MATHEMATICS PAPER II
8. SIMILAR TRIANGLES
KEY CONCEPTS ` Similar figures : The figures which have same shape are called similar figures. ` Similar polygons : Two polygons with same number of sides are said to be similar iff. i) ii)
all the corresponding angles are equal, and all the corresponding sides are in the same ratio (in proportion).
Note : All regular polygons with same number of sides are always similar. ` Similar triangles : Two triangles are said to be similar iff their i) ii)
corresponding angles are equal, or corresponding sides are in the same ratio (in proportion).
` Basic proportionality theorem [Thale's theorem] : "If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio".
` Converse of Basic proportionality theorem :
"If a line divides two sides of a triangle in the
same ratio, then the line is parallel to the third side".
` A.A.A. similarity criterion : "In two triangles, if the angles are equal then the sides opposite to the equal angles are in the same ratio (or proportion) and hence the two triangles are similar".
` A.A.similarity criterion : "If two angles of one triangle are respectively equal to the two angles of another triangle, then the two triangles are similar".
` S.S.S. similarity criterion : "If in two triangles, the sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the triangles are similar".
` S.A.S similarity criterion : "If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar".
`
The ratio of the areas of two similar triangles is equal to the squares of the ratio of their corresponding sides.
Notes :
i)
In two similar triangles, ratio of corresponding sides = ratio of corresponding altitudes. = ratio of corresponding medians. = ratio of perimeters = a : b, then ratio of their areas
ii)
2
= a
: b
2
If the ratio of areas of two similar triangles is A
1
: A , then the ratio of their corre-
2
sponding sides (or) altitudes (or) medians (or) perimeters =
X Class MATHEMATICS PAPER II
" "
.
272
`
If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then the triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
` Pythagoras Theorem : "In a right triangle, the square of hypotenuse is equal to the sum of the squares of the other two sides".
` Converse of Pythagoras Theorem : "In a triangle, if square of one side is equal to the sum of the squares of other two sides, then the angle opposite to the first side is a right angle and the triangle is a right angled triangle".
`
If
i)
DABC ~ DDEF Scale factor =
%& &' %' "# #$ "$
ii) Scale factor < 1 [reduced or diminished figure] = 1 [congruent figure] > 1 [enlarged figure]
1 MARK QUESTIONS 1. Write 'Basic Proportionality Theorem'. Sol.
See Key Concepts.
2. Write 'the converse of Basic Proportionality Theorem'. Sol.
See Key Concepts.
3. Write the conditions for the similarity of two polygons. Sol.
See Key Concepts.
4. Write the conditions for the similarity of two triangles. Sol.
See Key Concepts.
5. Define A.A. similarity criterion for two triangles. Sol.
See Key Concepts.
6. State S.S.S. similarity criterion for two triangles. Sol.
See Key Concepts.
7. Give any two examples for similar figures. Sol.
i) all equilateral triangles ii) all squares
iii) all circles.
8. 'All squares are similar'. why ? Sol.
All squares are similar because they are regular polygons.
9. Can you give any two examples from daily life where scale factor used ? Sol.
i) Blue prints of constructions
ii) to print different size photographs. iii) to make geographical
maps.
10. Write 'Pythagoras Theorem'. Sol.
See Key Concepts.
X Class MATHEMATICS PAPER II
273
11. State 'Converse of Pythagoras Theorem'. Sol.
See Key Concepts.
12. In DPQR, E, F are points on 12 and 13 respectively. Is &' & 23 ? If PE = 4cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm. Sol.
Given, in
DPQR, E, F are points on
12 13
respectively.
1
and PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm.
1& 2& &' & 23
So
DN
DN
1' 3'
'
&
DN
DN
3
2 [by converse of basic proportionality theorem].
13. Prove that a line joining the mid-points of any two sides of a triangle is parallel to third side. Sol.
Let D, E are the mid points of
Then, AD = DB
and AE = EC
Þ
respectively in
DABC. "
= 1
%
%& & #$
&
= 1
$
#
"% "& %# &$
So,
Þ
"% %# "& &$
Þ
"# "$
[ by converse of basic proportionality theorem]
Hence proved.
14. DABC ~ DDEF, BC = 3 cm, EF = 4 cm and ar[DABC] = 54 cm2. Find ar[DDEF]. Sol.
Given,
DABC ~ DDEF
BC = 3 cm, EF = 4 cm, ar[
We know,
BS <'%&'> BS <'"#$>
&' #$
BS <'%&'>
DABC] = 54 cm2. [by theorem]
\ ar[DDEF] = u
= 96 cm
2.
15. D ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2. Sol.
Given
DABC is a right angled isosceles triangle and ÐC = 90°
So, AC and AB
= BC
2
2
= AC
"
2
+ BC
2
= AC + AC
2
Q AC = BC]
[
2
= 2 AC
\ AB2 = 2AC2 X Class MATHEMATICS PAPER II
$
# 274
2 MARKS QUESTIONS 1. In D ABC, Sol.
Given, in
and
%& & #$
D ABC,
and
Q
%& & #$
"% , AC = 5.6 cm. Find AE. %# , AC = 5.6 cm
"% "% %# "# "& "%
%& & #$ "$ "# ,
"&
"
[by basic proportionality theorem]
&
%
$
#
\ AE = 3 × 0.7 = 2.1 cm.
2. Prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. Sol.
Let, in
D ABC, 'D' is the mid-point of
And
%&
drawn parallel to
"#
Þ AD = DB and
"% %#
= 1
#$
"
Then,
Þ
"& "% &$ %#
Þ
"& &$ = 1
Þ
AE = EC
Þ
E is mid-point of
Þ
%&
%
[by basic proportionality theorem]
' "% %#
§ ¨ ©
·¸
#
& $
¹
"$ .
bisects the third side
"$ .
Hence proved.
3. A flag pole 4m tall casts 6 m shadow. At the same time, a nearby building casts a shadow of 24m. How tall is the building ? Sol.
Given, Height of the flag pole, h = 4 m Length of shadow of flag pole, l = 6 m Length of shadow of building = L = 24 m Height of the building, H = ?
We know
) I M
[by theorem]
) H = 4 × 4 = 16 m
X Class MATHEMATICS PAPER II
275
4. Diagonals
"$
and
#% of a trapezium ABCD with "# & %$ intersect each other at the point
'O'. Using the criterion of similarity for two triangles, show that Sol.
Given, in trapezium ABCD,
"$ #%
and
From
"# & %$
intersecting at 'O'.
0
Q vertically opposite angles]
[
0"# 0
%$Q alternate interior angles]
[
\ D AOB ~ D COD [Q A.A. similarity criterion] So,
$
%
D AOB and D COD,
"0# $0%
0" 0# . 0$ 0%
0" 0# 0$ 0%
[
#
"
Q C.P.S.T.]
5. Show that the ratio of the medians of two similar triangles is equal to the ratio of their corresponding sides. Sol.
Let
D ABC ~ D DEF then
i)
"
ii)
"# #$ "$ %& &' %'
% # & $ '
and P, Q are the mid-points of
then
From
BP = PC =
BC,
EQ = QF =
EF
#$
and
&'
respectively,
#
1
$
&
2
'
D ABP and D DEQ,
# & #1 &2
\ So,
%
"
[
\ from (i)]
#$ &'
#1 "# &2 %&
=
[
#$ &'
Q P, Q are mid-points of
[
#$ &' ]
Q from (ii)]
D ABP ~ D DEQ [by S.A.S. similarity criterion] \
"1 "# #$ "$ %2 %& &' %'
\ from (ii)]
[
Hence proved.
6. Diagonals of a trapezium ABCD with
"# & %$
intersect each other at the point 'O'. If AB = 2
CD then find the ratio of areas of D AOB and D COD. Sol.
Given, in trapezium ABCD,
X Class MATHEMATICS PAPER II
"# & %$ 276
"$ #%
intersecting at 'O'.
and AB = 2 CD From
"0# $0%
0"# 0
%$0
Q vertically opposite angles]
[
Q alternate interior angles]
[
\ D AOB ~ D COD [Q A.A. similarity criterion]
Then,
$
%
D AOB and D COD,
"# $%
BS <'"0#> BS <'$0%>
[by theorem]
$%
=
\ AB = 2 CD]
[
%$ $%
=
#
"
= 4 : 1
$%
7. A ladder 25 m long reaches a window of building 20 m. above the ground. Determine the distance of the foot of the ladder from the building. Sol.
Let, in
D ABC,
AB = length of the ladder = 25 m AC = height of the window from the ground = 20 m
\
\
"# "$
C = 90°, BC =
"
=
=
=
N
#
= 15 m
N
$
Distance of the foot of the ladder from the building = 15 m
8. In the given figure, if Sol.
[by Pythagoras Theorem]
From the figure, in
From
From
D ABD,
D ACD,
"%#
"%$
"% A #$
Þ DABC, = 90°
= 90°
AB
2
Þ AB
2
\ AD
2
Þ AC
BD
Þ AB X Class MATHEMATICS PAPER II
2
2
$
"% A #$
2
\ AD From (1) and (2),
then prove that AB2 + CD2 = AC2 + BD2.
2
= AD
2
= AB
2
2
= AC
2
+ CD
CD
CD
2
= AC
2
% (1)
+ CD
2
= AC
BD
2
2
= AD
2
+ BD
2
(2)
#
"
2
+ BD
2
277
9. PQR is a triangle, right angled at 'P' and 'M' is a point on PM2 = QM. MR. Sol.
Given, in So,
D PQR, 1
= 90° and
23
such that 1. A 23 . Show that
1. A 23
3
D PMQ ~ D RMP [\ Theorem]
1.
2. Þ .3 1.
\ PM
2
.
[C.P.S.T.]
2
1
= QM. MR
4 MARKS QUESTIONS 1. State and prove 'Basic Proportionality Theorem'. 2. State and prove 'converse of basic proportionality theorem'. 3. State and prove 'Pythagoras Theorem'. 4. State and prove 'Converse of Pythagoras Theorem'.
3/ are respectively the medians of two similar triangles D ABC and D PQR. Prove $. "# that i) D AMC ~ D PNR ii) 3/ 12 .
5. $. and
6. Prove that if the areas of two similar triangles are equal, then they are congruent. 7. D, E, F are the mid-points of sides D ABC.
#$ $" "# of DABC. Find the ratio of areas of D DEF and
8. Equilateral triangles are drawn on the three sides of a right angled triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides. 9. Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal. 10. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. 11. If 'O' is any point inside a rectangle ABCD, then prove that OB2 + OD2 = OA2 + OC2. 12. ABC is a right triangle right angled at C. Let BC = a, CA = b, AB = c and 'P' be the length of . perpendicular from 'C' on 'AB'. Prove that i) PC = ab ii) 1 B C 13. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m. Find the distance between their tops. 14. In an equilateral triangle ABC, 'D' is a point on side #$ such that BD = 9 AD2 = 7 AB2.
BC Prove that
5 MARKS QUESTIONS 1. Construct a triangle of sides 4 cm, 5 cm and 6 cm. Then, construct a triangle similar to it whose sides are of the corresponding sides of the first triangle. X Class MATHEMATICS PAPER II
278
2. Construct an isosceles triangle whose base is 8 cm and altitude is 4 cm. Then, draw another triangle whose sides are
time the corresponding sides of the isosceles triangle.
3. Construct a triangle shadow similar to the given D ABC with its sides equal to
of the
corresponding sides of the triangle ABC.
OBJECTIVE QUESTIONS I. Fill in the blanks. 1.
If a line drawn parallel to one side of a triangle, then it divides other two sides in .......................
2.
If a line divides two sides of a triangle in the same ratio, then the line is ............. to the third side.
3.
In
D ABC, if 'D' and 'E' are the mid-points of
4.
In
D PQR, if
5.
In
D ABC, if AD : DB = AE : EC, then ..............................................
6.
If in
7.
Symbol for similarity is .......................................
8.
The relation 'similarity' is ....................................... relation.
9.
The polygons in which all sides and angles are equal then they are .......................................
D XYZ,
"#
and
"$
respectively, then AD : DB = ................
9: & 23 , then PX : PQ = .......................................... 12 & :;
and XP : XY = 3 : 5 and XQ = 6 cm, then QZ = .......................................
10.
The regular polygons with same number of sides are always .......................................
11.
All equilateral triangles are always similar because .......................................
12.
All isosceles triangles are .......................................
13.
Reduced or enlarged photographs of an object are .......................................
14.
Two polygons with the same number of sides are ......................................., if their corresponding angles are equal and corresponding sides are equal.
15.
The ratio of the corresponding sides of similar triangles is called .......................................
16.
If the scale factor is equal to one, then the similar figure is .............................. to the given figure.
17.
In
18.
If
D ABC ~ D DEF,
19.
If
D ABC ~ D DEF, scale factor
20.
D ABC, if
%& & #$ "
then
D ADE and D ABC are .......................................
= 50°,
'
= 70°, then
#
= .......................................
and DE = 9 cm then AB = .......................................
If the scale factor is greater than one, then similar figure is ....................................... comparing with the given figure.
21.
In a quadrilateral ABCD, if the diagonals are intersecting at 'O' such that
"0 $0 #0 %0
, then ABCD
is a ....................................... 22.
A person 1.65 m tall casts 1.8 m shadow, at the same instance, a lamp-post casts a shadow of 5.4m then height of lamp-post is .......................................
X Class MATHEMATICS PAPER II
279
23.
24. 25.
If
D ABC ~ D PQR and
"9
and
%:
$.
and
3/
are medians respectively then CM : AB = .......................................
are altitudes of two similar triangles
D ABC and D DEF then AX : AB = ..............................
If the ratio of the corresponding sides of two similar triangles is 3 : 2 then the ratio of their perimeters is .......................................
26.
If the ratio of the corresponding sides of two similar triangles is 1 : 2 then the ratio of their corresponding altitudes is .......................................
27.
If the areas of two similar triangles are equal then they are .......................................
28.
If the areas of two similar triangles is 1 : 2 then the ratio of their corresponding sides is ...............
29.
If the ratio of the corresponding altitudes of two similar triangles is 2 : 3 then the ratio of their areas is .......................................
D ABC, if
%& & #$
D ABC, if
%& & #$
D ADE] : Ar[D ABC] = .......................................
30.
In
31.
In
32.
If the perimeters of two similar triangles
2
36 cm 33.
then Ar[
and AD : DB = 2 : 3 then Ar[
then
D ADE ~ D ABC'. Here the criterion is ....................................... D ABC, D PQR are in the ratio 3 : 4 and Ar[D
ABC] =
D PQR] = .......................................
If the areas of two similar triangles are equal then their corresponding sides are in the ratio .......................................
34.
If the corresponding sides of two similar triangles are equal, and the area of one triangle is
2
64 cm , then the area of other triangle is .......................................
35.
#$ $" "#
If D, E, F are the mid-points of
of
D ABC respectively then the ar[D DEF] : ar[D ABC]
= ....................................... 36.
The ratio of the areas of equilateral triangles described on a side and diagonal of a square is .......................................
37.
Diagonals of trapezium ABCD with
"# & %$
intersecting at 'O'. If AB = 3 CD then ar[
D
COD] :
D AOB] = .......................................
ar[
#
38.
In a right triangle ABC,
39.
In
D ABC,
40.
In
D PQR, PQ2 + QR2 = PR2 then
i)
#
= 90° and
= 90° then AB
2
2
#% A "$
then
D ADB ~ .......................................
2
+ BC = AC . This is .......................................
D PQR is .......................................
ii) hypotenuse is ....................................... iii) right angled vertex is .......................................
iv)
2
= .......................................
41.
Pythagoras theorem was earlier stated by the Indian mathematecian is .......................................
42.
For a right triangle with integer sides, at least one of its measurements must be ............ number.
43.
In a square ABCD, AB
44.
In a rhombus ABCD, AC
2
2
: AC
2
X Class MATHEMATICS PAPER II
= .......................................
2
+ BD
= .......................................
280
45.
In
D ABC,
#
= 90° and
#% A "$ then BD2 = .......................................
46.
In
D PQR,
1
= 90° and
1. A 23
47.
In
D XYZ,
:
= 90° and
:1 A 9;
48.
D ABC is an isosceles triangle right angled at 'C' then AC : AB = .......................................
49.
If 'O' is any point in the interior of rectangle ABCD then OA
50.
In an obtuse angled triangle ABC, obtuse angle at A and
i)
AB
ii) AB iii) AB
51.
In
2 2 2
2
AC
2
BD + CD
2
#
D ABC,
then PQ
2
then YZ
2
= .......................................
= .......................................
2
2
+ OC
"% A #$
= .......................................
then
= ....................................... = ....................................... = .......................................
= 90°,
#% A "$ , AD = 4 cm and AC = 13 cm then BD = .......................................
ANSWERS I.
1) same ratio
2) parallel
3) 1 : 1
4) PY : PR
6) 4 cm
7) ~
8) equivalent
9) regular polygons
10) similar to each other
11) they are regular polygons
12) not similar to each other
13) similar
5)
%& & #$
14) congruent
15) scale factor
16) congruent
17) similar to each other
18) 60°
19) 13.5 cm
20) enlarged
21) trapezium
22) 4.95 m
23) RN : PQ
24) DY : DE
25) 3 : 2
26) 1 : 2
27) congruent
28)
29) 4 : 9
30) 4 : 25
31) A.A.A. similarity criterion (or) S.A.S. similarity criterion (or) S.S.S. similarity criterion)
32) 48 cm
2
2
33) 1 : 1
37) 1 : 9
38)
34) 64 cm
D ABC or D BDC
40) i) right triangle
41) Boudhayan
2
44) AB
2
+ BC
2
+ CD + DA
46) QM. QR
50) i) BD
2
CD
2
ii) AC
2
2
(or) 4AB
CD
13
iii) Q
iv) 90°
43) 1 : 2
47) XZ . ZP
2
36) 1 : 2
39) Pythagoras theorem
ii)
42) even
35) 1 : 4
(or) 4BC
48)
2
2
(or) 4CD
2
iii) AC
2
(or) 4 DA
2
2
49) OB
2
+ BD
45) AD. DC
2
+ OD
51) 6 cm
h h h h h X Class MATHEMATICS PAPER II
281
9. TANGENTS AND SECANTS TO A CIRCLE
KEY CONCEPTS `
The locus of points in a plane equidistant from a fixed point is called called
`
circle. The fixed point is
centre of the circle and fixed distance is called radius of the circle.
According to Salman, who is a Bengali-American mathematician, there can be 3 possible ways of presenting a circle and a line given on a plane. i)
If the line and circle have no-common point, then the line is called
ii) If the line intersects the circle at two points then the line is called
non-intersecting line. secant of the circle.
iii) If the line intersects the circle at only one point then the line is called
1
2
2
1 1
2
12 JT4FDBOU
12 OPOJOUFSTFDUJOHMJOF `
tangent to the circle.
12 JT5BOHFOU to touch'.
The word tangent is taken from the Latin word 'tangree' which means ' introduced by Danish Mathematician
It was
Thomas Fineke in 1583. point of contact.
`
The common point of the tangent and circle is called
`
The tangent at any point of a circle is
`
The line containing the radius through the point of contact is also called
perpendicular to the radius through the point of contact. normal to the centre
at the point.
0
1 `
"
2
O is the centre of circle of radius OA. AP is a tangent drawn from an external point P. The length
of tangent AP =
01 0"
2 0 X Class MATHEMATICS PAPER II
1 282
`
The lengths of tangents drawn from an external point to a circle are
equal.
" 0
1 #
`
The centre of the circle lies on the
1" 1#
bisector of the angle between two tangents drawn from a
point outside it.
`
The line segment joining any two points on a circle is called
diameter.
through the centre of circle is called
`
chord. The largest chord passing
In two concentric circles, the chord of bigger circle, that touches the smaller circle is
bisected
at the point of contact with the smaller circle.
0 " `
1
#
If two tangents AP and BP are drawn to a circle with centre O from an external point P then
"1# 0"# 0#" " 1
0 # `
If the circle touches all the sides of a quadrilateral internally the sum of two pairs of opposite sides are equal AB + CD = BC + DA.
$
1
%
4
+ " `
,
#
A chord divides the circle into two parts. i) Major segment ii) Minor segment.
1 .BKPS 4FHNFOU "
X Class MATHEMATICS PAPER II
.JOPS 4FHNFOU 2
#
283
semicircles.
`
Diameter of circle divides it into two
`
The area of sector which makes an angle x° at the centre of circle is
Yq
×
pr2, where r is radius
of circle.
0 S
Y¡ S
#
" `
Area of segment of circle = The area of sector formed with the segment The area of triangular part of sector.
0 "
1
#
`
Area of segment APB = Area of sector OAPB Area of
`
The radius of circumscribing circle of a regular hexagon is equal to the length of its side.
`
The radius of a circle is r. The length of the tangent drawn from an external point at a distance 'd' is l. Then
d2 = l2 + r2
(or)
l2
D OAB.
= d2 r2 (or) r2 = d2 l2
`
Area of Major segment = Area of circle Area of Minor segment.
`
To find the quantities r, l and d, when any two of them are given, the following pythagoras triplets are useful to say the answer directly. a) 3, 4, 5
b) 5, 12, 13
c) 6, 8, 10
d) 7, 24, 25
e) 8, 15, 17
f) 9, 12, 15
g) 12, 35, 37
h) 15, 20, 25
i) 16, 30, 34
j) 20, 21, 29 etc.
1 MARK QUESTIONS 1. PA, PB are two tangents drawn from an external point of a circle with centre O. What is the value of 1"0 1#0 ? Sol.
180°
2. Radius of a circle is 5 cm. The angle between the tangents drawn from an external point is 60°. Find the distance between the centre of circle and intersection point of tangents ? " "0 Sol. From D AOP, sin 30° = 10 ¡ 1 0 ¡ Þ PO = 10 cm 10 # X Class MATHEMATICS PAPER II
284
3. The length of tangent from a point to a circle of radius 3 cm is 4 cm. What is the distance between centre of that point ? Sol.
r = 3 cm, l = 4 cm
" 0
d =
M
S
1
= 5cm
4. From a point Q, the length of tangent is 24 cm, and the distance of Q from the centre is 25 cm. What is the radius of circle ? Sol.
l = 24 cm, d = 25 cm, r =
E S
= 7 cm
5. If AP, AQ are two tangents to a circle with centre O such that 102 = 110°. Find 1"2 ? 1 Sol. 0 1 " 2 = 360°
"
= 360
0
1
0
2
= 360 290 = 70°
¡
"
2
6. PA, PB are two tangents and angle between them is 80°. Find 10" 10# ? Sol.
"10
= 40°,
0"1
= 90°
" ¡
0
1
# \
10"
10#
= 180 (90 + 40) = 50°
7. PA, PB are two tangents with angle at their point of intersection is 70°. What is the value of "10 and #10 ? Sol.
35°
8. Two concentric circles of radii 5 cm and 3 cm are drawn. Find the length of the chord of larger circle which touches the smaller circle ? Sol.
a = 5 cm, b = 3 cm
Length of chord =
B C
= 8 cm
9. What is the angle in major segment ? Sol.
acute angle
X Class MATHEMATICS PAPER II
285
10. What is the angle in minor segment ? Sol.
obtuse angle
11. The perimeter of a circle is 440 cm. Find the side of square inscribed in the circle ? Sol. pd = 440 d = 440 ×
= 140 cm
Diagonal of square = 140 °
Side =
u
= cm
2 MARKS QUESTIONS 1. Draw a circle and two lines parallel to a given line such that one is a tangent and the other is a secant to the circle ? 2. Prove that the lengths of tangents drawn from an external point to a circle are equal ? 3. The length of minute hand of a clock is 14 cm. Find the area swept by the minute hand in 10 minutes ? 4. A chord of a circle of radius 10 cm, subtends a right angle at the centre. Find the area of minor segment and major segment ? 5. If a chord subtends an angle 120° at the centre of circle of radius 12 cm. Find the area of corresponding minor arc ? 6. Find the area of shaded region in the fig, if ABCD is a square of side 7 cm, and APD, BPC are semicircles.
DN
%
$
1
#
"
7. AB, CD are respectively area of two concentric circles of radii 21 cm, and 7 cm with centre O. If "0# = 30°. Find the area of shaded region.
0
$ "
% #
8. Construct a tangent to a circle at a given point when the centre of the circle is known. X Class MATHEMATICS PAPER II
286
4 MARKS QUESTIONS 1. In the adjacent figure radius of circle is 21 cm and
"0#
= 120°. Find the area of segment AYB ?
" 0
: #
2. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct pair of tangents and measure their lengths. Verify by using pythagoras theorem. 3. Prove that parallelogram circumscribing a circle is a rhombus. 4. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Verify by actual calculation. 5. Draw a circle with the help of a bangle. Take a point outside the circle and construct a pair of tangents from this point to the circle. Measure them and conclude it.
OBJECTIVE QUESTIONS I. Fill in the blanks. 1.
The no.of tangents drawn through any point of a circle is ..................................................
2.
A tangent to a circle intersects it in .................................................. point.
3.
A line intersecting a circle in two points is called a ...................................................
4.
The no.of tangents can be drawn to a circle is ..................................................
5.
The common point of a tangent to a circle and the circle is called ..................................................
6.
If the line drawn through the end point of radius of circle is perpendicular to the radius then it is called .................................................. to the circle.
7.
The tangents drawn on either sides of diameter of circle are .................................................. to each other.
8.
The maximum no.of parallel tangents drawn to a circle ..................................................
9.
The no.of tangents can be drawn from an interior point to a circle is ..................................................
10.
The no.of tangents drawn from an external point to a circle is ..................................................
11.
It a circle touches all sides of a quadrilateral ABCD, internally then AB + CD = ......................................
12.
The parallelogram circumscribing a circle is ..................................................
X Class MATHEMATICS PAPER II
287
13.
The angle between the tangents drawn from an external point to a circle is always ............................
14.
The angle between the tangent to a circle and radius at the point of contact is ...........................
15.
Radius of circle is 6 cm. The length of tangent drawn to the circle through a point 10 cm from the centre is ..................................................
16.
The radius of circle is 7 cm. The area of sector making an angle 60° with centre of circle is ..................................................
17.
The length of tangent to a circle of radius 4 cm is 3 cm. The distance from the centre of circle to the intersection point of tangent and external point is ..................................................
18.
Two concentric circles of radii a and b (a > b) are given. The chord AB of larger circle touches the smaller circle at C. The length of AB is ..................................................
¡
Y¡
0
19.
In the adjacent figure value of x = ..................................................
20.
The length of tangent drawn from an external point of a circle at a distance of 6 cm to a circle of radius 3 cm is ..................................................
21.
The no.of tangents can be drawn parallel to a secant of circle is ..................................................
ANSWERS I.
1) 1
2) 1
3) secant of circle
4) infinite
5) point of contact
6) tangent
7) parallel
8) 2
9) 0
10) 2
11) BC + AD
12) Rhombus
13) acute angle
14) 90°
15) 8 cm
16) 25.66 cm
17) 5 cm
18)
19) 120°
20)
DN
2
B C
21) 2
h h h h h
X Class MATHEMATICS PAPER II
288
10. MENSURATION
KEY CONCEPTS ` Cube :
2
a)
Lateral surface area = 4a
b)
Total surface area = 6a
c)
Volume = a
d)
Diagonal =
(a = side)
2
3
B
` Cuboid : l = length, b = breadth, h = height a)
Lateral surface area = 2h(l + b)
b)
Total surface area = 2(lb + bh + hl)
c)
Volume = lbh
d)
Diagonal =
M
C I
` Cylinder : r = Radius of the base, h = height
S
pr2
a)
Area of base =
b)
Perimeter of base = 2 r
c)
Lateral surface area = 2 rh
d)
Total surface area = 2
e)
Volume =
p
p
pr(r + h)
pr2h
` Cone : r = Radius of the base, h = height, l = slant height S
I
a)
l=
b)
Lateral surface area =
c)
Total surface area =
d)
Volume =
prl
pr(l +r)
2 pr h
` Sphere : r = Radius of the sphere a)
Total surface area / curved surface area = 4
b)
Volume =
` Hemisphere :
pr2
3 pr p
2
a) Lateral surface area = 2 r
p
2
b) Total surface area = 3 r
c) Volume =
`
3 pr
Volume of Embankment of ring =
X Class MATHEMATICS PAPER II
p(R2 r2) h. 289
1 MARK QUESTIONS 1. Find the volume of a sphere of radius 2.1 cm 2. Find the total surface area of a cuboid with measurements 5 cm × 4 cm × 3 cm. 3. A right circular cylinder has base radius 14 cm and height 21 cm. Find it's curved surface area. 4. Find the volume of hemisphere of radius 3.5 cms.
2 MARKS QUESTIONS 1. Find the volume and the total surface area of a hemisphere of radius 3.5 cms. Sol.
Radius of hemisphere = r = 3.5 cms. Total surface area of hemisphere = curved surface area + area of circle.
pr2 + pr2 = 3pr2
= 2
= 3 ×
Volume of hemisphere =
3 pr = u
2
× 3.5 × 3.5 = 111.5 cm
3
× 3.5 × 3.5 × 3.5 = 89.8 cm
2. An oil drum is in the shape of a cylinder having the following dimensions. Diameter is 2m. and height is 7 m. If the painter charges ` 3 per m3, to paint the drum, find the charge to be paid to the painter. Sol.
Radius of cylinder = r = 2m Height of cylinder = h = 7 m. Lateral surface area of cylinder = 2
prh
= 2 ×
= 88 m Charges per 1 m
2
for painting
2 Charges for 88 m
for painting
Amount to be paid to painter
=
× 2 × 7
2
`3
= 3 × 88 = =
N
N
` 264.
` 264
3. A jokers cap is in the form of right circular cone whose base radius is 7 cm and height is 24 cm. Find the area of the sheet required to make 10 such caps. Sol.
Radius of the cap r = 7 cm Height of the cap h = 24 cm
\ Slant height of the cap (l)
X Class MATHEMATICS PAPER II
VÊ { Ó
I
=
S
=
=
ÇÊV =
= 25 cm.
290
Lateral surface area of cone =
prl =
× 7 × 25 = 550 sq.cm.
Sheet required to make 10 caps = 10 × lateral surface area of cone = 10 × 550 = 5500 sq.cms.
4. A sports company was ordered to prepare 100 paper cylinders for shuttle cocks. The required dimensions of the cylinder are 35 cm length / height and its radius is 7 cm. Find the required area of thin paper sheet needed to make 100 cylinders. Sol.
Radius of cylinder = r = 7 cm Height of cylinder = h = 35 cm
pr(r + h)
Total surface area of cylinder = 2
= 2 ×
DN
× 7 × (7 + 35) DN
= 44 × 42 = 1848 sq.cms
\ Area of
thin paper sheet needed to make 100 cylinders = 100 × 1848 = 184800 sq.cms.
5. A self help group wants to manufacture jokers caps (conical caps) of 3 cm radius and 4 cm height. If the available colour paper sheet is 1000 cm2, then how many caps can be manufactured from that paper sheet ? Sol.
Radius of cone = r = 3 cm Height of cone = h = 4 cm
Slant height of cone = l =
S
I
=
=
Lateral surface area of cone =
prl =
VÊ {
= 5 cm
× 3 × 5 =
ÎÊV
sq.cm.
Area of colour paper = 1000 sq.cms.
Number of caps can be manufactured from the given colour sheet =
"SFB PG DPMPVS TIFFU -BUFSBM TVSGBDF BSFB PG DPOF =
= 21 caps
6. A solid iron rod has a cylindrical shape. Its height is 11 cm and base diameter is 7 cm. Then find the total volume of 50 rods. Sol.
Diameter of cylinder = d = 7 cm
Radius of cylinder = r =
X Class MATHEMATICS PAPER II
E
cm
291
Height of cylinder = h = 11 cm Volume of the cylinder =
=
pr2h
u u
× 11 =
cm
V Ê£ £
3
ÇÊV
Volume of 50 rods = 50 ×
3
= 21,175 cm
7. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the capsule is 14 mm. and the width is 5 mm. Find its surface area. £{Ê Êx
Sol. Diameter of both cylinder and hemispheres = d = 5mm
radius = r =
E
mm
Length of the capsule = h = 14 mm. Total surface area of capsule
= Lateral surface area of cylinder + 2 × curved surface area of hemisphere
2
prh + 2 × 2pr
= 2
pr(h + 2r) = u
= 2
u
=
§ · u ¨ u ¸ © ¹
u u
= 298.57 sq.mm
8. Two cubes each of volume 64 cm3 are joined end to end together. Find the surface area of the resulting cuboid. Sol.
3
Volume of cube = a
Side of cube = a =
= 64 cm
3
= 4
3
cm
3
= 4 cm.
By joining two cubes end to end resulting cuboid measurements are l = 8 cm, b = 4 cm, h = 4 cm
nÊV
V Ê{ {ÊV ³ {ÊV
X Class MATHEMATICS PAPER II
V {Ê
292
Lateral surface area of cuboid = 2h (l + b) = 2 × 4 (8 + 4) = 96 cm
2
= 2 (lb + bh + lh)
Total surface area of cuboid
= 2(8 × 4 + 4 × 4 + 8 × 4) = 2 × 80 = 160 cm
2
9. Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 7 cm. Sol.
Diameter of largest right circular cone that can be cut out of a cube having edge 7 cm = d = 7 cms
Ç Ó
radius = r =
cm
height = h = 7 cms
Volume of the cone =
=
DN
2 pr h
DN
u u u u 3
= 89.83 cm .
10. A metallic sphere of radius 4.2 cm. is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder. Sol.
Radius of metallic sphere = r
= 4.2 cms
Radius of cylinder
= 6 cms
1
= r
2
Volume of sphere = Volume of cylinder
pr 3 = pr22h 1 Height of cylinder = h =
S S S S
u u u uu
= 2.74 cm
11. A 20 m deep well with diameter 7 m. is dug and the earth from digging is evenly spread out to form a platform 22 m. by 14 m. Find the height of the platform. Sol.
Deep (height) of dug (cylinder) = h = 20 m
Diameter = d = 7m,
radius = r =
N
E m.
Volume of the sand from dug
=
=
pr2h
N
u u u = 770 m3
I
Cuboid length = l = 22m, cuboid breadth = b = 14m. Volume of sand = volume of cuboid
3
770 m
= lbh
N
N
22 × 14 × h = 770
Height of platform = h =
X Class MATHEMATICS PAPER II
u
= 2.5 m.
293
4 MARKS QUESTIONS 1. A toy is in the form of a cone mounted on a hemisphere the diameter of the base and the height of the cone are 6 cm and 4 cm respectively. Determine the surface area of the toy ? 2. A storage tank consists of a circular cylinder with a hemisphere stuck on either end. If the external diameter of the cylinder be 1.4 m. and its length be 8 m. Find the cost of painting it on the outside at rate of ` 20 per m2. 3. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm and its base radius is of 3.5 cm, find the total surface area of the article.
4. An iron pillar consists of a cylindrical portion of 2.8 m height and 20 cm. in diameter and a cone of 42 cm. height surmounting it. Find the weight of the pillar if 1 cm3 of iron weighs 7.5 g. 5. A toy is made in the form of hemisphere surmounted by a right cone whose circular base is joined with the plane surface of the hemisphere. The radius of the base of the cone is 7 cm.
of the hemisphere. Calculate the height of the cone and the surface area of the toy correct to 2 places of decimal.
and its volume is
6. A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of right circular cone mounted on a hemisphere is immersed into the tub. The radius of the hemisphere is 3.5 cm and height of cone outside the hemisphere is 5 cm. Find the volume of water left in the tub. 7. Spherical marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm., which contains some water. Find the number of marbles that should be dropped into the beaker, so that water level rises by 5.6 cm. 8. Metallic spheres of radius 6 cm., 8 cm. and 10 cm. respectively are melted to form a single solid sphere. Find the radius of the resulting sphere. 9. A well of diameter 14 m. is dug 15 m. deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 7 m to form an embankment. Find the height of the embankment. 10. How many silver coins, 1.75 cm in diameter and thickness 2 mm., need to be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm ? X Class MATHEMATICS PAPER II
294
OBJECTIVE QUESTIONS I. Fill in the blanks. 1.
Ratio of volume of cone and cylinder with same radius and height is ............................................
2.
The lateral surface area of cylinder is equal to the curved surface area of cone. If the radius be same find the ratio of the height of the cylinder and slant height of cone.
A.
Lateral surface area of cylinder = curved surface area of cone.
SS I
SS M
I M \ h : l = 1 :2 3.
Find the volume of right circular cone with radius 6 cm and height 7 cm.
A.
r = 6 cm, h = 7 cm
Volume of right circular cone =
4.
2 u u uu pr h =
= 264 sq.cm.
The radii of two cones are in the ratio 2 : 1 and their volumes are equal. Then the ratio of their heights is ............................................
5.
If d is the diameter of a sphere, then it's volume is ............................................
6.
The horizontal cross section of a cylinder is ............................................
7.
The vertical cross section of a cylinder is ............................................
8.
The vertical cross section of right circular cone is ............................................
9.
The diagonal of the cuboid with dimensions 9 cm × 12 cm × 20 cm is ............................................
10.
The maximum volume of a cone that can be curved out of a solid hemisphere of radius 'r' is ............................................
11.
The ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube is ............................................
12.
The volume of a ball of radius 1 cm is ............................................
13.
If the volume of hemisphere is 18
14.
The surface area of a sphere 4 cm in diameter is ............................................
15.
The slant height of a conical vessel of radius 6 cm and height 8 cm is ............................................
16.
If the total surface area of a hemisphere is 115.5 cm , then it's radius is ............................................
17.
The volume of a cone with radius of the base 7 cm and height 24 cm is ...........................................
18.
The curved surface area of a cone with radius 3 cm and height 4 cm is ............................................
19.
The curved surface area of cylinder with radius 7 cm and height 10 cm is ..........................................
20.
The volume of a cylinder with radius of the base 3.5 cm and height 12 cm is ........................................
p cm3, then the radius of the hemisphere is ......................................
2
X Class MATHEMATICS PAPER II
295
21.
The volume of earth dug from 10 m deep well with a diameter 7 m is ............................................
22.
The volume of metallic sphere having diameter 6 cm is ............................................
23.
If the slant height and height of a cone are 25 cm and 20 cm then radius is ............................................
24.
If the radius and height of a cone are 9 cm and 40 cm respectively, then its slant height is ............................................
25.
A sector of radius 7 cm and central angle 45° is folded into a cone along its radius. Then the slant height of the cone so forms is ........... and its curved surface area is ............................................
ANSWERS I.
1) 1 : 3
2) 1 : 2
3) 264 sq.cm
6) circle
7) rectangle
8) isosceles triangle
10)
SS
14) 16
18)
11) 6 :
p cm2
22) 288
cm
2
p
15) 10 cm
2
= 47.14 cm
p cm3
23) 15 cm
12)
16)
cm
3
4) 1 : 4
5)
9) 25 cm
3
= 4.19 cm
13) 3 cm
3
cm = 3.5 cm
19) 440 cm
2
24) 41 cm
SE
17) 1232 cm
20) 132 cm
25) 7cm,
3
3
21) 1540 m
cm
2
2
= 19.25 cm
h h h h h
X Class MATHEMATICS PAPER II
296
11. TRIGONOMETRY
KEY CONCEPTS `
The word 'Trigonometry' is derived from the Greek roots 'tri' means three ; 'gonia' means 'an angle' and 'metron' means 'measure'. Thus 'Trigonometry' means three angle measure.
`
Hipparchus (140 B.C.), a Greek mathematician established the relationship between the sides and angles of a triangle.
` Trigonometrical ratios : Let us consider D ABC in which
#
= 90°, let
#"$
=
q,
#$"
= 90
q
$ oT # q=
0QQPTJUF TJEF UP T #$ )ZQPUFOVTF "$
ii) cos
q=
"EKBDFOU TJEF UP T "# )ZQPUFOVTF "$
iii) tan
q=
0QQPTJUF TJEF UP T #$ "EKBDFOU TJEF UP T "#
iv) cot
q=
"EKBDFOU TJEF UP T "# 0QQPTJUF TJEF UP T #$
i)
sin
v) cosec
vi) sec
q=
q=
T
"
)ZQPUFOVTF "$ 0QQPTJUF TJEF UP T #$
)ZQPUFOVTF "$ "EKBDFOU TJEF UP T "#
Relation between the trigonometrical ratios : cosec q, sec q and cot q are respectively reciprocals of sin
q, cos q and tan q.
tan
q=
TJO T DPT T
ii) sec
q=
DPT T
i)
iii) cosec
q=
DPU T
TJO T
X Class MATHEMATICS PAPER II
297
` Trigonometric ratios from 0° to 90° q
0c
SD
SD
SD
SD
0°
30°
45°
60°
90°
sin
q
0
1
cos
q
1
0
tan
q
0
cosec
q
¥
sec
q
1
cot
q
¥
1
2
1
2
¥
1
¥
0
` Trigonometric Identities : 2
a) sin
Þ
2
i)
sin
ii)
cos
b) sec
Þ
q + cos2 q = 1
2
DPT T
q = 1 sin2 q (or) cos q =
TJO T .
q tan2 q = 1
UBO T
2
q = sec2 q 1 (or) tan q =
TFD T
ii)
tan
i)
ii)
q cot2 q = 1 2
cosec
cot
£ q < 90°
q = 1 + tan2 q (or) sec q =
sec
2
for 0°
2
i)
c) cosec
Þ
2
q = 1 cos2 q (or) sin q =
2
for 0° < A
£ 90°
q = 1 + cot2 q (or) cosec q =
q = cosec2 q 1 (or) cot q =
DPU T
DPTFD T
` Trigonometric Ratios of Complementary Angles : Two angles are said to be complementary angles if their sum equals to 90° sin (90
q) = cos q
cos (90
q) = sin q
X Class MATHEMATICS PAPER II
298
q) = cot q
tan (90
q) = sec q
cosec (90 sec (90
q) = cosec q
cot (90
q) = tan q
`
The maximum and minimum values of sin
q
`
The maximum and minimum values of tan
q are + ¥, ¥
& cos
q are 1 and 1
1 MARK QUESTIONS 1. Is sin x = Sol.
does exist for some value of angle x ?
The value of sin x always lies between 1, and 1.
Here sin x =
which is greater than 1.
So it does not exist.
2. Is tan A is product of tan and A ? Think and write. Sol.
The symbol tan A is used as an abbreviation for 'the tangent of the angle a'. tan A is not the product of 'Tan' and A. 'tan' seperated from 'A' has no meaning.
3. What can you say about cosec 0° = Sol.
? Is it defined ? Why ? TJO q
Sin 0° = 0
cosec 0 =
TJO q
= not defined.
Reason : Division by 0 is not allowed. Hence
is indeterminate.
4. sec 0° = 1, Why ? Sol.
sec 0° =
DPT q
= 1
5. Is it right to say that sin (A + B) = sin A + sin B ? Justify your answer. Sol.
Let A = 30° and B = 60° LHS sin (A + B) = sin (30 + 60) = sin 90° = 1
RHS : sin 30° + sin 60° =
Hence LHS
¹ RHS.
So it is not right to say that sin(A + B) = sin A + sin B.
X Class MATHEMATICS PAPER II
299
6. Evaluate Sol.
TFD ¡ DPTFD ¡
.
cosec A = sec (90° A)
Þ \
cosec 55° = sec (90 55°) = sec 35°
TFD q
TFD ¡ DPTFD ¡
TFD q
= 1
7. If sin A = cos B then prove that A + B = 90° Sol.
Given that sin A = cos B (1) We know cos B = sin (90 B) we can write (1) as sin A = sin(90 B) If A, B are acute angles, then A = 90 B
Þ
A + B = 90°
8. Express sin 81° + tan 81° interms of trigonometric ratios of angles between 0° and 45° Sol.
We can write sin 81° = cos (90 81°) = cos 9° tan 81° = tan (90 9) = cot 9° Then sin 81° + tan 81° = cos 9° + cot 9°
9. If sec q + tan q = p then what is the value of sec q tan q ? Sol.
Given that sec
q + tan q = p 2
We know that sec
Þ
(sec
Þ
sec
q tan2 q = 1
q + tan q) (sec q tan q) = 1
q tan q =
TFD T UBO T
. Q
10. Evaluate (sec2 q 1) (cosec2 q 1) Sol.
2
(sec
q 1) (cosec q 1) 2
= tan
=
2
ª « « ¬
q × cot q
UBO
2
Tu
UBO
DPTFD
T
' DPU T
' TFD T UBO T ª « ¬
T
º » DPU T »¼
º UBO T »¼
= 1
11. If sin C = Sol.
, then find cos A.
$
From the Adjacent figure In
D ABC, sin C =
#
= 90°
"# 0QQPTJUF TJEF UP $ "$ )ZQPUFOVTF
Now cos A =
"# "EKBDFOU TJEF UP " "$ )ZQPUFOVTF
X Class MATHEMATICS PAPER II
#
" 300
, find sec x.
12. If tan x = Sol.
Given tan x =
UBO Y
Now sec x =
· ¸ © ¹ §
¨
=
=
2 MARKS QUESTIONS 1. If tan A = Sol.
, then find the other trigonometric ratios of angle A.
Given tan A =
0QQPTJUFTJEF "EKBDFOUTJEF
Hence tan A =
Î
{
For angle A, opposite side = BC = 3k (where k is any positive integer) Adjacent side = AB = 4k (where k is any positive integer)
\
In
D ABC by Pythagoras theorem AC
2
2
= AB
Þ
AC
\
sin A =
L
=
cos A =
cot A =
2
2
= (3k)
2
+ (4k)
= 9k
2
+ 16k
2
= 25 k
2
= 5k = Hypotenuse
#$ L "$ L
$# L "$ L
cosec A =
sec A =
+ BC
TJO "
DPT "
UBO "
X Class MATHEMATICS PAPER II
301
2. If ÐA and ÐP are acute angles such that sin A = sin P, then prove that ÐA = ÐP. Sol.
Given sin A = sin P
Þ
#$ 23 "$ 12
(from the adjacent figure)
#$ 23 "$ 12
\ Let
= k
+
*
,
......... (1)
By using Pythagoras theorem AB
2
= AC
"# 13
2
BC
2
; PR
"$ #$ 12 23
=
2
2
= PQ
2
"$ L "$ 12 L 12
=
(From (1))
"$ L
12 L
=
"$ 12
"$ "# #$ 12 13 23
Hence,
then
QR
D ABC ~ D PQR
\ ÐA = ÐP
3. In a right angle triangle ABC, right angle is at B, if tan A =
, then find the value of
sin A cos C + cos A sin C Sol.
Given, tan A =
0QQPTJUFTJEF UP " #$ "EKBDFOUTJEF UP " "#
Þ
BC =
k (where k is any +ve integer)
AB = 1k (where k is any +ve integer) Now in AC
2
D ABC, By Pythagoras theorem
= AB
2
+ BC
2=
(k)
X Class MATHEMATICS PAPER II
2
+ 3k
2
= 4k
2
302
L
AC =
= 2k, Hypotenuse.
0QQPTJUF TJEF UP " #$ )ZQPUFOVTF "$
sin A =
"EKBDFOU TJEF UP " "# )ZQPUFOVTF "$
cos A =
0QQPTJUF TJEF UP $ "# )ZQPUFOVTF "$
sin C =
"EKBDFOU TJEF UP $ #$ )ZQPUFOVTF "$
cos C =
\ sin A cos C + cos A sin C =
u u
=
= 1
4. Evaluate : i) 2 tan2 45° + cos2 30° sin2 60° ii) Sol.
i)
TFD ¡
UBO
¡
TJO ¡ DPT ¡ 2 tan
2 45°
= 2(1)
2
+
2 30°
+ cos
§ ¨ ©
· ¸ ¹
2 60°
sin
§ ¨ ©
· ¸ ¹
= 2
TFD q UBO q ii)
TJO q DPT q
=
§ · §· ¨ ¸ ¨© ¸¹ ©¹
=
5. If sin (A B) = Sol.
sin = 30°
A B = 30°
cos (A + B) =
Þ
= 1
, cos (A + B) = , 0° < A + B £ 90°, A > B, find A and B.
Since sin (A B) =
Þ
(1)
= cos 60°
A + B = 60°
X Class MATHEMATICS PAPER II
(2)
303
From (1) & (2)
A + B = 60° A B = 30° 2A
= 90°
q
ÞA=
q
= 45°
Now A + B = 60°
Þ B = 60° 45° = 15°.
45° + B = 60°
6. Evaluate : sin 60° cos 30° + sin 30° cos 60° What is the value of sin(60° + 30°). What can you conclude ? Sol.
Sin 60° cos 30° + sin 30° cos 60°
u u
=
=
=
= 1
and sin(60° + 30°) = sin 90° = 1 From the above we have sin(60° + 30°) = sin 60° cos 30° + sin 30° cos 60° If A = 60°, B = 30° then sin (A + B) = sin A cos B + cos A sin B.
7. If cos 7A = sin(A 6°) where 7A is an acute angle, find the value of A. Sol.
Given cos 7A = sin(A 6°)
Þ
sin(90° 7A) = sin (A 6°)
Q cos q = sin 90 q)
(
Since (90 7A) & (A 6°) are both acute angles,
\
90° 7A = A 6°
Þ
7A + A = 90 + 6
Þ
8A = 96°
A =
= 12°
8. If A, B and C are interior angles of triangle ABC, then show that TJO Sol.
In a
D ABC,
#$ " DPT .
ÐA + ÐB + ÐC = 180°
On dividing by 2 we get
" #$ = 90° #$ " Þ = 90 Taking sin ratio on both sides
sin
Þ
" ·¸ ¹ DPT "
§# $· ¨ ¸ © ¹
= sin
# $· ¸ © ¹
=
§
sin ¨
§ ¨ ©
X Class MATHEMATICS PAPER II
304
9. Show that tan 48° tan 16° tan 42° tan 74° = 1. Sol.
Q tan (90 q) = cot q]
tan 48° = tan(90 42°) = cot 42°
[
tan 16° = tan (90 74°) = cot 74° LHS = tan 48° tan 16° tan 42° tan 74° = cot 42° . cot 74° . tan 42° . tan 74°
=
UBO q UBO q UBO q UBO q
= 1
10. Show that cot q + tan q = sec q . cosec q. Sol.
q + tan q
LHS = cot
=
DPT T TJO T TJO T DPT T
=
DPT T TJO T TJO T DPT T
=
TJO T DPT T
=
TJO T DPT T
= cosec
Q sin2 q + cos2 q = 1]
[
q. sec q
11. In D PQR with right angle at Q, the value of ÐP is x. PQ = 7 cm and QR = 24 cm. Find sin x and cos x. Sol.
In
PR
2
= PQ
PR =
i)
1
D PQR, ÐQ =90° 2
+ QR
sin x =
ii) cos x =
2
= 7
2
+ 24
2
= 49 + 576 = 625
DN
= 25 cm
23 13 12 13
2
Y
DN
3
4 MARKS QUESTIONS 1. Given cot q =
. then evaluate
TJO T TJO T
. DPT T DPT T
2. Prove that
DPT T = cosec q + cot q. DPT T
3. Prove that
TJO T TJO T = sec q + tan q.
X Class MATHEMATICS PAPER II
305
4. Evaluate (1 + tan q + sec q) (1 + cot q cosec q). 5. Evaluate (sin q + cos q)2 + (sin q cos q)2. 6. Show that (cosec q cot q)2 =
DPT T . DPT T
7. Simplify sec A(1 sin A) (sec A + tan A). 8. Prove that (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A. 9. If cosec q + cot q = k. then prove that cos q =
L . L
10. If sec q + tan q = p then find the value of sin q in terms of p. 11. For which value of acute angle
DPT T DPT T = 4 is true ? For which value of TJO T TJO T
0° £ q £ 90° above equation is not defined ?
OBJECTIVE QUESTIONS I. Fill in the blanks. q is ................................................
1.
The maximum value of sin
2.
If A = 30° then sin 2A = ................................................
3.
If cos 2
4.
tan 135° = ................................................
5.
6.
q = sin 4q, here 2q and 4q are acute angles then the value of q is ................................................
TJO " TJO "
In
= ................................................
$
D ABC, ÐB = 90°, ÐC = q from the figure tan q = ................................................. " 2
7.
3 sin
8.
sin(A B) =
9.
sec A.
10.
45° + 2 cos
TJO q DPT q
2
#
60° = ................................................
, cos(A + B) = , then ÐA = ................................................
TJO "
= ................................................
= ................................................
11.
cos 0° + sin 90° +
12.
sin
13.
The value of tan 1° . tan 2° . tan 3° , ............ tan 89° = ................................................
q=
T
cosec 60° = ................................................
D B , cos q = , then cot q = ................................................ E C
X Class MATHEMATICS PAPER II
306
14.
UBO UBO
15.
UBO q UBO q
16.
= ................................................
%
In the figure the value of cos
2
"
= ................................................
2
is ................................................
&
17.
9 sec
18.
If sin
q = cos q, then the value of q = ................................................
19.
If tan
q=
20.
tan
2
A 9 tan
f
2
I
¡
T
#
A = ................................................
, then the value of cos 2
60 + 2 sin
q = ................................................
45° = ................................................
ANSWERS I.
1) 1
6)
2)
11) 4
16)
7) 2
12)
D B
17) 9
3) 15°
4) 1
5) cos A
8) 45°
9) 1
10) 1
13) 1
14)
18) 45°
19)
15) 0
20) 4
h h h h h
X Class MATHEMATICS PAPER II
307
12. APPLICATIONS OF TRIGONOMETRY
KEY CONCEPTS ` Line of sight : It is the line drawn from the eye of an observer to the object viewed. ` Angle of elevation :
It is the angle formed by the line of sight with the horizontal, when the
object viewed is above the horizontal level. In this case, we have to raise our head to look at the object.
PCTFSWFSTFZF
U HI O J T JP G BU P W F O MF -J GF P F HM "O
)PSJ[POUBMMFWFM
` Angle of depression : It is the angle formed by the line of sight with the horizontal, when the object viewed is below the horizontal level. In this case, we have to lower our head to look at the object.
0CTFSWFSTFZF HO JE MJ V#
)PSJ[POUBMMFWFM "OH MFP GEF QSF TTJP MJ O O F P G T JH I U
0CKFDU
2 MARKS QUESTIONS 1. A ladder of length x metre is leaning against a wall making angle q with the ground. Which trigonometric ratio would you like to consider to find the height of the point on the wall at which the ladder is touching ? 2. Length of the shadow of a 15 metre high pole is metres at 7 Oclock in the morning. Then, what is the angle of elevation of the Sun rays with the ground at the time ? 3. A boat has to cross a river. It crosses the river by making an angle of 60° with the bank of the river due to the stream of the river and travels a distance of 600 m to reach the another side of the river. What is the width of the river ? X Class MATHEMATICS PAPER II
308
4 MARKS QUESTIONS 1. A statue stands on the top of a 2 m tall pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the statue. 2. From the top of a building, the angle of elevation of the top of a cell tower is 60° and the angle of depression to its foot is 45°. If distance of the building from the tower is 7 m, then find the height of the tower. 3. A wire of length 18 m had been tied with electric pole at an angle of elevation 30° with the ground. Because it was covering a long distance, it was cut and tied at an angle of elevation 60° with the ground. How much length of the wire was cut ? 4. Two poles of equal heights are standing opposite to each other on either side of the road, which is 120 feet wide. From a point between them on the road, the angle of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles. 5. The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m, find the height of the tower from the base of the tower and in the same straight line with it are complementary. 6. The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of
metre, find the speed of the jet plane. 7. A straight highway leads to the foot of a tower. Ramaiah standing at the top of the tower observes a car at an angle of depression 30°. The car is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point. 8. Two men on either side of a temple of 30 metre height observe its top at the angles of elevation 30° and 60° respectively. Find the distance between the two men. 9. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground by making 30° angle with the ground. The distance between the foot of the tree and the top of the tree on the ground is 6 m. Find the height of the tree before falling down.
OBJECTIVE QUESTIONS I. Fill in the blanks. 1.
The ratio of the length of a rod and its shadow is 1 :
. Then the angle of elevation of sun is
..............................................
X Class MATHEMATICS PAPER II
309
2.
If the angle of elevation of a tower from a distance of 100 mts from its foot is 60°, then the height of the tower is ..............................................
3.
In
"
D ABC, ÐB = 90°, if AB = 12m, ÐC = 30° then, BC = ..................................
N $
¡
# #
4.
In
D ABC, ÐA = 90°, if AB = m, AC = 100m then q = .........................
$ 5.
T
"
The length of the shadow of a tower is equal to its height. Then the angle of elevation of the sun is ..............................................
ANSWERS I.
1) 30°
2)
N
3)
4) 60°
5) 45°
h h h h h
X Class MATHEMATICS PAPER II
310
13. PROBABILITY
KEY CONCEPTS ` Probability : The theoretical probability of an event E, written as P(E) is defined as
/VNCFS PG PVUDPNFT GBWPVSBCMF UP & /VNCFS PG BMM QPTTJCMF PVUDPNFT PG UIF FYQFSJNFOU
P(E) =
Where we assume that the outcomes of the experiment are equally likely. ` Mutually Exclusive Events :
Two or more events of an experiment, where occurance of an
event prevents all other events are called Mutually Exclusive Events.
` Sample space : The set of all possible outcomes in an experiment is called Sample space. Ex : Sample space for the throw of a dice
S = {1, 2, 3, 4, 5, 6}
`
The probability of a sure event (certain event) is one.
`
The probability of an impossible event is zero.
`
The probability of an event E is a number P(E) such that 0
` Elementary events :
£ P(E) £ 1.
An event having only one outcome is
called an Elementary event. The
sum of the probability of all the elementary events of an experiment is 1.
`
For any event E, P(E) +
1 &
= 1, where
&
stands for not E
E and
&
are called complementary events.
E and
&
are also disjoint sets.
1 MARK QUESTIONS 1. If P(E) = 0.05 what is the probability of 'not E' ? Sol.
Given P(E) = 0.05 Probability of not E' denoted by
\
1 &
= 1 P(E)
&
Q P(E) +
[
1 &
= 1]
= 1 0.05 = 0.95.
2. What is the probability of drawing out a red king from a deck of cards? Sol.
Total no.of deck of cards = 52 Let E be the event of drawing out a red king
\
n(E)
= 2
P(E)
=
/PPG PVUDPNFT GBWPVSBCMF UP & 5PUBM OPPG BMM QPTTJCMF PVUDPNFT
=
X Class MATHEMATICS PAPER II
311
3. Can Sol.
be the probability of an event ? Explain.
can't be the probability of any event.
Since, the probability of any event (E) should lie between 0 and 1 i.e 0
£ P(E) £ 1
4. Define mutually exclusive events with example. Sol.
Two events are mutually exclusive if the occurrence of one event prevents the occurence of another event.
Ex : When a coin is tossed getting a head and getting a tail are mutually exclusive.
5. Sangeeta and Reshma, play a tennis match. It is known that the probability of Sangeetha winning the match is 0.62. What is the probability of Reshma winning the match ? Sol.
Let S and R are the events of winning the match of Sangeetha and Reshma.
Given P(S) = 0.62
The probability of Reshma's winning
P(R) = 1 P(S)
= 1 0.62
= 0.38
2 MARKS QUESTIONS 1. Find the probability that number selected at random from the numbers 3, 4, 5, 6, ...... 25. is prime. Sol.
Prime numbers between 3 and 25 are E = {3, 5, 7, 11, 13, 17, 19, 23} n(E) = 8 n(S) = 23
P(E) =
O & O 4
2. A bag contains 40 balls out of which some are red, some are green and remaining are black. If the probability of drawing a red ball is number of black balls ? Sol.
and that of green ball is then what is the
Let R, G, B are the events of drawing Red balls, Green balls, Black balls respectively.
Given
P(R) =
P(G) =
X Class MATHEMATICS PAPER II
312
P(B) = 1
· ¸ ¹
= 1
§ · ¨ ¸ © ¹
= 1
=
P(B) =
§ ¨ ©
=
O #
O 4
O #
u Þ n(B) =
= 10
(OR)
P(R) =
u
P(G) =
u
No.of Red and Green balls = 22 + 8 = 30 No.of black balls = 40 30 = 10
3. The probability of getting bad egg in a lot of 400 is 0.035, then find the number of bad eggs in the lot ? Sol.
Let E be the event of getting bad egg.
Þ
O &
O 4
P(E)
=
0.035
=
n(E)
= 400 × 0.035
O &
= 14
4. Two dice are thrown at the same time. Find the probability that the sum of two numbers appearing on the top of the dice is more than 9 ? Sol.
Let E be the event of getting the sum of two numbers appearing on the top of the dice more than 9.
\
E = {(6, 6), (6, 5), (6, 4), (5, 6), (5, 5), (4, 6)} n(E) = 6 n(S) = 6 × 6 = 36
P(E) =
O & O 4
X Class MATHEMATICS PAPER II
313
5. What is the probability that a non-leap year selected at random will contain 53 Sundays ? Sol.
Non-leap year contains 365 days. i.e. 52 weaks and one day left. That day may be either Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday
\
n(S) = 7
\
n(E) = 1
\
P(E) =
O & O 4
Let E be the event of getting 53
rd
Sundays out of Seven days.
4 MARKS QUESTIONS 1. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? (ii) white ? (iii) not green? 2. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) a king of red colour (ii) a face card (iii) a red face card (iv) the jack of hearts. 3. A die is thrown twice. What is the probability that (i) 5 will not come up either time ? (ii) 5 will come up at- least once? 4. A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? Suppose the bulb drawn in previous case is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective? 5. Three unbiased coins are tossed together. Find the probability of getting i) all heads ii) two heads iii) one head iv) at least two heads. 6. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 and these are equally likely outcomes. What is the probability that it will point at (i) 8 ? (ii) an odd number ? (iii) a number greater than 2 ? (iv) a number less than 9?
OBJECTIVE QUESTIONS I. Fill in the blanks 1.
The definition of probability was given by ...................................................
2.
The probability of an event that cannot happen is ...................... such an event is called ...................
3.
The sum of the probabilities of all the elementary events of an experiment is ...........................
4.
A coin is tossed 1000 times. Head occurs 625 times. The probability of getting a tail is ...................
X Class MATHEMATICS PAPER II
314
5.
The probability of getting the letter M in the word "MATHEMATICS" ...................................................
6.
Two coins are tossed simultaneously. The probability of getting exactly one head ............................
7.
The probability that a non-leap year should have 52 Mondays is ...................................................
8.
P(E) =
9.
A and B are mutually exclusive events with P(A) =
10.
then
1 &
= ...................................................
P(B), then P(A) = ...............................................
A mathematical Textbook contains 250 pages. A page is selected at random. The probability of selecting the page number is a perfect square is ...................................................
11.
Estimating the probability of an event without conducting an experiment is called .......................
12.
In a foot ball match Ronaldo makes 4 goals from 10 penalty kicks. Then the probability of converting a penalty kick into goal is ...................................................
ANSWERS I.
1) Pierre Simon Laplace in 1795
4)
9)
12)
5)
10)
2) 0, impossible event
6)
7)
3) one
8)
11) Theoretical probability (or) Classical probability
h h h h h
X Class MATHEMATICS PAPER II
315
14. STATISTICS
KEY CONCEPTS `
Father of statistics was Sir Ronald.A.Fisher.
Arithematic mean (or) mean (or) Average : Mean is obtained, when sum of the values of all the observations is divided by number of observations.
Arithematic mean
Y
Y
4VN PG WBMVFT PG BMM PCTFSWBUJPOT /VNCFS PG PCTFSWBUJPOT
6Y
J
O
Mean of grouped data is calculated by : a) The direct method : Y
6G Y 6G J
J
J
b) The assumed mean method : Y
B
6G E 6G J
J
J
c) The step deviation method : Y
§ 6GJ VJ · ¨ ¸uI © 6GJ ¹
B
` Mode : The value among the observations which occurs most frequently is called Mode.
Modal Class : In a group of frequency distribution a class with maximum frequency is called Modal Class.
Mode (z) = M
§ · G G ¨ ¸uI © G G G ¹
l = lower boundary of modal class h = size of the modal class interval f = frequency of modal class
1
f = frequency of class preceding the modal class
0
f = frequency of class succeding the modal class
2
` Median : Mid value of given observations is called Median.
Calculation of Median of ungrouped data : First arrange the observations in ascending order. If number of observations n is odd.
Median =
§ O · ¨ ¸ © ¹
UI
observation.
§ O· ¨ ¸ ©¹
UI
If n is even, median is average of
and
§ O · ¨ ¸ © ¹
UI
observations.
Formula of Median of grouped data :
Median M = M
§ ¨ ¨ ©
O · DG ¸ ¸uI G ¹
X Class MATHEMATICS PAPER II
316
l = lower boundary of median class n = number of observations cf = cumulative frequency of class preceding the median class. f = frequency of median class h = size of median class
` Cumulative frequency curve (or) (ogive curve) :
First prepare cumulative frequency table,
then draw a graph by taking cumulative frequency on Y-axis and upper (or) lower limits of corresponding class intervals on X - axis. Join the points by a free hand smooth curve. Then cumulative frequency curve or ogive curve is obtained.
Ogives are two types : Less than ogive : A graph is drawn by taking upper limits of class interval on X-axis and less than cumulative frequency on Y-axis and join the points by a free hand smooth curve than less than ogive is obtained.
Upper than ogive : A graph is drawn by taking lower limits of class interval along the X-axis and greater than cumulative frequency on Y-axis and join the points by free hand smooth curve, then more than ogive is obtained.
Calculation of median from cumulative frequency curve : First locate
O
value on the Y-axis.
From this point draw a line parallel to the X-axis cutting the curve at a point. From this point draw a perpendicular to X-axis. Foot of this perpendicular determines median of the data.
DZO FV RF S' F WUJ BM V N V $
.FEJBO 6QQFSMJNJUT Calculation of median from less than ogive & more than ogive : Draw both less than ogive and greater than ogive curves on the same axis. The two ogives will intersect each other at a point. From this point, if we draw a perpendicular to the X-axis, the x-coordinate of the point at which it cuts X-axis gives us the median.
ZD FOV R FS 'F WJ UB MV N V $
-JNJUT X Class MATHEMATICS PAPER II
.FEJBO
317
1 MARK QUESTIONS 1. Find Mode of observations 5, 6, 9, 6, 12, 3, 6, 11, 6, 7. Sol.
By arranging given observation in ascending order 3, 5, 6, 6, 6, 6, 7, 9, 11, 12. Most frequent term is 6
\
Mode = 6
2. A doctor observed that the pulse rate of 4 students is 72, 3 students is 78, 2 students is 80. Find the mean of pulse rate of above students. Sol.
Total pulse rate of 9 students
= 4 × 72 + 3 × 78 × 2 × 80 = 228 + 234 + 160 = 682
Mean of pulse rate of students =
= 75.77
3. Sita secured 23, 24, 24, 22, 20 marks in a test. Find mean marks. Sol.
AM =
4VN PG PCTFSWBUJPOT /PPG PCTFSWBUJPOT
=
= 22.6
4. The wickets taken by a bowler in 10 cricket matches are as follows : 2, 6, 4, 5, 0, 2, 1, 3, 2, 3. Find mode of data. Sol.
Let us arrange observations in order = 0, 1, 2, 2, 2, 3, 3, 4, 2 is the number of wickets taken by the bowler in maximum number of matches (3 times). So mode of data is 2.
5. Find the possible measures of central tendency from the following graph. Sol.
Less than ogive and more than ogive are intersect each other. X-coordinate of point of intersection is median.
\
Median = 12.5
6. If another observation is added to the data 0, 1, 2, 2, 2, 3, 3, 4, 5, 6 does the mode change (or) not ? Sol.
Given observations are 0, 1, 2, 2, 2, 3, 3, 4, 5, 6 Mode = 2 If other than '3' is added, the mode remaining same.
X Class MATHEMATICS PAPER II
318
7. Point of intersection of less than ogive and more than ogive curves is (15.5, 20). Find median. Sol.
Point of intersection of less than ogive and greater than ogive curves is (15.5, 20). X-coordinate of point of intersection = Median
\ Median = 15.5
8. The abscissa of point of intersection of less than ogive curve and more than ogive curve of a grouped data gives... Sol.
Median
9. Mean of 9, 11, 13, P, 18, 19 is P. Find the value of P ? Sol.
Mean =
P
=
P
=
4VN PG PCTFSWBUJPOT /PPG PCTFSWBUJPOT
1 1
6P = 70 + P 5P = 70
P
=
= 14.
10. Median of Sol.
Y Y Y Y
Y is 8. Find x ?
By arranging given observations in ascending order.
Y Y Y Y
Y Median
Y
= 8
\x=3×8 x = 24
11. Find A.M of a d, a, a + d ? Sol.
Mean (A.M) =
=
=
4VN PG PCTFSWBUJPOT /PPG PCTFSWBUJPOT B E BB E B
= a.
X Class MATHEMATICS PAPER II
319
12. Find lower limit of model class of the following data ?
$* ' Sol.
High frequency = 15 Modal class = 10 15 Lower limit = 10
20 15
og i
th an
og iv
th an
or e
e
Le ss
M
ve
13. Find the median from the following graph.
10 5
Sol.
5 10 15 20 25 30
Less than ogive more than ogive are intersecting at point P(20, 15) X-coordinate of P = 20.
\
Median = 20
14. a, b, c are mid values of a data and corresponding frequencies are a, b, c respectively. Find AM. Sol.
x = a, b, c
i
f = a, b, c
i
A.M =
=
BuB CuC DuD BCD
B C D B CD
15. Find the mean of first n odd natural numbers. Sol.
Sum of first n odd natural numbers = n
Median of first n odd numbers =
O O
2
= n.
16. Mean of first n odd natural numbers is Sol.
O , Find n.
Mean of first n odd numbers = n
\ O \
O
n = 81.
X Class MATHEMATICS PAPER II
320
17. 35 is removed from the data 30, 34, 35, 36, 37, 38, 39, 40. Find the increasing median. Sol.
By arranging given observations in ascending order.
30, 34, 35,
Median =
38, 39, 40
= 36.5
If 35 is removed then the data is 30, 34, 36,
, 38, 39, 40
\ Median = 37 Increase in median = 37 36.5 = 0.5
18. Median of 24, 25, 26, x + 2, x + 3, 30, 31, 34 is 27.5. Find x ? Sol.
Given observations are 24, 25, 26,
Median =
YY
= 27.5
Y
= 27.5
Y Y
30 31, 34
2x + 5 = 55 2x = 55 5 2x = 50 x = 25
19. Find the median of first 10 prime numbers. Sol.
First 10 prime numbers are 2, 3, 5, 7,
Median =
, 17, 19, 23, 29
= 12
20. Mode of 64, 60, 48, x, 43, 48, 43, 34 is 43. Find the value of x + 3. Sol.
By arranging given observations in ascending order 34, 43, 43, 38, 38, 60, 64, x If Mode = 43, then x must be 43. Since 43 is most frequent observation in this data
\
x = 43 x + 3 = 43 + 3 = 46
2 MARKS QUESTIONS 1. Find mean of the following frequency distribution.
.JE WBMVF PG O DMBTT JOUFSWBM 'SFRVFODZ O Sol.
x = 1, 2, 3, 4, .... n
i
X Class MATHEMATICS PAPER II
321
f = 1, 2, 3, 4, .... n
i
Sxifi = 12 + 22 + 32 + ......... + n2 =
Sfi
O O O
Sn2 =
= 1 + 2 + 3 + ..... + n
=
A.M =
Sn =
O O
O O O
O O
6Y G 6G
J J J
O
2. Mean of 17, 4, 8, 6, 15 is m and median of 8, 14, 10, 5, 7, 20, 19, n is (m 1) find m, n. Sol.
Mean of 17, 4, 8, 6, 15
=
4VN PG PCTFSWBUJPOT /PPG PCTFSWBUJPOT
=
=
= 10
\ m = 10 Median of 8, 14, 10, 5, 7, 5, 20, 19, n = m 1 = 10 1 = 9 By arranging them in ascending order 5, 5, 7, 8,
O , 10, 14, 19, 20
Median = n median = 9
\n
= 9
\ m = 10 ; n = 9
3. Mean of x,
Sol.
Mean of x,
\x+
Y
3
Mean of x ,
is m. Find mean of x3, . Y Y Y
Y =
Y
= m
= 2m (1)
Y
Y is
Y
=
=
X Class MATHEMATICS PAPER II
ª Y «¬
º Y »¼
ª§ · «¨ Y ¸ «¬© Y¹
Y
§ ·º ¨ Y ¸» Y© Y ¹ »¼ 322
4. Sol.
Y G
J
J
=
3 [(2m) 3(2m)]
=
3 [8m 6m]
=
2 2 2m(4m 3) = m (4m 3)
, If mean is 3.55 find K.
Mean =
6G Y 6G J
J
=
u u u , u u u ,
=
, ,
=
, ,
J
, ,
= 3.55
183 + 3K
= 50 × 3.55 + 3.55K
183 + 3K
= 177.5 + 3.55K
183 177.5 = 3.55K 3K 5.5 = 0.55K
\K
=
K = 10
5. Median of 4, 6, a, 9, 10, 19 is 7.5. Find 'a'. Sol.
By arranging given observations in ascending order 4, 6,
B , 10, 19.
n = 6
UI § · ¨© ¸¹
UI
Median is average of
rd
3
\ Median =
B a + 9
,
4
observation.
th
= 7.5
= 15
a
= 15 9
a
= 6.
6. Mode of 6, 3, 5, 6, 7, 5, 8, 7, 6, 2k + 1, 9, 7, 13 is y. Find k. Sol.
By arranging given observations in ascending order 3, 5, 5, 6, 6, 6, 7, 7, 7, 8, 9, 13, (2k + 1)
X Class MATHEMATICS PAPER II
323
As mode is 7, (2k + 1) must be equal to 7. 2k + 1 = 7 2k = 6 k = 3
7. Median of 10, 12, 14, 16, 18 is 14. Find median of 13, 15, 17, 19, 21. Sol.
Given observations are 10, 12,
, 16, 18
\ Median = 14 By adding '3' to each given observations = 10 + 3, 12 + 3, 14 + 3, 16 + 3, 18 + 3 = 13, 15,
, 19, 21
\ Median = 17
Note : Median of a data is m. By adding 'n' to each observation new median = m + n. 8. In 100 numbers, fours are 20, fives are 40, sixes are 30, remaining are tens. Find mean. Sol.
Total numbers = 100 Fours = 20 Fives = 40 Sixes = 30 Ten = 100 (20 + 40 + 30) = 100 90 = 10
Mean =
=
u u u u
= 5.6
9. AM of 10 numbers is 7. AM of 15 numbers is 12. Find their common AM ? Sol.
AM of 10 numbers = 7 Sum of 10 numbers = 10 × 7 = 70 AM of 15 numbers = 12 Sum of 15 numbers = 15 × 12 = 180 Sum of 25 numbers = 70 + 180 = 250
Mean of 25 numbers =
= 10
10. Find median of 20, 29, 38, 33, 42, 38, 43, 25. Sol.
By arranging them in ascending order 20, 25, 29, 33, 38, 38, 42, 43. n = 8 (even)
Median = average of
X Class MATHEMATICS PAPER II
O UI § O ·UI
¨© ¸¹
terms
324
= average of
§· ¨ ¸ ©¹ th
= average of 4
=
UI
, 5
§
¨
©
th
UI
· ¸ ¹
terms
terms
= 35.5
11. Find mode of 2, 6, 4, 5, 0, 2, 1, 3, 2, 3. If another observations is added to data, does the mode changes (or) not ? Sol.
Given observations are 2, 6, 4, 5, 0, 2, 1, 3, 2, 3 By arranging ascending order = 0, 1, 2, 2, 2, 3, 3, 4, 4, 6 '2' is most frequent term
\ Mode = 2 By adding '3' to this data, mode is also equal to 3. So except 3, another observation is added, mode does not change.
12. Mean of 30 students is 42. Marks of two students are zero. Find mean of remaining students. Sol.
No.of students = 30 Mean = 42 Total marks = 30 × 42 = 1260 Marks of two students = 0 Total marks of remaining students = 1260 0 = 1260 Remaining number of students = 30 2 = 28
Mean of remaining students =
= 45.
13. Find median of the following data.
.BSLT /PPG 4UVEFOUT Sol.
.BSLT /PPG 4UVEFOUT $'
Total number of observations = 17 (odd)
Median
\ Median class \ Median X Class MATHEMATICS PAPER II
=
§ O · ¨ ¸ © ¹
=
§ · ¨ ¸ © ¹
UI
term
= 9
th
term
= 20 9 14 = 20
325
14. Find mean of the following data.
Y G
J
J
Sol.
Mean =
6G Y 6G J
J
=
u u u u u
=
=
J
= 7.42
15. Mode is 120.8, Mean 128 calculate median. Sol.
Mode
Median
= 3 median 2 mean
=
=
=
.PEF .FBO u
= 125.6
16. Write formulae for median. Write words for symbols. Sol.
Median m = M
§ ¨ ¨ ©
O · DG ¸ ¸uI G ¹
l = lower boundary of median class
n = number of observations
c = cumulative frequency of class preceding the median class.
f
f = frequency of median class
h = size of median class
17. Find mean of a 2d, a d, a, a + d, a + 2d. Sol.
Mean =
4VN PG PCTFSWBUJPOT /PPG PCTFSWBUJPOT
=
B E B E B B E B E
=
B
= a
X Class MATHEMATICS PAPER II
326
18. For a given data l = 40, f1 = 7, f0 = 3, f2 = 6, n = 15, Find mode. Sol.
Mode (z) = M
§ · G G ¨ ¸uI © G G G ¹
= 40 +
§ · ¨ ¸ u © ¹
= 40 +
× 15
× 15
= 40 + 12 = 52
19. For a given data a = 200, Sfixi = 106, Sfi = 45, n = 20. Calculate mean by stepwise deviation method. Sol.
Y
B
§ 6G Y · ¨ J J ¸uO © 6GJ ¹
= 200 +
§ · u ¸ ¨ © ¹
= 200 47.11 = 152.89
20. Find the value of x, if mode of following distribution is 25.
7BMVF Y Y 'SFRVFODZ G J
J
Sol.
Mode of this data = 25 (given) So, 25 must have high frequency (four times)
\ x = 25.
21. Write Algorithm of calculation median from less than ogive and more than ogive curves. Sol. Step I : Draw less than ogive type and greater than type cumulative curves (ogive curves) on the graph paper.
Step II : Mark the point of intersection of two curves drawn in step I. Let this point is P. Step III : Draw a perpendicular PM, from this point P
to X - axis.
X - coordinate of M gives median.
$'
1
N -JNJUT X Class MATHEMATICS PAPER II
327
22. Difference between mode and median of a data is 24. Find difference between median and mean. Sol.
Mode Median = 24 Mode = 3 Median 2 Mean Mode Median = 2 Median 2 Mean 24 = 2(Median Mean)
Median Mean =
= 12
4 MARKS QUESTIONS 1. Draw "OGIVE CURVE" of the following frequency distribution table.
$MBTTFT 'SFRVFODZ 2. The following distribution gives the daily incomes of 50 workers of a factory.
%BJMZ *ODPNF *O SVQFFT
/PPG XPSLFST Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive. 3. During the medical check up of 35 students of a class their weights were recorded as follows :
8FJHIU JO LH /PPG 4UVEFOUT -FTT UIBO -FTT UIBO -FTT UIBO -FTT UIBO -FTT UIBO -FTT UIBO -FTT UIBO -FTT UIBO Draw a less than type of ogive for the given data. Hence obtain the median weight from the graph verify the result by using the formula. 4. The following tables give production yield per hectare of wheat of 100 farmers of a village.
1SPEVDUJPO :JFME 2VJ)FD
/PPG 'BSNFST Change the distribution to a more than type distribution and draw its ogive. X Class MATHEMATICS PAPER II
328
5. The distribution below gives the weight of 30 students of a class. Find the median weight of the students.
8FJHIU JO LH /PPG 4UVEFOUT 6. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
.POUIMZ DPOTVNQUJPO /PPG DPOTVNFST 7. If the median of 60 observations given below is 28.5 find the values of x and y.
$MBTT *OUFSWBM 'SFRVFODZ Y Z 8. The median of the following data is 525. Find the values of x and y, If the total frequency is 100.
$MBTT *OUFSWBM 'SFRVFODZ Y Z 9. A survey regarding the heights (in cm) of 51 girls of class X of a school was conducted and data was obtained as shown in table. Find their median.
)FJHIU JO DN /PPG (JSMT -FTT UIBO -FTT UIBO -FTT UIBO -FTT UIBO -FTT UIBO -FTT UIBO 10. The following table shows the ages of the patients admitted in a hospital during a year.
"HF JO ZFBST /PPG 1BUJFOUT Find the mode and the mean of the data give below. Compare and interpret the two measures of centeral tendency. X Class MATHEMATICS PAPER II
329
11. The following data gives the information on the observed life times (in hours) 225 electrical components.
-JGF 5JNF JO IPVST
'SFRVFODZ Determine the modal life times of the components. 12. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
3VOT /PPG #BUhTNFO
Find the mode of the data. 13. The marks distribution of 30 students in a mathematics examination are given in the adjacent table. Find the mode of this data. Also compare and interpret the mode and the mean.
$MBTT *OUFSWBM /PPG 4UVEFOUT 14. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
-JUFSBDZ SBUF JO /PPG $JUJFT 15. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency 'F'.
%BJMZ 1PDLFU "MMPXBODF JO 3T
/PPG $IJMESFO G
16.
1FSDFOUBHF PG 'FNBMF 5FBDIFST /PPG 4UBUFT 65 Find mean by a) Direct method b) Assumed mean method c) Step deviation method and verify the results.
X Class MATHEMATICS PAPER II
330
OBJECTIVE QUESTIONS I. Fill in the blanks. 1.
A.M. of first 'n' natural numbers is ..........................................................
A.M. =
6O O O O O
O
2.
Mean of 3, 4, 5, 6, 7 is ..........................................................
3.
The mode of 20, 30, 20, 30, 40, 10, 50 is ..........................................................
4.
The mode of first 'n' natural numbers is ..........................................................
5.
The median of first 'n' natural numbers is ..........................................................
6.
If A.M. = 39, Mode = 34.5, the median is ..........................................................
7.
The mean of 2, 3, 3, 2, 3, 1, 0, ..........................................................
8.
Median of 13, 23, 12, 18, 26, 19, 14 is ..........................................................
9.
Class interval of 10-20, 20-30 is ..........................................................
10.
The median of natural numbers from 1 to 9 is ..........................................................
11.
The observation which occurs more frequently in a data is ..........................................................
12.
Which central tendency is used on all observations ..........................................................
13.
If the AM of 3, 5, 9, x, 11 is 7 then x = ..........................................................
14.
Mode of 21, 16, 21, 18, 14, 21, 18 is ..........................................................
15.
Formula to find A.M in deviation method ..........................................................
16.
Formula of mode for a grouped data (z) = ..........................................................
17.
Find mode of 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6 is ..........................................................
18.
Find mode of 5, 6, 9, 10, 6, 12, 3, 6, 11, 10, 4, 6, 7 is ..........................................................
19.
The mid value of the class 40-50 is ..........................................................
20.
The class interval of 1-8, 9-16, 17-24 is ..........................................................
21.
The median of first 10 prime numbers is ..........................................................
22.
Mean of first odd natural numbers ..........................................................
23.
Which measure of central tendency can be determined graphically .....................................................
24.
Emperical relation between mean, median and mode is ..........................................................
25.
The median of a given frequency distribution can be determined graphically ...............................
26.
Construction of cumulative frequency table is useful in determining the ........................................
27.
For a given data of 60 observations, the less than ogive and the more than ogive interest at (66.5, 30) The median of data is ..........................................................................................
X Class MATHEMATICS PAPER II
331
28.
If each observation of a data is increased by 'a' then mean ..........................................................
29.
If the mean of x , x , x , ................. x
1
2
3
n
Y
is
then the mean of
Y Y Y Y
O B B B B
is
........................... 30.
The word ogive is derived from the French word ..........................................................
31.
The mid value of the class is used in ..........................................................
32.
Formula for median of frequency distribution ..........................................................
33.
While drawing ogives ........................ are taken on X-axis and ....................... are taken on Y-axis.
34.
The x-coordinate of point of intersection of two ogives indicates ..........................................................
ANSWERS I.
1)
§ O · ¨ ¸ © ¹
§ O · ¨ ¸ © ¹
2) 5
3) 20 and 30
4) none
5)
6) 37.5
7) 2
8) 18
9) 10
10) 5
11) mode
12) A.M.
13) 7
14) 21
15) B
6G E 6G J
J
16) M
G G § · ¨ ¸uI G G G ¹ ©
17) No mode
18)16
21) 12
22) n
23) median
24) mode = 3 median 2 mean
25) ogive
26) median
27) 66.5
28) increases by a
29)
J
19) 45
20) 8
31) Arithematic mean
32)
§Y· ¨ ¸ ©B¹
M
§ ¨ ¨ ©
30) ogee
O · DG ¸ uI ¸ G ¹
33) Boundaries, cumulative frequencies
34) median
h h h h h
X Class MATHEMATICS PAPER II
332
