= The point of intersection with thex- axis is (,
o)
1 Total4
10.12 Slope of the line passing through the points (-2,5) and (3, 8) =
Slope of the line passing through the points (5, -2) and (8,3) r
=
i
Since the slopes of the two line are not the same, the lines are not parellel. To frame the equation of the line passing through a point (x, y) and slope .
2
Eg : Equation of the line passing through the point (0,0) and having slope is •
y-O —o
3 3 = 5 ,y= 5 x ThtaI4
i0.•1
Slope of the line passing through the points (2,5) and (3, 5) 10 2 -i 1 -5 3-2 If (x, y) is a point on the line passing thorugh the point (4,6) and parallel to the line passing through the points (2,5) and (3, 5), then, x-4
y-6=2x-8. 2x - y —2=0 Ifx =3, then2x3—y=2,y =4 i
(3,4) is a point on this line. Similarly we can fmd other points by changing the values of the x- coordinate. 1 'Jbtal4 10.14 If (x, y) is the point of intersection, then - y-6 1
x-22
2y-12=x-2 2y—x=101) y-2 -1 x-6
2
2y-4=x+6 x+2ylO—(2) x+2y101) x+2ylO—(2) 4)' =20 (1)+(2) y= 5 ffweputy =5 inequation(1),thenX + 2 x 5 = 10 _r=0 The point of intersection is (0,5)
TotaI4 10.15 Slope of line AR
5-1
= j--- =
6
=2
-
-- = = 2 Slope of line BC= 9-54
1
and Clie on the same Since the slopes of the lines AR and BCare equal, the points A, B 1 line I It is not possible to construct a triangle by joining these three points. Total 3 10.16
0 Angles of PR are 30 ,600 , 90 0 . since OA = 1
AQ=,OQ=
1 Angle of AOBR are 450450 900 Since OB = 1, 1 1 OR =T2 , BR =
1
1
,
), slope of OA =
-
2
=
1
tan300._
1 1
SlopeofOB.
0. 1 =i=1
=1
tan45°=1 . . . .. . . 1 Slopes of the lines and the tangent values of the angles made by the lines with the x-axis are equal. 10J7..
.
Total 5
.
Slope ofthetangentthioughA
10-5 5 = 124 =
1
Since the tangents passing through the end points of the diameter are parallel, the slope of the tangent through Bis .
1
If (y)isanypointonthetangenttIiroughB, then x-8
8
840=5x40 8y=5x .. 5x
y=i .
1
.
.
. .
. .
..
. .
.
Ifx=16,.theny= 10 (16, 10) is another point on the tangent Thtal4
C(10,12)
L4,12)
10.18
1 b8, 4)
A(2, 4)
To draw the rough figure. Since they-coordinates of points on AR and CD are equal, these lines are parallel to the x-axis. . 1 AR and CD are parallel. Slope of AD=
124
=
=
1
=4
Slope, of BC =
1 1
Since the slopes of AD and BCare equal, they are parallel. Since the opposite sides are parallel, ABCDis a parallelogram
IbUd
10.19
.
. 4
6-2
Slope of the line
8—
.
=
.
.
.
.
.
If (x, y) is any point on the line, then y-2 x-5
-
4 3
6=4x-20 Equation of the line is 4x- 3y - 14 If the x-coordinate of a point on this line is 2, then 4x2-3y--14=0 .
. .
3y -
..
.
.
-6 (2, -2) is another point on the line. To find the co-ordinates of any other point.
...
.
..
1
-
10.20
•
3x-6y+1.0=0, Ifx=1,then3xl-6y+1O=O 6y=13
•
13 y= 6is a point on the line.
(1,
1
Ifx=4,then3x4-6y+100 t5y=22 22 (4,
is a point on the line. (13
22 '1
" 1= Slope of the line = 'Ii '-'
(1-4)
1 k = 9- x -1 =-1 6 3 2. _
1
Total 3
If (1) and(x2,y2) are any two point on the line 4x± 2y— 9
=
0,
then 411+2y1-9=0and4x2+2y2-90. .
4x1 +2y1 -9=4x2 +2y2 -9 4(x1 - x2) + 2(y1 - y2) =0 2(y1 - y2) = —4(x1 - x2) Y1Y2
••
1 1
•
1
•
•
-
- 2 2 . . Slope of the line = --2 If (x, y) is any point on the line having slope —2, ,-x2
.
.
1
x-4
y-7=-2x+8 2x+y— 15=.0.is the required equation 10.22
1
.
(a) SlopéofthelinèAB=
1
.
=
1
•
.
(b) If we take (x, y) as a point on the line, then •
•.
.
y -4 -i-
=l,y-4=x-2
.
.
Equationoftheline;x—y+2=0
177
.
.
1
1, then 1-y+20;y =3 If Another point (1, 3) (c) For any point (x,y) on this line, x-y + 2 = 0 ie,y=x+2. ie, they-coordinate is obtained by adding 2 to the x- coordinate.
1 1 1
TotaI5 10.23 7-5
Slope of the line = 3-2 = =2 If we take a point (x, y) on the line, then x-3 y-7= 2x-6 Equationofthelineis2x-y+10(l) Since(y)isapointonthiSlifle, 2x-y+ 1=0 ffx=x+1,theny(y+2) 2(x +1)_+2)+12X+2_y2+1 =2x - y+ 1 =0 Therefore(x+ 1, y+ 2)-is a point on the line Thtal4 10.24 Consider the point as F(x1 y). Then 2x-3y1 +70—(l) 3x1 +2y1 -90—(2) 4x1 -6y1 +140—(3) (1)x(2), ,
9x1 +6y1 -2704) (2)x(3), -13=0 (3)+(4),13x1
• •
x1 =1 lfweputx1 = I in equation(1), then 2x1-3y1 +7=0 3y1 =9,y1 =3 ThepointofinterseCtionOftheliflesis (1,3) if•wetake a.point on the line passing through Fas (x, y), then x-12 2y-6=x-1 Equationisx-2y+5O
•
•. (1)
TotaI5
11. Statistics 11.1
Weight (kg.)
Number of Children
.
.
Class average
Total weight
32.5
975
30-35.
3
35-40
8
37.5
300.0
40-45
12
42.5
510.0
45-50
9
47.5
427.5
50-55
6
52.5
315.0
55-60
.
2
.
Total i - ui
-
115.0
40
1765.0
.u11suucuug tue laDle
1
For finding the class average
1
For finding the total weight
1
For finding mean = 44.1
1
Total 4 11.2
Class average
Number of days
Total temperature
2.5
2
5.0
7.5
.
.
.
125 17.5
.
22.5 275
Total
.
7
87.5
10
175.0
6
135.0
2
55
30
.
22.5
.
480.0
.
For constructing the table
1 1 1
For finding the class average For finding the total temperature Forfindrngmean
480 .30 .
--
= 16°
1
Total 4 11.3
Height (cm.) Lessthanl4O Less than 145 Less than 150 Lessthanl55 Lessthanl6O Lëssthanl65
Nberofchildren . . .
.
.3 8... .: 22 34 40. H 42 -
.
.
2
•1
ynumber
1
=21
= 42
I
21-8 150-145 22-8 13
x-1455X
14
1
x= 145+4.64 = 149.64
Thtal 5
11.4
Less than 250 Less than 350 Lessthan45O Less than 550 Lessthan6SO Lessthan75O. 850
9 .....,.-. 16 34 66 .86 94 ,. . 98 100
. .
.
.
.
Less than 950
.
2.
.
100 :.
50-34 550-450
-
66-34
x_450.1OOX
.
.
.
.
.
.
16 32 .
x=450+50=SOO
.
.
TABI .............
11.5 [
L L
Central angle (in degrees) Lessthan4S. Lessthan9O. Less than Less than 180 Less than 225 Lessthan27O Less than 315 .. Lessthan360
..
[
L
.
.
..
.
.
.
.
•.
.
.
L
Numberof. sectors ...•1 . -. 4 . 14 26. 37
.
..
.
.
. .2
.
.45
. .
. .
.
.
.
. .
.
4.8. 50.
.
.
.
.
..
.
.
I• •
50 y= --=25
I
To know that fory= 14, x= 135 and fory=26, x=180. --
x-135 25-14 180-135 - 26-14
x-135=45x
1
11 12
x=135+45x.1=145+41.25=17625
.1
12
Wall 11.6 Score.
Number of children
5 6. 7 8.
..
1. 3 10 12 9 .5
:
10 Total
Total score
•. .
.
40
5. 18. 70 96 .81 50 320
io construct the table To find the total score
1
To find, mean = - =8
1
1
40
lbtal3 11.7
Daily wages (in Es.)
425 400 450 325 475 500 300 Total
Number.of workers
•
.
.
•
.
.
..
•
.
6 6. 4 4 2 2 1 25
Total wages (in Es.)
••
.
.
2550 2400 .1800 1300 .950 1000 300 10300
-
I
To construct the table To find the total wages 1 Toflndmean 10300 mean= 25 =412
1
.
.: .....
- -.
lbtal3
SAMPLE QUESTION PAPER
Mathematics 1ime: 2 Hours
TotalScore:80
General Instructions • Write the answer only after reading every question. • The first 15minutes are alloted as cool-off time. • Give explanations to the steps leading to the answer wherever necessaiy • If any pair of questions have an OR between them, only one of them need be answered. • Numbers like it, ..j ..J, need not be simplifeid with approximate values, unless specified. The algebraic form of an arithmetic sequence is 7n +4
(2)
a) What is the remainder got when the numbers in this sequence are divided by 7? b)Prove that 95 is a term of this sequence. In the figure below, LA = 900, LB= 950, LD= 1000.
C
suppose we draw a circle with BD as diameter: Find whether the points A and C are within the circle, on the circle or outside the circle. (2)
El
A
The sides of, a rectangle are parallel to the coordinate axes and the coordinates of two opposite vertices are (2, 4) and (5, 10). Find the coordinates of the other two vertices.
(2)
The 3' and 5th terms of an arithmetic sequence are in the ratio 7: 11. What is the ratio of the 10th and 16th terms.
(3)
5. D
In the figure below, 0 is the centre of the circle and A, B, C, D are points on it. Prove that LOAD + LOCD ± LABC= 1800. (3) FA II
From the four corners of a square tin-sheet, small squares of sides 6 centimetres are cut off and the tabs folded upwards to form an open box. The box can contain 15 litres of water. What was the length of the sides of the original square.
(3)
OR
A quadrilateral has two diagonals and a pentagon has five diagonals. Is there a polygon with 50 diagonals? Give reasons for your answer. The distance between two buildings is 60 metres. A boy standing right in the middle between them sees the top of one building at an angle of elevation of 450• and the top of the other building at an angle of elevation of 60°. Calculate the height of the (3) buildings. A circular sector of central angle 144 ° is bent around to form a cone. What is the ratio . ( 3) of the base-radius and slant height of the cone? Draw convenient axes of coordinate in the answer-sheet and mark the points with . (3) . coordinates (3, —2), (0, 4), (-5, 3), (-4, —2) Paper slips bearing numbers from 0 to 9 are put in a box and slips with numbers ito 9 are put in another box. A slip is drawn from each box without looking and their sum
found. . . . What is the probability of getting the sum 1?
.
.
What is the probability of getting the sum 18? What is the probabilty of getting the sum.between 1. and 18?
estloll Bank
What is the remainder got when the polynomialp(x) = 2 - 3 + 5x— 8 is divided by x - 2 7 What number added to p(x) gives a polynomial which has x - 2 as a factor? (3) How many points are there on the x-axis which are at a distance of 5 from the point (4, 4)? 'Which are they? (3) The table below shows 50 households classified according to their daily income:(3) Income Number
1 1
200 250 250 300 300 350 -
5
-
1
8
350 400 400 450 450 500 10 8 7
-
-
-
-
1
12
Calculate the mean income. 14. To dig a well, the first metre costs 1000 rupees and every metre afterwards costs 300 rupees more. A well l6metres deep, was dug.
(3) How much more than the cost of the first metre was the cost of the last metre?
What was the total cost of digging? 15. Draw a circle with two angles 55°, 800 and circumradius 3.5 centimetres. Mea- sure and write down the lengths of their sides.
(4)
16.
D
U
7
In the figure, a circle is drawn
C
touching the sides of a square. The lines PQ QR, RS SZ' Tt 1
UV, E'W are all tangents to the
circle: Prove that the sum of the perimeters of the triangles AP'BRQ, CTSand DUVis equal to the perimeter of the square. (4) I
FA
17. Draw LABC with AB= 6cm, AC= 7 cm and LA = 70 0. Draw its incircie. Measure (4) and write down its radius.
Draw a circle of radius 3 centimetres and draw tangents to it from a point 8 centimetres from its centre. Measure and write down the lengths of these tangents. a) Prove that if 22_ 1 is a factor of the polynomial ax3 + bx3 + cr + d = 0,then (4)
a+ c=b+d= 0
b) Which of the polynomials, 2x3 - 3x3— 2x-3; 2x3— 3x3— 2x+ 3 has x3— 1 as a
factor The table below shows the classification of the students in a class according to their (4)
heights in centimetres: Height Number
130 135 135 140 140 145 -
3
-
'9
-
13 .•
155- 160
145 - 150 150 - 155 8
2
5
Calculate the median height A tank has a large tap to fill it and a small tap toempty it. The time needed to empty the tank (using the small tap) is 4minutes more than the time needed to fill it (using the large tap). When both taps were opened, it took 80miñutes to fill the tank. What would be the time needed to fill the tank, using the large tap?
.
(5)
A triangular sheet has one side 20 centimetres long and the two angles on this side are (5)
550 and 85° How long is the circumradius of the triangle? What are the lengths of the other two sides of the triangle? (sin55° = 0.8192,cos55° = 0.5736, sin85° = 0.9962, cos85° = 0.0872) OR
A triangular traffic island has two of its sides. 20 metres and 25 metres and the angle between them is 55° What is the area of the island? What is the length of its third side? (sin55° = 0.8 192, cos55° = 0.5736)
j1 QuestiOhBai1k1 t•1 22. The lateral edges of a square pyramid are 26 centimetres long and its base diagonals (5) are 20 centimetres long What is the height of the pyramid? What is its slant height? Compute its volume. Compute its lateral surface area. OR
A solid metal hemisphere is melted and recast into a cone of the same radius. What is the ratio of the base diameter and height of the cone? Find out which of these solids have greater surface area. 23. What is the slope of the line joining the points with coordinates (-4, -3) and (4, 9).What is the equation of this line? Find the coordinates of the points where this line cuts the axes of coordinates. (5)
9
Scheme of Valuation Scoring Imiators
Q.No
1
Remainder when divided by 7 is 4
1.
13 7r 95 4 Remainder obtained on dividing 95 by 7 =4 95 is a term of the sequence .
OR
7n+4=95;7iz=95- 4= 91 n= 2.
91
=
1
Cis a point outside the circle
1
=
Second corner
(5, 4) =
2d=llk-
•
1
(-2, 10)
..
101hterm=5thterm+ 5d=1lk+10k=21k
•
.:.
+ 6d=21k+12k=33k
.
1
•
1
lOthterm: 16 1 term=21k:33k=7: 11
•
OR •
x3 .=7.. 11
10
16 =
7
1
.
x37:11=7:11 •
2
.......... .
..
16thterm=10thterm
2
.
7k=4k
d=2k
.
1
.
3rd term = 7k; 5th term = Ilk
4
2
13; a counting number
A is a point on the circle
First corner
3.
1
.
.
1 1
3
uestion Bankj1j
Q.No
Scoring Indicators
Join.OD LOAD =LODA
1
LOCD=LODC
+ LODC = LOAD + LOCD
LADC— LODA
1
LABC+/OAD +LOCD =LABC+LADC = 180°
One side of the square
=x
=
1
1
3
V
6(x— 12)'= 15000 sq. cm .
1
(x-12)2 =2500
1
x-12=.50
1
62 cm. OR If the number of vertices n(n-1)
a, then -.
50
..V.
1
,z2 -3n-100=0
1
b2 -4ac=9+400=409
1
Discriminant is not a perfect square
3
The polygon doesn't have 50 diagonals B D
C
.45 30m.
60 0
30m.
A
CDr30 m.
1
AB=Ojth.
1
3
H
Q.No 8
Scoring Indicators Radius of the cone, r =x R 360
Sub 1
r=R
1
1 9.
To draw the
.t
y axes
3
1
For marking 4 points 10.
Ththl
4x =2
3
Total number of number pairs = 10 x 9 = 90 The number pair with sum 1 is (0, 1) Probability of getting sum 1 =
1
The number pair with sum 18 is(9, 9) Probability of getting sum 18 =
1
The probability of getting the sum between 1 and 18 = 1 -
11.
2
=
88
1
P2)=2x23 -3x22 +5x2-8
1
= 16-12+10-8-6
1
3
Remainder when Px) is divided by (x - 2) = 6 The number added to make (x - 2). as a factor = —6 12.
1
3
If the point is (x, 0), then (x— 4)2 + (4 - 0)2 = 52
1
(x-4)2 =9 x-4=3or-3 x=7orl
1
(x, 0) = (7, 0) or (1, 0)
1
3
Q.No
13,
I
Sub Total Score Score
ScorIng lkfficators Class average
225
Frequency 5
275
325
375
425
475
8
12
10
8
7
3750
3400
3325
1125 2200 3900
fx
17700
Mean= 17700 =354 50
14.
For finding the class average
1
For findingJr
1
For finding mean
1
3
The amount required to dig 1 metre =1000,1300,1600,... 16thterm=1000±300x15=5500
1
The excess amount required to dig the 16 1 metre = 5500 - 1000 = 4500
1
Total expenditure =
1
[1000 + 5500]
= 8 x 6500 = T52000
15.
16.
1
To mark three points A, B, Con the circle so that the central angles are 1100, 160°, 90°
2
Draws MBC
1
Measures and writes the measures of An, BCand AC
1
The points where the sides AB and AD touch the circle are KandL
PWtouches the circle at M FM=PK •WM=WL
1
PW= PM+WM=PK+WL AP+PW+AW=AP+PK+WL+AW=A/(+AL =AK+AL 0K+OL
1
4
4
Q.No
Sub Score
Scoring Ind icators
Total Score
Perimeter of zI4PW= Diameter of the circle = One side of the square
1
Perimeter of 4 triangles = Perimeter of the square 17
-
1
To construct L\ABC
1
To draw the angle bisectors.
1
To locate a point where the circle touches any one side of the triangle.
1
To draw the incircle and write the inradius by measuring
1
4
4
OR 1
To draw a circle with centre 0 and radius 3 cm. To mark the point P at a distance of 8 cm from the centre
18
To draw a circle with OP as diameter
1
To draw tangents from P to the circle
1
and write the lengths by measuring
1
X2
4
—1=(x-1)(x+l)
(a)P(x)=ax3 +b+ cx+d P1)= 01a+b+c+d=O
1
i91 1 )= OPa - b+c—d=O
1
a+c=b+d 1
(b)a+b+c+dPa+c+a+c=O :.a+c=O
1
b-i-d=O
The polynomial having 19
N
- 1) as a factor = 2
-3
135
140
145
150
155
160
3
12
25
33
38
40
=20.
x-140 145-140
(2
Median=x. -
20-12 25-12
- 2x+3
1
4
Q.No
Sub Score
-
Scoring indicators
-
x-140 •
8
-
5Th
-
-
x143
- -
-
For writing
1
-
For writing the cumulative frequencyf
1.
To find the median class
1
To find the median 20
1
-
Time required for the larger tap
=
Time required for the smaller tap
x =
x+480
x
+
4
-
-
=
1
- -
-.
-
+4x=320 +
4x+ 4
=
4
1
Quantitiy of water remaining in 1 minute
x
Total Score
1 324
(x+2) 2 =324
-
-
-
x+2=18 x=16 21
-
Angle opposite to 20 cm side Circum diameter -
=
40°
20 sin40
=
-
b sin55
20 sin40
-
OR
-
-
-
-
-
--
-.
1
20 Xsin85 sin 40 -
•
-
Let the side opposite to 85° angle be c =
.2
- -
-
20 xsin55 sin40
5
1
Let the side opposite to 55° angle be b.
•
1
-
•
-
-
-
•
1
5
Scoring Indicators
Q.No (a) Area
1
-x20x25xsin5O
2011 /50
CD, CD sin5O== --:
1
.CD=2Oxsin5O
1
AD= 20xcos50
1
BD=25—AD
BC= In
22
1
CD2 +BD2
another method
d=9+c2-2bccosA
2
=202 +252 -2x20x25Xcos50
1
=1025-l000xcos5O
1
a =
1
41025--1000xcos5O
Lateral edge = 26 cm. Base diagonal =20 cm. I
Base edge = 10./bm. height, h =
4262
slant height, / = J576 + 50
Volume =
102
1
= 24 cm.
..J242 + (5 ,/)2
-i626
1
x(10,J)2 x24= 1600cu.cm.
1
Lateralsurfacearea=20X 10,Jx ,J
= 20.,,[1252 sq.cm.
1
5
Q.No
HIIIQust0 BankHjH Sub Total
Scorm Indicators
OR
h=2r,
height : base diameter = 1: 1
1
Surfacearea of the hemisphere
3 7r 12
1
Slant height of the cone = '1h2 + r2 =,j=,jr .
1
Surface area of the cone =
it
rl+
1 Base areas are the same. But ,j , ,> 27i The cone has more area 23.
(a)Slope= 9-3
.
3
12
=
(b) If (x, y) is a point on this line, then
•1 1 1
y-9_3 x-42
2(y-9) =3(x-4) 2y-18=3x-12 2y=3x+6
- 3x - 6 = 0
i
5
11111
11111
.
Scoring Indicators
Q.No
Thta1
When it intersects the x- axis, y = 0 -3x+6•=0,x=-2 1
The point intersecting the x- axis = (-2, 0) When it intersects the y- axis, x = 0 2y=6,y=3 The point intersecting the y- axis
1
(0, 3)
5
'V
-
-
I
I
