INTRODUCTION TO TRIGONOMETRY 119
th AC _ PR
n AB PQ
Theref AC, MB Lk sa 1
erefore, PR = PQ." y (1)
Now, using Pythagoras theorem,
BC = J AB?- AC'
and QR = PQ' - PR? :
BC VAB?-AC? /k?PQ?-k?PR? -_k./PQ?- PR? k am
So, Ay = Rw o.oo ee 2
* QR ~ /pg?-PR? -,/PQ?- PR? «PQ? - PR?» | 2)
From (1) and (2), we have Ce
AC _AB_BC
PR PQ. QR ~ . Then, by using Theorem 6.4, A ACB ~ A PRQ and therefore, Z B= ZQ. A
Example 3 : Consider A ACB, right-angled at C, in
which AB = 29 units, BC = 21 units and Z ABC =8
(see Fig. 8.10). Determine the values of |
{ ' ° 29
(i) cos? 6 + sin' 6, \_ ' . (ii) cos? 6 - sin? 0. :
Solution : In A ACB, we have c (\N B
21
_- 2 2 2 2
AC = AB -- BC = ¥(29)° - (21) Fig. 8.10
= f(29 - 21)(29 + 21) = ,/(8)(50) = V400 = 20units
S Sine="AB 29°C ABO
Now.


cos? 6 - sin? 0. : Solution : In A ACB, we have c (\N B 21 _- 2 2 2 2 AC = AB -- BC = ¥(29)° - (21) Fig. 8.10 = f(29 - 21)(29 + 21) = ,/(8)(50) = V400 = 20units S Sine="AB 29°C ABO Now. () cos?@ + sina = {20 * (21) _ 20+ 27 _ 400+ 441 _,
ow, (i) cos' + sin'@ = 9 29 992 ga)
Gi) cos? - sin? = | 2 * (20) _ (21+ 20)(21-20) 41
and (ii) cos* 9 - sin' 6 = 29 29 992 841°
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120 MATHEMATICS
Example 4: In a right triangle ABC, right-angled at B, A
if tan A = 1, then verify that
2 sinAcosA= 1. BC . Solution : In A ABC, tan A= AB = 1 (see Fig 8.11)
B Cc
Le., BC = AB
Fig. 8.11
Let AB = BC = k, where k is a positive number. Now, AC = ./AB'+ BC? i
= (k)? + (k)? =k V2. |
ae Beet AB
Therefore, sin A= AC V2. and cosA= AC 2
So 2 sin A cos A= (+5) (=| = 1 which is the required value
V2)\V2)
Example 5 : In A OPQ, right-angled at P, Q
OP = 7 cm and OQ - PQ = 1 cm (see Fig. 8.12). Determine the values of sin Q and cos Q. Solution : In A OPQ, we have ~
< OQ? = OP? + PQ? Le., "ad + PQ) = OP? + PQ? (Why?)
Le., 1 + PQ? + 2PQ = OP? + PQ?
ie, _ 14+2PQ= 7 (Why?)
P 7 oO
i.e., PQ = 24cm and OQ =1+4+PQ=25cm em
Fig. 8.12
5 | ond 24
0, sinQ= 5 an cos Q= 55
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INTRODUCTION TO TRIGONOMETRY 121
EXERCISE 8.1
1. InAABC, right-angled at B, AB = 24cm, BC =7 cm. Determine :
(i) sin A, cos A P
(i) sin C,cosC
2. In Fig. 8.13, find tan P-cotR.
3 12 cm 13 cm
3. Ifsin A= ra calculate cos A and tan A.
4, Given 15 cot A = 8, find sin A and sec A.
13 8 R
5. Given sec 0= DD' calculate all other trigonometric ratios. Fig. 8.13
6. If Z A and Z B are acute angles such that cos A = cos B, then show that Z A = ZB.
7 1+ sin 6)(1 - sin 0 .
7. IfcotO= -> evaluate: (i) (1+ sin 6)(1= sin 6), (ii) cot? @
8 (1 + cos 8)(1 - cos) oo ;
1- tan? A , : |
8. If 3 cot A=4, check whether 7--__73,_ =cos? A - sin?A or not.
1+tan'A -
1
9. In triangle ABC, right-angled at B, if tan A= BR find the value of:
(i) sin A cos C+cos A sin C
(i) cos A cos C- sin A sinC a
fr
10. In APQR, right- angled at Q, PR + QR = 25 cm and PQ =5 cm. Determine the values of
sin P, cos P and tan P." .
11.


value of: (i) sin A cos C+cos A sin C (i) cos A cos C- sin A sinC a fr 10. In APQR, right- angled at Q, PR + QR = 25 cm and PQ =5 cm. Determine the values of sin P, cos P and tan P." . 11. State whether the following are true or false. Justify your answer. (i) The value of tan A is always less than 1.
12 >
(i) sec A= 5 for some value of angle A. (itt) cos A is the abbreviation used for the cosecant of angle A. (tv) cot A is the product of cot and A.
. 4
(v) sin6= 3 for some angle 0.
8.3 Trigonometric Ratios of Some Specific Angles
From geometry, you are already familiar with the construction of angles of 30°, 45°,
60° and 90°. In this section, we will find the values of the trigonometric ratios for these
angles and, of course, for 0°. Reprint 2025-26


122 MATHEMATICS
Trigonometric Ratios of 45° C
In A ABC, right-angled at B, if one angle is 45°, then
the other angle is also 45°, ie., Z A= ZC = 45°
(see Fig. 8.14). So, BC =AB (Why?)
A B
Now, Suppose BC = AB = a. Fig. 8.14
Then by Pythagoras Theorem, AC" = AB? + BC? = a? + a? = 2a',
and, therefore, AC = ayJ2- :
Using the definitions of the trigonometric ratios, we have :
in 45° side opposite to angle 45° BC a 1. ,
in = -. Sa FF Cr
° hypotenuse AC a2 2: -
45° = side adjacent toangle 45° AB a | 1.
cos = hypotenuse . AC aV2 2
side opposite to angle 45° BC a
tan 45° = = - = - = 1
side adjacent to angle 45° AB a
In od I
Also, cosec 45° = -- = v2, sec 45° = -- =y2, cot 45° = --- =].
sin 45 y cos 45 tan 45°
SS . Trigonometric Ratios of 30° and 60° A
Let us now calculate the trigonometric ratios of 30° [|
and 60°. Consider an equilateral triangle ABC. Since
each angle in an equilateral triangle is 60°, therefore, 30°
ZA=ZB=2ZC=60°.


cot 45° = --- =]. sin 45 y cos 45 tan 45° SS . Trigonometric Ratios of 30° and 60° A Let us now calculate the trigonometric ratios of 30° [| and 60°. Consider an equilateral triangle ABC. Since each angle in an equilateral triangle is 60°, therefore, 30° ZA=ZB=2ZC=60°. Draw the perpendicular AD from A to the side BC B A60° Cc
(see Fig. 8.15). _?P
Now A ABD = AACD (Why?) Fig. 8.15
Therefore, - BD= DC
and ZBAD= Z CAD (CPCT)
Now observe that:
AABD is aright triangle, right- angled at D with Z BAD = 30° and Z ABD = 60°
(see Fig. 8.15). Reprint 2025-26


INTRODUCTION TO TRIGONOMETRY 123
As you know, for finding the trigonometric ratios, we need to know the lengths of the
sides of the triangle. So, let us suppose that AB = 2a.
1
Then, BD= 5 BC =a
and AD? = AB? - BD? = (2a)? - (a)' = 3a',
Therefore, AD= av3
Now, we have : .
130° eee Had coe 392 = AD 3 _ 93
sin 30°= -- =-=-, cos = = =
AB 2a 2 AB 2a 2
o_ BPL a bt 4
tan 30° = AD a3 3° | . Al sore =, cee3q22 a
so, cosec = Gn30 sec = cos 30° 3 :
cot 30° = ---= 3. oe -
tan 30° . . Similarly, ;
moor= AD - 23 53 oo. 692 = 4, tan 60° B
sin 60° = -- = -- = -., cos = -, tan = /3,
AB 2a 2°.» 2 3
60° <= see 60° = 2 and cot 60° =
cosec = 3) sec = 2 and cot = 3B
ML . Trigonometric Ratios of 0° and 90°
Let us see what happens to the trigonometric ratios of angle C
A, if itis made smaller and smaller in the right triangle ABC
(see Fig. 8.16), till it becomes zero.


= cosec = 3) sec = 2 and cot = 3B ML . Trigonometric Ratios of 0° and 90° Let us see what happens to the trigonometric ratios of angle C A, if itis made smaller and smaller in the right triangle ABC (see Fig. 8.16), till it becomes zero. As Z A gets smaller and
smaller, the length of the side BC decreases.The point C gets
closer to point B, and finally when Z A becomes very close A ; B
to 0°, AC becomes almost the same as AB (see Fig. 8.17). Fig. 8.16
Cc.
" ot A of 4
° Cc oe . ' oo 7
Y " AC "ot "od wot
4 ° a . iC
A BA B A B A B A B A B
Fig. 8.17
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124 MATHEMATICS
When Z A is very close to 0°, BC gets very close to 0 and so the value of
BC
sinA= AC is very close to 0. Also, when Z A is very close to 0°, AC is nearly the
AB
same as AB and so the value of cos A= AC is very close to 1. This helps us to see how we can define the values of sin A and cos A when
A = 0°. We define : sin 0° = 0 and cos 0° = 1. Using these, we have : °
sin 0° 1 ar .
tan 0° = --- =0, cot 0° = --->: which is not defined. (Why?)
cos 0° tan 0° .
1 ra ae
sec 0° = --- =] and cosec 0° = ----; which is again not defined.(Why?)
cos 0° sin 0° eo
Now, let us see what happens to the trigonometric ratios of Z A, when it is made
larger and larger in A ABC till it becomes 90°. As Z A gets larger and larger, Z C gets
smaller and smaller. Therefore, as in the case above, the length of the side AB goes on
decreasing. The point A gets closer to point B.


when it is made larger and larger in A ABC till it becomes 90°. As Z A gets larger and larger, Z C gets smaller and smaller. Therefore, as in the case above, the length of the side AB goes on decreasing. The point A gets closer to point B. Finally when Z A is very close to 90°,
Z C becomes very close to 0° and the side AC almost coincides with side BC
(see Fig. 8.18). a :
Cc Cle ¢ C c
Li] J : A, ol] od
A B a8 A a B ae
, . Fig. 8.18
When Z C is very close to 0°, Z A is very close to 90°, side AC is nearly the
same as side BC, and so sin A is very close to 1. Also when Z A is very close to 90°,
Z Cis very close to 0°, and the side AB is nearly zero, so cos A is very close to 0. So, we define : sin 90° = 1 and cos 90° = 0. Now, why don't you find the other trigonometric ratios of 90°? We shall now give the values of all the trigonometric ratios of 0°, 30°, 45°, 60°
and 90° in Table 8.1, for ready reference. Reprint 2025-26


INTRODUCTION TO TRIGONOMETRY 125
Table 8.1
za | oe | so | ase | oo |
; 1 1 v3
sinA 0 7 V2 > 1
V3 1 1
cos A 1 > V2 7 0
A 0 4 1 Not defined
tan B V3 ot define
A | Not defined 2 = 1
cosec ot define V2 B
A 1 = 2 Not defined
sec B J2 ot define
A Not defined 1 + 0
cot ot define V3 B
Remark : From the table above you can observe that as Z A increases from 0° to
90°, sin A increases from 0 to 1 and cos A decreases from | to 0.
/
Let us illustrate the use of the values in the table above through some examples. Example 6 : In A ABC, right-angled at B, A
AB = 5 cm and Z ACB = 30° (see Fig. 8.19). Determine the lengths of the sides BC and AC.
. 5
Solution : To find the length of the side BC, we will"
choose the trigonometric ratio involving BC and the -~
given side AB.


at B, A AB = 5 cm and Z ACB = 30° (see Fig. 8.19). Determine the lengths of the sides BC and AC. . 5 Solution : To find the length of the side BC, we will" choose the trigonometric ratio involving BC and the -~ given side AB. Since BC is the side adjacent to angle B c
C and AB is the side opposite to angle C, therefore Fig. 8.19
- AB anc
BC tan
+ 5 30° a
1.¢., BC tan = 3
which gives BC= 5/3 cm
Reprint 2025-26


126 MATHEMATICS
To find the length of the side AC, we consider in 30° AB Why? sin = AC (Why?)
; 1S
1.¢., 2 = AC
i.e., AC = 10cm
Note that alternatively we could have used Pythagoras theorem to determine the third
side in the example above,
ice., AC = JAB? + BC? = ,/5?+ (5V3)? cm = 10cm. Example 7 : In A PQR, right-angled at ~ So
Q (see Fig. 8.20), PQ = 3 cm and PR = 6 cm. * PrP. Determine Z QPR and Z PRQ. - -
Solution : Given PQ=3cmandPR=6cm.. cm 6cm PQ. Therefore, PR = sinR Q R Fig. 8.20 nredel
or oye 62
So, "ZPRQ = 30°
and therefore, Z QPR = 60°. (Why?)
You may note that if one of the sides and any other part (either an acute angle or any
side) of a right triangle is known, the remaining sides and angles of the triangle can be
determined. no Ss 1 1
Example 8 : If sin (A- B) = 7° 008 (A+ B)= 2" 0° <A+B< 90°, A> B, find A
andB. . |
: 1 Solution : Since, sin (A - B) = ~~ therefore, A- B = 30° (Why?) (1)


the triangle can be determined. no Ss 1 1 Example 8 : If sin (A- B) = 7° 008 (A+ B)= 2" 0° <A+B< 90°, A> B, find A andB. . | : 1 Solution : Since, sin (A - B) = ~~ therefore, A- B = 30° (Why?) (1) 1
Also, since cos (A + B) = 2? therefore, A+ B =60° (Why?) (2)
Solving (1) and (2), we get: A= 45° and B = 15°. Reprint 2025-26


INTRODUCTION TO TRIGONOMETRY 127
EXERCISE 8.2
1. Evaluate the following :
(i) sin 60° cos 30° + sin 30° cos 60° (ii) 2 tan? 45° + cos? 30° - sin? 60°
(iti cos 45° (iv) sin 30° + tan 45° - cosec 60°
ii 3 3 wv a
sec 30° + cosec 30 sec 30° + cos 60° + cot 45°
5 cos" 60° + 4 sec" 30° - tan" 45° '
(v) +2 fo} 2 fe}
sin* 30° + cos" 30
2. Choose the correct option and justify your choice : a
@ 2 tan 30° . . ,
i) -,-= ",
1+ tan? 30° . on
(A) sin60° (B) cos 60° . (©) tan 60° _* © sin30°
_ I- tan" 45° _ . °
(i) 1+ tan? 45° . ,
(A) tan 90° ®) 1 © sin45° @) 0
(iii) sin2A=2 sin Ais true whenA=. ,
(A) o@ f° ) (B) 30° - O48 (D) 60°
NL .
2 tan 30° :
() T= tan? 30°
(A) cos 60°, (B) sin 60° (C) tan 60° (D) sin 30°
a , 1
3. Iftan(A+B)= ./3 and tan(A--B)= B®" <A+B<90°; A>B, find A and B.
4. State whether the following are true or false. Justify your answer.
@) sin(A+B)=sinA+sinB. (ii) The value of sin 6 increases as 0 increases.


cos 60°, (B) sin 60° (C) tan 60° (D) sin 30° a , 1 3. Iftan(A+B)= ./3 and tan(A--B)= B®" <A+B<90°; A>B, find A and B. 4. State whether the following are true or false. Justify your answer. @) sin(A+B)=sinA+sinB. (ii) The value of sin 6 increases as 0 increases. (iit) The value of cos 6 increases as 9 increases. (iv) sin @=cos 9 for all values of 6. (v) cot Ais not defined for A= 0°. Reprint 2025-26


128 MATHEMATICS
8.4 Trigonometric Identities A
You may recall that an equation is called an identity
when it is true for all values of the variables involved. Similarly, an equation involving trigonometric ratios
of an angle is called a trigonometric identity, if it is
true for all values of the angle(s) involved. In this section, we will prove one trigonometric
identity, and use it further to prove other useful C , B
trigonometric identities. Fig. 8.21
In A ABC, right-angled at B (see Fig. 8.21), we have:
AB? + BC? = AC? © dd)
Dividing each term of (1) by AC', we get | . AB? BC? AC? a |
- + - = - ;
AC? AC? ACH . ABY (BC) (ACY -
1.¢., Sn] Tian] HR lazn -
AC AC LAC
1.¢., (cos A) + (sin A)? = 1 :
Le., cos? A + sin? A = 1° ° (2)
This is true for all A such that 0° < A < 90°. So, this is a trigonometric identity. Let us now divide (1) by AB. We get
AB' BC? AC"
--_ + -_- = ~--
_ , AB? AB* - AB?
_. {ABY (2c) (Ac)
Or, . ssi tlao] ERE la
AB AB AB
ie., 1+ tan'A = sec?


A < 90°. So, this is a trigonometric identity. Let us now divide (1) by AB. We get AB' BC? AC" --_ + -_- = ~-- _ , AB? AB* - AB? _. {ABY (2c) (Ac) Or, . ssi tlao] ERE la AB AB AB ie., 1+ tan'A = sec? A (3)
Is this equation true for A = 0°? Yes, it is. What about A = 90°? Well, tan A and
sec A are not defined for A = 90°. So, (3) is true for all A such that 0° < A < 90°. Let us see what we get on dividing (1) by BC". We get
AB' BC? AC? Re) Re = pe
BC' BC BC
Reprint 2025-26


INTRODUCTION TO TRIGONOMETRY 129
AB) (BC) (Acy
1.¢., sa] tion] =lan
BC BC BC
i.e., cot? A + 1 = cosec? A (4) Note that cosec A and cot A are not defined for A = 0°. Therefore (4) is true for
all A such that 0° < A< 90°, Using these identities, we can express each trigonometric ratio in terms of other
trigonometric ratios, i.e., if any one of the ratios is known, we can also determine the
values of other trigonometric ratios. Let us see how we can do this using these identities. Suppose we know that ] .
tan A= B Then, cot A= /3. . . -
2° V3
Since, sec? A= 1+ tan?A= 1414, sec A= -=, andcosA= V3.
393 0°. VB 2
Again, sin A= /1-cos?A =,/1- , = = Therefore, cosec A = 2. Example 9 : Express the ratios cos A, tan A and sec A in terms of sin A.
fo N
Solution: Since. cos? A + sin' A= 1, therefore,
cos? A= 1-sin? A, ie, cos A= +1 -sin?A
This gives i cosA= ,/l-sin? A (Why?)
H aan SA sin A and sec A 1 1
ence, taa A = -- = =
cosA J/1-sin'?A cos A Jl-sin?


A. fo N Solution: Since. cos? A + sin' A= 1, therefore, cos? A= 1-sin? A, ie, cos A= +1 -sin?A This gives i cosA= ,/l-sin? A (Why?) H aan SA sin A and sec A 1 1 ence, taa A = -- = = cosA J/1-sin'?A cos A Jl-sin? A
Example 10 : Prove that sec A (1 - sin A)(sec A+ tan A) = 1. Solution :
LHS A (1 -sin A)(sec A A 1 (1 - sin A) _1_,snA
= sec A(1 - sin A)(sec A + tan A) = cos A cos A cosA
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130 MATHEMATICS
(l1--sin A)(1+sin A) 1-- sin? A
~ cos' A cos' A
cos' A | 1 RH
~ cos? A = RHS
tA- A A-1l
Example 11 : Prove that aoe ore
cotA+cosA cosecA+l1
LA A OS - cos A :
Solution : LHS = SA =c0s A _ sim A |
cot A+ cos A +cos A -
sin A .
cos Af 1] [o-1] . A 4
sim sin " _
_ ; _ = cosec - = RHS
+
cos A| --+1 - +1 cose
snA .} _\snA
in @-cos 6 +1 1 ,
Example 12 : Prove that "ee = --_---> using the identity
sin6+cos@-1 sec @-tan@
sec? 6 = | + tan' 6. | - .


" _ _ ; _ = cosec - = RHS + cos A| --+1 - +1 cose snA .} _\snA in @-cos 6 +1 1 , Example 12 : Prove that "ee = --_---> using the identity sin6+cos@-1 sec @-tan@ sec? 6 = | + tan' 6. | - . Solution : Since we will apply the identity involving sec 9 and tan 9, let us first
convert the LHS (of the identity we need to prove) in terms of sec 6 and tan @ by
dividing numerator and denominator by cos 0.
sin@-cos@+1 tan@-1+sec@
LHS = +. =.
sin6@+cos@-1 tan@+1-sec@
_ (tan @+secO)-1 _ {(tan 0+ sec 9) - 1} (tan 6 - sec 0)
~ (tan @-sec@)+1 {(tan 6 - sec 0) + 1} (tan @- sec 6)
: (tan? @ - sec" ) - (tan 6 - sec 0)
= {tan 8 - sec 8 + 1} (tan®@ - sec 8)
_ -1- tan 6+ sec 8
~ (tan 0 - sec 6+ 1) (tan 6 - sec 6)
Reprint 2025-26


INTRODUCTION TO TRIGONOMETRY 131
-1 _ 1
~ tan@-sec@ sec @- tan @
which is the RHS of the identity, we are required to prove. EXERCISE 8.3
1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
2. Write all the other trigonometric ratios of Z Ain terms of sec A. .
3. Choose the correct option. Justify your choice.
@) 9sec?A-9 tan? A= ,
(A) 1 (B) 9 © 8 Oo |
(ii) (1 + tan 6 + sec 9) (1 + cot 8 -cosec 8) = - . . (A) 0 (B) 1 ~ ©2 oe (D) -1
(ii) (secA+tanA)(1-sinA)= a -
(A) sec A (B) sinA ~ ©) cosec A : (D) cos A
-- L+tan? A | ; :
(iv) 1+cot? A
(A) sec? A ®) -1 - ©}cot?A (D) tan? A
4. Prove the following identities, where the angles involved are acute angles for which the
expressions are defined.


(A) sec A (B) sinA ~ ©) cosec A : (D) cos A -- L+tan? A | ; : (iv) 1+cot? A (A) sec? A ®) -1 - ©}cot?A (D) tan? A 4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined. Ne x
@ ( 6 10)? 1- cos 8 (i) cos A , Ltsin A > sec A
i) (cosec 8-cot 6)? = ii) ----_ + ---- =
1 +cos 0 1+sinA cos A
(iii) _tan 6 + cot 8 =1+ sec 6 cosec 9
1-cot®@ 1-tan9
[Hint : Write the expression in terms of sin 9 and cos 8]
. l+sec A _ sin? A . oo. (iv) secA L-cosA {Hint : Simplify LHS and RHS separately]
(v) cos A = sin A+) _ cosec A + cot A, using the identity cosec? A= 1+ cot? A.
cos A + sin A -1
1+sinA in 9-2 sin? (vi) avs =secA+tanA (vii) Sinyaesin o 2 sin' ® = tan 0
V1 -sin A 2 cos" 6 - cos 6
(viii) (sin A + cosec A)'+ (cos A+ sec A)? =7 + tan?A+cot?A
Reprint 2025-26


132 MATHEMATICS
x) (cosec A - sin A)(sec A - cos A) = a
(ix) tanA + cot A
[Hint : Simplify LHS and RHS separately]
2 2
[+tan°A | {l-tanA ;
@) 1+ cot?A 1-cot A} ~ ar A
8.5 Summary
In this chapter, you have studied the following points : .
1. Inaright triangle ABC, right-angled at B,
sin A= side opposite to angle AY cos A = side adjacent to angle A
hypotenuse hypotenuse .
side opposite toangle A
tan A= .
side adjacent to angle A "
2. cosee A = ---: sec A= -/_: tan A= -1-> tan A= SA :
sin A cosA cotA . cos A
3. Ifone of the trigonometric ratios of an acute angle is known, the remaining trigonometric
ratios of the angle can be easily determined. .
4, The values of trigonometric ratios for angles 0°, 30°, 45°, 60° and 90°.
5. The value of sin A or cos A never exceeds 1, whereas the value of sec A (0° < A< 90°) or
cosec A (0° <A < 90°) is always greater than or equal to 1.
in2 2 = - :
6. sin? A+cos?A=1, ~.
sec? A-tan*A=1 for.0°<A<90°, °
cose'? A=1+cot?A for 0°<A<90°
Reprint 2025-26


(a) ees
nade hing
PRA: oe
I i
foes =
1062CHO9
9.1 Heights and Distances . :
In the previous chapter, you have studied about trigonometric ratios. In this chapter,
you will be studying about some ways in which trigonometry is used in the life around
you. oo ' "
Let us consider Fig. 8.1 of prvious chapter, which is redrawn below in Fig. 9.1.
we) = a
X o a hye ' . ' a :
ro . : mY * .
yee ; I yee :
" " ;
ae Anseetdteratin i; ao Anse ofeeration ad
Fig. 9.1
In this figure, the line AC drawn from the eye of the student to the top of the
minar is called the line of sight. The student is looking at the top of the minar. The
angle BAC, so formed by the line of sight with the horizontal, is called the angle of
elevation of the top of the minar from the eye of the student. Thus, the line of sight is the line drawn from the eye of an observer to the point
in the object viewed by the observer. The angle of elevation of the point viewed is
Reprint 2025-26


134 MATHEMATICS
the angle formed by the line of sight with the horizontal when the point being viewed is
above the horizontal level, i.e., the case when we raise our head to look at the object
(see Fig. 9.2),
Object
g
ay
be got
© ot
yw CPi
peel. ae
Foro hoses ces ececc + Horizontal level "
o)
re :
Fig. 9.2 . Now, consider the situation given in Fig. 8.2. The girl sitting on the balcony is
looking down at a flower pot placed on a stair of the temple. In this case, the line of
sight is below the horizontal level. The angle so formed by the line of sight with the
horizontal is called the angle of depression. , ° - Thus, the angle of depression of a point on the object being viewed is the angle
formed by the line of sight with the horizontal when the point is below the horizontal
level, i.c., the case when we lower our head to look at the point being viewed
(see Fig. 9,3), ae - : ' .
ee
d Angle of
A das depress;
a a rh Pression
| , (eo) Object
Fig. 9.3 Now, you may identify the lines of sight, and the angles so formed in Fig. 8.3.


our head to look at the point being viewed (see Fig. 9,3), ae - : ' . ee d Angle of A das depress; a a rh Pression | , (eo) Object Fig. 9.3 Now, you may identify the lines of sight, and the angles so formed in Fig. 8.3. Are they angles of elevation or angles of depression? Let us refer to Fig. 9.1 again. If you want to find the height CD of the minar
without actually measuring it, what information do you need? You would need to know
the following: (i) the distance DE at which the student is standing from the foot of the minar
Reprint 2025-26


SOME APPLICATIONS OF TRIGONOMETRY 135
(ii) the angle of elevation, Z BAC, of the top of the minar
(iii) the height AE of the student. Assuming that the above three conditions are known, how can we determine the
height of the minar? In the figure, CD = CB + BD. Here, BD = AE, which is the height of the student. To find BC, we will use trigonometric ratios of Z BAC or Z A. In A ABC, the side BC is the opposite side in relation to the known Z A. Now,
which of the trigonometric ratios can we use? Which one of them has the two values
that we have and the one we need to determine? Our search narrows down to using
either tan A or cot A, as these ratios involve AB and BC. Theref as?e tase hich Ivi ld gi BC erefore, tanA= Ap ° COtA= GG' Which on solving woul give us BC. By adding AE to BC, you will get the height of the minar. - we Now let us explain the process, we have just discussed, by solving some problems. Example 1 : A tower stands vertically on the ground.


Ap ° COtA= GG' Which on solving woul give us BC. By adding AE to BC, you will get the height of the minar. - we Now let us explain the process, we have just discussed, by solving some problems. Example 1 : A tower stands vertically on the ground. From a point on the ground,
which is 15 m away from the foot of the tower, the angle of elevation of the top of the
tower is found to be 60°. Find the height of the tower. Solution : First let us draw a simple diagram to
represent the problem (see Fig. 9.4). Here AB d
represents the tower, CB is the distance of the point fi
from the tower and Z ACB is the angle of elevation. es
We need to determine the height of the tower, i.c., xy
AB. Also, ACB is a triangle, right- angled at B. AN
To solve the problem, we choose the trigonometric A
ratio tan 60° (or cot 60°), as the ratio involves AB Wi
mo ve Lid
and BC. ' \
N tan 60° A
ow, an = Be WYN
" cO =
AB <- 15m->B
ie. 3 =
15 Fig. 9.4
ie. AB = 15V3
Hence, the height of the tower is 15/3 m. Reprint 2025-26


136 MATHEMATICS
Example 2 : An electrician has to repair an A
electric fault on a pole of height m. She needs | === {== Fto reach a point 1.3m below the top of the pole -*
to undertake the repair work (see Fig. 9.5). What B™
should be the length of the ladder that she should ' S
use which, when inclined at an angle of 60° to
the horizontal, would enable her to reach the
required position? Also, how far from the foot e
of the pole should she place the foot of the a:
ladder? (You may take ./3 = 1.73) et . Solution : In Fig. 9.5, the electrician is required to ne °
reach the point B on the pole AD. t | _ 60
. D \ Cc
So, BD =AD - AB = (5 - 1.3)m =3.7 m. . Fig. 9.5
Here, BC represents the ladder. We need to find. FC ,
its length, i.e., the hypotenuse of the right triangle BDC. _
Now, can you think which trigonometic ratio should we consider? It should be sin 60°. | oe
BD 3.7 - V3 So, BC" sin 60° o Bc 4 cy . Theref \ BC STK? _ 4.08 ( ) erefore, = ->=- =4.28 m (approx.


of the right triangle BDC. _ Now, can you think which trigonometic ratio should we consider? It should be sin 60°. | oe BD 3.7 - V3 So, BC" sin 60° o Bc 4 cy . Theref \ BC STK? _ 4.08 ( ) erefore, = ->=- =4.28 m (approx. OB PP
1.¢e., the length of the ladder should be 4.28 m. N " DC t 60° a
ow, _. BDS co "Bh
1 | DC 37 2.14
Le, | | = BR =? m (approx.)
Therefore, she should place the foot of the ladder at a distance of 2.14 m from
the pole. Reprint 2025-26


SOME APPLICATIONS OF TRIGONOMETRY 137
Example 3: An observer 1.5 m tall is 28.5 m away A
from a chimney. The angle of elevation of the top tn
of the chimney from her eyes is 45°. What is the uIL
height of the chimney? TIT
|
Solution : Here, AB is the chimney, CD the veer
observer and Z ADE the angle of elevation (see oll
Fig. 9.6). In this case, ADE is a triangle, right-angled 45°. tne, | Hea E
at E and we are required to find the height of the D rfl
chimney. f a
Cc B We have AB= AE+BE=AE+1.5 Fig. 9.6 7
and DE= CB =28.5m " To determine AE, we choose a trigonometric ratio, which involves both AE and
DE. Let us choose the tangent of the angle of elevation, ~
N tan 45° = A oo . . ow, an = DE- 5 |
i. BE |
nee ~ 28.5 . os
Therefore, a AE = 28.5
4 "
So the height of the chimney (AB) = (28.5 + 1.5) m= 30 m. Example 4 : From a point P on the ground the angle of elevation of the top of a 10 m
tall building is 30°.


~ 28.5 . os Therefore, a AE = 28.5 4 " So the height of the chimney (AB) = (28.5 + 1.5) m= 30 m. Example 4 : From a point P on the ground the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation
of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the
distance of the building from the point P. (You may take ./3 = 1.732)
Solution : In Fig. 9.7, AB denotes the height of the building, BD the flagstaff and P
the given point. Note that there are two right triangles PAB and PAD. We are required
to find the length of the flagstaff, i.e, DB and the distance of the building from the
point P, i.e., PA. Reprint 2025-26


138 MATHEMATICS
Since, we know the height of the building AB, we Dg
will first consider the right A PAB. -
AB =: We have tan 30°= -- eo
. 1 10 Zk RS
a Baap pL poe ae A
Therefore, AP = 10V3 Fig. 9.7 :
i.e., the distance of the building from P is 10V3 m= 17.32 m. . Next, let us suppose DB = x m. Then AD = (10 + x) m. . - _ _ AD _10tx
Now, in right A PAD, tan 45° = AP | 10/3 -
there _ lO+x oe
erefore, =e.
10V3
ie., ~x= 10 (v3~-1) =7.32
co
So, the length of the flagstaff is 7.32 m. . A
Example 5 : The shadow of a tower standing
on a level ground is found to be 40 m longer
when the Sun's altitude is 30° than when it is
60°. Find the height of the tower.


=7.32 co So, the length of the flagstaff is 7.32 m. . A Example 5 : The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun's altitude is 30° than when it is 60°. Find the height of the tower. Solution : In Fig. 9.8, AB is the tower and
BC is the length of the shadow when the p ~\ /\
Sun's altitude is 60°, i.e., the angle of | <--- 40m --+C B
elevation of the top of the tower from the tip Fis. 9.8
of the shadow is 60° and DB is the length of 1B 7
the shadow, when the angle of elevation
is 30°. Now, let AB be A m and BC be x m. According to the question, DB is 40 m longer
than BC. Reprint 2025-26


SOME APPLICATIONS OF TRIGONOMETRY 139
So, DB = (40+x)m
Now, we have two right triangles ABC and ABD. In AABC tan 60° A n , an = Be h
or, V3 = (1)
x In AABD 30° AB n , tan = Bp ' . 1 h 5
Le., V3 > x+40 ©
From (1), we have h= xJ3 . . Putting this value in (2), we get (xV/3)V3 =x+40,ie., 3x=x440
iLe., x= 20 oO, / : So, h= 20V3_ - OO [From (1)]
Therefore, the height of the tower is 20/3 m. Example 6 : The angles of depression of the top and the bottom of an 8 m tall building
from the top of a multi-storeyed building are 30° and 45°, respectively. Find the height of the multi- storeyed building and the distance between the Q------------------y--g--,P
two buildings. 30°§5
Solution : In Fig. 9.9, PC denotes the multi- storyed building and AB denotes the 8 m tall
building. We are interested to determine the height
of the multi-storeyed building, i.c., PC and the B D
distance between the two buildings, i.e., AC. Look at the figure carefully. Observe that PB is a transversal to the parallel lines PQ and BD. A c
Therefore, ZQPB and ZPBD are alternate angles, and SO are equal. Fig. 9.9


of the multi-storeyed building, i.c., PC and the B D distance between the two buildings, i.e., AC. Look at the figure carefully. Observe that PB is a transversal to the parallel lines PQ and BD. A c Therefore, ZQPB and ZPBD are alternate angles, and SO are equal. Fig. 9.9 So Z PBD = 30°. Similarly, Z PAC = 45°,
In right A PBD, we have Reprint 2025-26


140 MATHEMATICS PD 1 -- = c= BD = PD BD = (an 30 B or V3
In right A PAC, we have PC 45° 1 AC > tan =
L.e., PC = AC
Also, PC = PD + DC, therefore, PD + DC = AC. :
Since, AC = BD and DC = AB = 8 m, we get PD + 8 = BD= PDV3 (Why?) -
This gives PD = --- = ---->_ = 4 (v3 + 1)m. V3-1 (V3 +1)(V3-1) =, :
So, the height of the multi-storeyed building is {4(v3 + 1) + stm = 4(3 + V3)m
and the distance between the two buildings is also 4(3 + V3) m. Example 7 : From a point on a bridge across a river, the angles of depression of
the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge
is at a height of 3 m from the banks, find the width of the river.
oe N
Solution ; In Fig 9.10, A and B ° P
represent points on the bank on . St TAEE PANY
opposite sides of the river, so that RE am NG -
AB is the width of the river.


m from the banks, find the width of the river. oe N Solution ; In Fig 9.10, A and B ° P represent points on the bank on . St TAEE PANY opposite sides of the river, so that RE am NG - AB is the width of the river. Pisa (QuE2F SREY J E SES
point on the bridge at a height of 3 A DB
m, i.e., DP = 3 m.* We are ;
interested to determine the width Fig. 9.10
of the river, which is the length of
the side AB of the D APB. Now, _AB = AD + DB
In right AAPD, Z A= 30°. S tan 30° "
0, an = AD
Reprint 2025-26


SOME APPLICATIONS OF TRIGONOMETRY 141
1 3
Le., Fe = zn or AD= 3V3 m
V3 ~ AD N3
Also, in right A PBD, Z B = 45°. So, BD = PD =3 m. Now, AB = BD+AD=3 + 3V¥3 =3(1+ V3) m. Therefore, the width of the river is 3 (3 + 1)m,
EXERCISE 9.1 | 1. Acircus artist is climbing a 20 m long rope, which is A
tightly stretched and tied from the top of a vertical -
pole to the ground. Find the height of the pole, if 20m
the angle made by the rope with the ground level is
30° (see Fig. 9.11). , 2. A tree breaks due to storm and the broken part BO (™. Cc
bends so that the top of the tree touches the ground -_.
making an angle 30° with it. The distance between . Fig. 9.11
the foot of the tree to the point where the top
touches the ground is 8 m. Find the height of
the tree. / oy 3. A contractor plans to install two slides for the children to play in a park.


30° with it. The distance between . Fig. 9.11 the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. / oy 3. A contractor plans to install two slides for the children to play in a park. For the children
below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and
is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have
a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What
should be the length of the slide in each case? 4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m
away from the foot of the tower, is 30°. Find the height of the tower. 5. A kite is flying ata height of 60 m above the ground. The string attached to the kite is
temporarily tied to a point on the ground. The inclination of the string with the ground
is 60°. Find the length of the string, assuming that there is no slack in the string. 6.


flying ata height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string. 6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of
elevation from his eyes to the top of the building increases from 30° to 60° as he walks
towards the building. Find the distance he walked towards the building. 7. From a point on the ground, the angles of elevation of the bottom and the top of a
transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower. Reprint 2025-26


142 MATHEMATICS 8. A statue, 1.6 m tall, stands on the top of a 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 pedestal. 9. The angle of elevation of the top of a building from the foot of the tower is 30° and the
angle of elevation of the top of the tower from the foot of the building is 60°. If the tower
is 50 m high, find the height of the building. 10. Two poles of equal heights are standing opposite each other on either side of the road,
which is 80 m wide. From a point between them on the road, the angles 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. OA 11. A TV tower stands vertically on a bank Gs : /foa
of a canal. From a point on the other peony
bank directly opposite the tower, the
angle of elevation of the top of the
tower is 60°.


poles and the distances of the point from the poles. OA 11. A TV tower stands vertically on a bank Gs : /foa of a canal. From a point on the other peony bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m ,
away from this point on the line joing .
this point to the foot of the tower, the A / Pa
angle of elevation of the top of the' _ D 20m C u
tower is 30° (see Fig. 9.12), Find the . B
height of the tower and the width of : Fig. 9.12
the canal. . 12. From the top of a7 m high building, the angle of elevation of the top of a cable tower is
60° and the angle of depression of its foot is 45°. Determine the height of the tower. 13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of
depression of two ships are 30° and 45°. If one ship is exactly behind the other on the
same side of the lighthouse, find the distance between the two ships. 14.


tower. 13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. 14. A1.2 mtall girl spots a balloon moving wo @)
with the wind in a horizontal line at a ® wa
height of 88.2 m from the ground. The C a
angle of elevation of the balloon from oot a
the eyes of the girl at any instant is oS be 88.2 m
60°. After some time, the angle of < 600."
elevation reduces to 30° (see Fig. 9.13). i ANB bead : Y
Find the distance travelled by the
balloon during the interval. Fig. 9.13 15. Astraight highway leads to the foot of a tower. A man standing at the top of the tower
observes a car at an angle of depression of 30°, which is approaching the foot of the Reprint 2025-26


SOME APPLICATIONS OF TRIGONOMETRY 143
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. 9.2 Summary In this chapter, you have studied the following points : 1. @ The line of sight is the line drawn from the eye of an observer to the point in the
object viewed by the observer. (i) The angle of elevation of an object viewed, is the angle formed by the line of sight
with the horizontal when it is above the horizontal level, i.e., the case when we raise
our head to look at the object. (iii) The angle of depression of an object viewed, is the angle formed by the line of sight
with the horizontal when it is below the horizontal level, i.e., the case when we lower
our head to look at the object. °
2. The height or length of an object or the distance between two distant objects can be
determined with the help of trigonometric ratios. oe
CJ
Reprint 2025-26


[a] an [a]
aa
rhetaene
[en] sor
1062CH10
10.1 Introduction -
You have studied in Class IX that a circle is a collection of all points ina plane
which are at a constant distance (radius) from a fixed point (centre). You have
also studied various terms related to a circle like chord, segment, sector, arc etc. Let us now examine the different situations that can arise when a circle and a line
are given in a plane. , :
So, let us consider a circle and a line PQ. There can be three possibilities given
in Fig. 10.1 below:
P. . P S) - P
AS
. B
Q : Q Q
(i) (ii) (ili)
: Fig. 10.1
In Fig. 10.1 (4), the line PQ and the circle have no common point. In this case,
PQ is called a non-intersecting line with respect to the circle. In Fig. 10.1 (i), there
are two common points A and B that the line PQ and the circle have. In this case, we
call the line PQ a secant of the circle. In Fig. 10.1 (iii), there is only one point A which
is common to the line PQ and the circle. In this case, the line is called a tangent to the
circle. Reprint 2025-26


CIRCLES 145 You might have seen a pulley fitted over a well whichis used Zax
in taking out water from the well. Look at Fig. 10.2. Here the rope @
on both sides of the pulley, if considered as a ray, is like a tangent |
to the circle representing the pulley. (6) Is there any position of the line with respect to the circle
other than the types given above? You can see that there cannot
be any other type of position of the line with respect to the circle. Fie. 10.2
In this chapter, we will study about the existence of the tangents 18. iN
to a circle and also study some of their properties. 10.2 Tangent to a Circle . In the previous section, you have seen that a tangent* to a circle is a line that
intersects the circle at only one point. - To understand the existence of the tangent to a circle at a point, let us perform
the following activities: . ou
Activity 1 : Take a circular wire and attach a straight wire AB at a point P of the
circular wire so that it can rotate about the point P in a plane.


of the tangent to a circle at a point, let us perform the following activities: . ou Activity 1 : Take a circular wire and attach a straight wire AB at a point P of the circular wire so that it can rotate about the point P in a plane. Put the system on a table
and gently rotate the wire AB about the point P to get different positions of the straight
wire [see Fig. 10.3(i)]. . , In various positions, the wire intersects the AN A
circular wire at P and at another point Q, or Q, or 6
Q,, etc. In one position, you will see that it will KL '
intersect the circle at the point P only (see position An | G
A'B' of AB). This shows that a tangent exists at . ys
the point P of the circle. On rotating further, you X -
can observe that in all other positions of AB, it will R, Yrs
intersect the circle at P and at another point, say R, B BY RS
or R, or R,, etc. So, you can observe that there is
only one tangent at a point of the circle. Fig. 10.3 (i)


observe that in all other positions of AB, it will R, Yrs intersect the circle at P and at another point, say R, B BY RS or R, or R,, etc. So, you can observe that there is only one tangent at a point of the circle. Fig. 10.3 (i) While doing activity above, you must have observed that as the position AB
moves towards the position A' B', the common point, say Q,, of the line AB and the
circle gradually comes nearer and nearer to the common point P. Ultimately, it coincides
with the point P in the position A'B' of A'"B"'. Again note, what happens if 'AB' is
rotated rightwards about P? The common point R, gradually comes nearer and nearer
to P and ultimately coincides with P. So, what we see is: The tangent to a circle is a special case of the secant, when the two end
points of its corresponding chord coincide. *The word 'tangent' comes from the Latin word 'tangere', which means to touch and was
introduced by the Danish mathematician Thomas Fineke in 1583. Reprint 2025-26


146 MATHEMATICS
Activity 2 : On a paper, draw a circle and a p" secant PQ of the circle. Draw various lines parallel to the secant on both sides of it. You P will find that after some steps, the length of p' the chord cut by the lines will gradually decrease, i.e., the two points of intersection of the line and the circle are coming closer and closer [see Fig. 10.3(41)]. In one case, it becomes zero on one side of the secant and in \ \Q""
another case, it becomes zero on the other side Q of the secant. See the positions P'Q' and P'"'Q" \
of the secant in Fig. 10.3 (ii). These are the Q
tangents to the circle parallel to the given secant ° PQ. This also helps you to see that therecannot = * Fig. 10.3 (i) be more than two tangents parallel to a given - ou secant, - _ This activity also establishes, what you must have observed, while doing
Activity 1, namely, a tangent is the secant when both of the end points of the
corresponding chord coincide.


therecannot = * Fig. 10.3 (i) be more than two tangents parallel to a given - ou secant, - _ This activity also establishes, what you must have observed, while doing Activity 1, namely, a tangent is the secant when both of the end points of the corresponding chord coincide. The common point of the tangent and the circle is called the point of contact
[the point A in Fig. 10.1 (iii)]and the tangent is said to touch the circle at the
common point. so : , Now look around you. Have you seen a bicycle
or acart moving? Look at its wheels. All the spokes
of a wheel are along its radii. Now note the position
of the wheel with respect to its movement on the
ground. Do you see any tangent anywhere? (See Fig. 10.4). In fact, the wheel moves along a line
which is a tangent to the circle representing the wheel. Also, notice that in all positions, the radius through
the point of contact with the ground appears to be at
right angles to the tangent (see Fig. 10.4). We shall Fig. 10.4
now prove this property of the tangent. Theorem 10.1 : The tangent at any point of a circle is perpendicular to the
radius through the point of contact.


point of contact with the ground appears to be at right angles to the tangent (see Fig. 10.4). We shall Fig. 10.4 now prove this property of the tangent. Theorem 10.1 : The tangent at any point of a circle is perpendicular to the radius through the point of contact. Proof : We are given a circle with centre O and a tangent XY to the circle at a
point P. We need to prove that OP is perpendicular to XY. Reprint 2025-26


CIRCLES 147
Take a point Q on XY other than P and join OQ (see Fig. 10.5). The point Q must lie outside the circle. (Why? Note that if Q lies inside the circle, XY
will become a secant and not a tangent to the
circle). Therefore, OQ is longer than the radius
OP of the circle. That is,
OQ > OP. Since this happens for every point on the
line XY except the point P, OP is the y ~~. xX
shortest of all the distances of the point O to the P Q
points of XY. So OP is perpendicular to XY. ; ,
(as shown in Theorem A1.7.) Fig. 10.5 . Remarks . ",
1. By theorem above, we can also conclude that at any point on a circle there can be
one and only one tangent. tO °
2. The line containing the radius through the point of contact is also sometimes called
the 'normal' to the circle at the point. EXERCISE 10.1
1. How many tangents can acirclehave?
2. Fillinthe blanks: _./
(i) A tangent to a circle intersects it in _____ point (s). (i) A line intersecting a circle in two points is calleda_-____--
(ii) A circle can have _________ parallel tangents at the most.


circle at the point. EXERCISE 10.1 1. How many tangents can acirclehave? 2. Fillinthe blanks: _./ (i) A tangent to a circle intersects it in _____ point (s). (i) A line intersecting a circle in two points is calleda_-____-- (ii) A circle can have _________ parallel tangents at the most. (iv) The common point of a tangent to a circle and the circle is called _________.
3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at
a point Q so that OQ = 12 cm. Length PQ is:
(A) 12cm (B) 13cm (C) 85cm (D) J119 cm.
4, Draw a circle and two lines parallel to a given line such that one is a tangent and the
other, a secant to the circle.
10.3 Number of Tangents from a Point on a Circle
To get an idea of the number of tangents from a point on a circle, let us perform the
following activity:
Reprint 2025-26


148 MATHEMATICS
Activity 3 : Draw a circle on a paper. Take a point P inside it. Can you draw a tangent to the A" circle through this point? You will find that all the lines through this point intersect the circle in SX) two points. So, it is not possible to draw any tangent to a circle through a point inside it [see Fig. 10.6 (i)]. Next take a point P on the circle and draw
tangents through this point. You have already (i) °
observed that there is only one tangent to the
circle at such a point [see Fig. 10.6 (ii)]. Finally, take a point P outside the circle and
try to draw tangents to the circle from this point. ~
What do you observe? You will find that you P
can draw exactly two tangents to the circle "
through this point [see Fig. 10.6 (iii)]. tO We can summarise these facts as follows: - Case 1: There is no tangent to a circle passing . through a point lying inside the circle. (ul) Case 2 : There is one and only one tangent toa circle passing through a point lying on the circle. T,
% / Case 3 : There are exactly two tangents to a circle through a point lying outside the circle.


. through a point lying inside the circle. (ul) Case 2 : There is one and only one tangent toa circle passing through a point lying on the circle. T, % / Case 3 : There are exactly two tangents to a circle through a point lying outside the circle. In Fig. 10.6 (iii), T,and T, are the points of P
contact of the tangents PT, and PT,
respectively. Hs 2 T The length of the segment of the tangent ,
from the external point P and the point of contact (iii)
with the circle is called the length of the tangent Fig. 10.6
from the point P to the circle. Note that in Fig. 10.6 (iii), PT, and PT, are the lengths of the tangents from P to
the circle. The lengths PT, and PT, have a common property. Can you find this? Measure PT, and PT,. Are these equal? In fact, this is always so. Let us give a proof
of this fact in the following theorem. Reprint 2025-26


CIRCLES 149 Theorem 10.2 : The lengths of tangents drawn from an external point to a circle are equal. Q Proof : We are given a circle with centre O, a point P lying outside the circle and two tangents PQ, PR on the circle from P (see Fig. 10.7). We P are required to prove that PQ = PR. For this, we join OP, OQ and OR. Then Z OQP and Z ORP are right angles, because R these are angles between the radii and tangents, ' and according to Theorem 10.1 they are right Fig. 10.7 angles. Now in right triangles OQP and ORP, ,
OQ = OR (Radii of the same circle)
OP=OP _. . (Common) Therefore, A OQP = A ORP -- . (RHS) This gives PQ = PR _ . 7 (CPCT) Remarks : oe 1. The theorem can also be proved by using the Pythagoras Theorem as follows: PQ' = OP? - OQ' = OP? - OR' = PR? (As OQ = OR) which gives PQ = PR. 7 " 2. Note also that Z OPQ = Z OPR. Therefore, OP is the angle bisector of Z QPR, i.e., the centre lies on the bisector of the angle between the two tangents. Let us take some examples.


= PR? (As OQ = OR) which gives PQ = PR. 7 " 2. Note also that Z OPQ = Z OPR. Therefore, OP is the angle bisector of Z QPR, i.e., the centre lies on the bisector of the angle between the two tangents. Let us take some examples. Example 1 : Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact. c Solution : We are given two concentric circles C, and C, with centre O and a chord AB of the A larger circle C, which touches the smaller circle C, at the point P (see Fig. 10.8). We need to prove that AP = BP. B Let us join OP. Then, AB is a tangent to C, at P Fig. 10.8 and OP is its radius. Therefore, by Theorem 10.1, OP L AB Reprint 2025-26


150 MATHEMATICS
Now AB is a chord of the circle C, and OP L AB. Therefore, OP is the bisector of the
chord AB, as the perpendicular from the centre bisects the chord,
1.¢., AP = BP
Example 2 : Two tangents TP and TQ are drawn
to a circle with centre O from an external point T. p
Prove that Z PTQ = 2 Z OPQ. P
Solution : We are given a circle with centre O, TS
an external point T and two tangents TP and TQ
to the circle, where P, Q are the points of contact
(see Fig. 10.9). We need to prove that Oo
Z PTQ= 2 Z OPQ Fig. 10.9 -
Let ZPTQ=9 : OO
Now, by Theorem 10.2, TP = TQ. So, TPQ is an isosceles triangle."
Loe. , od
Therefore, Z TPQ = Z TQP= 5 (180° - 8) = 90° - 5°
Also, by Theorem 10.1, ZOPT=90° |
. . 1
So, Z OPQ = Z OPT - Z TPQ = 50° [90° 50)
f° ) :
NL ~] 1
. = -§9=- ZPT
2 2 Q
This gives Z PTQ= 2 Z OPQ
Example 3 : PQ is a chord of length 8 cm of a P
circle of radius 5 cm.


OPQ = Z OPT - Z TPQ = 50° [90° 50) f° ) : NL ~] 1 . = -§9=- ZPT 2 2 Q This gives Z PTQ= 2 Z OPQ Example 3 : PQ is a chord of length 8 cm of a P circle of radius 5 cm. The tangents at P and Q
intersect at a point T (see Fig. 10.10). Find the Sem
length TP. 8 cm
Solution : Join OT. Let it intersect PQ at the T RTO
point R. Then A TPQ is isosceles and TO is the
angle bisector of Z PTQ. So, OT L PQ
and therefore, OT bisects PQ which gives O
PR =RQ=4cm.
5 5 DD Fig. 10.10
Also, OR = .OP* - PR? =,/5°- 4 cm =3cm. Reprint 2025-26


CIRCLES 151
Now, 2ZTPR+ Z RPO = 90° = Z TPR + Z PTR (Why?)
So, ZRPO=Z2Z PTR
Therefore, right triangle TRP is similar to the right triangle PRO by AA similarity.
i, TP RP. TP 4 20
This gives PO = Ro' Le. = 3 or TP = 3 cm. Note : TP can also be found by using the Pythagoras Theorem, as follows:
Let TP =x and TR=y. Then . P= y+16 (Taking right A PRT) d)
V?4+52= (y +3) (Taking right A OPT) © (2)
Subtracting (1) from (2), we get "
32 16 . _
25= 6y-7 or y= G3 - |
2 . Therefore, ee () +16= 206+) =" irom (1)
2002 |
or x= 3 :
Co' EXERCISE 10.2
In Q.1 to 3, choose the correct option and give justification.
1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from
the centre is 25 cm. The radius of the circle is T
(A) Tom (B) 12cm : <\
(C) 15cm . > (D) 24.5cm
2.


and give justification. 1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is T (A) Tom (B) 12cm : <\ (C) 15cm . > (D) 24.5cm 2. In Fig. 10.1 1, if TP and TQ are the two tangents \ ww \
to acircle with centre O so that Z POQ = 110°, Q
then Z PTQ is equal to
(A) 60° - (B) 70°
©) 80° (D) 90° Fig. 10.11
3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other
at angle of 80°, then Z POA is equal to
(A) 50° (B) 60°
(C) 70° (D) 80°
Reprint 2025-26


152 MATHEMATICS
4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
5. Prove that the perpendicular at the point of contact to the tangent to a circle passes
through the centre.
6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4
cm. Find the radius of the circle.
7, Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the
larger circle which touches the smaller circle.
8. Aquadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that
AB+CD=AD+BC °
R Cc
D x P Y ,
0 Op
A | xX] y
P Bo - QB
Fig. 10.12 Fig. 10.13
9, In Fig. 10.13, XY and X'Y' are two parallel tangents to a circle with centre O and
another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove
that Z AOB = 90°. .) °
10. Prove that the angle between the two tangents drawn from an external point to a circle
is supplementary to the angle subtended by the line-segment joining the points of
contact at the centre.
11. Prove that the parallelogram circumscribing a A
circle is arhombus.
12.


= 90°. .) ° 10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre. 11. Prove that the parallelogram circumscribing a A circle is arhombus. 12. Atriangle ABC is drawn to circumscribe a circle
of radius 4 cm such that the segments BD and
DC into which BC is divided by the point of
contact D are of lengths 8 cm and 6 cm
respectively (see Fig. 10.14). Find the sides AB
and AC.
13. Prove that opposite sides of a quadrilateral ¢ B
. wpe . <-6 cm->D<-38 cm -->
circumscribing a circle subtend supplementary
angles at the centre of the circle. Fig. 10.14
Reprint 2025-26


CIRCLES 153
10.4 Summary
In this chapter, you have studied the following points :
1. The meaning of a tangent to a circle.
2. The tangent to a circle is perpendicular to the radius through the point of contact.
3. The lengths of the two tangents from an external point to a circle are equal. CO Reprint 2025-26


ia [al s5 jG 'a F t Sh meee Tai 1062CH12
11.1 Areas of Sector and Segment of a Circle . Q
You have already come across the terms sector and i. qialor
segment of a circle in your earlier classes. Recall __ "
that the portion (or part) of the circular regionenclosed | :
by two radii and the corresponding arc is called a
sector of the circle and the portion (or part) of the
circular region enclosed between a chord and the
: . : A B
corresponding arc is called a segment of the circle. P
Thus, in Fig. 11.1, shaded region OAPB is a sector Fig. 11.1
of the circle with centre O. Z AOB is called the
angle of the sector. Note that in this figure, unshaded region OAQB is also a sector of
the circle. For obvious reasons, OAPB is called the minor sector and
OAQB is called the major sector. You can also see that angle of the major sector is
360° - Z AOB. ,
Now, look at Fig. 11.2 in which AB is a chord Q
of the circle with centre O. So, shaded region APB is Mai
a segment of the circle.


is called the major sector. You can also see that angle of the major sector is 360° - Z AOB. , Now, look at Fig. 11.2 in which AB is a chord Q of the circle with centre O. So, shaded region APB is Mai a segment of the circle. You can also note that Segm ont
unshaded region AQB is another segment of the circle 0 B
formed by the chord AB. For obvious reasons, APB °
is called the minor segment and AQB is called the Ss
major segment. SX,
&
Remark : When we write 'segment' and 'sector' x $
we will mean the 'minor segment' and the 'minor ;
sector' respectively, unless stated otherwise. Fig. 11.2
Reprint 2025-26


AREAS RELATED TO CIRCLES 155
Now with this knowledge, let us try to find some O
relations (or formulae) to calculate their areas. Let OAPB be a sector of a circle with centre
O and radius r (see Fig. 11.3). Let the degree
measure of Z AOB be 0. A
You know that area of a circle (in fact of a
circular region or disc) is nr'. In a way, we can consider this circular region to A B
be a sector forming an angle of 360° (i.e., of degree P "
measure 360) at the centre O. Now by applying the Fig. 11.3
Unitary Method, we can arrive at the area of the
sector OAPB as follows: ; ,
When degree measure of the angle at the centre is 360, area of the
sector = mr? . oe ,
So, when the degree measure of the angle at the centre is 1, area of the
nw a oe "
sector = ~~: oe "
360 . , .


, When degree measure of the angle at the centre is 360, area of the sector = mr? . oe , So, when the degree measure of the angle at the centre is 1, area of the nw a oe " sector = ~~: oe " 360 . , . Therefore, when the degree measure of the angle at the centre is 0, area of the
;
Tr 0 2 :
tor = -x6=-=-xmMr', .
sector 360 360 / :
Thus, we obtain the following relation (or formula) for area of a sector of a
circle: QO ) ;
Area of the sector of angle @=--- x nr?
rea of the sector o angle 360 ,
where r is the radius of the circle and 0 the angle of the sector in degrees. Now, a natural question arises : Can we find Q
the length of the arc APB corresponding to this
sector? Yes. Again, by applying the Unitary
Method and taking the whole length of the circle
(of angle 360°) as 2mr, we can obtain the required a
. 0
length of the arc APB as 360 x 2nr, yn
9 A™__.- B
P
=. --x2
So, length of an are of a sector of angle 0 360 xX 20 , Fig, 11.4
Reprint 2025-26


156 MATHEMATICS
Now let us take the case of the area of the
segment APB of a circle with centre O and radius r
(see Fig. 11.4). You can see that :
Area of the segment APB = Area of the sector OAPB - Area of A OAB
Q 2
= --~- Xr -areaof AOAB
360
Note : From Fig. 11.3 and Fig. 11.4 respectively, you can observe that:
Area of the major sector OAQB = ar - Area of the minor sector OAPB
and Area of major segment AQB = n° - Area of the minor segment APB
Let us now take some examples to understand these concepts (or results). Example 1 : Find the area of the sector of a circle
with radius 4 cm and of angle 30°. Also, find the area
of the corresponding major sector (Use m = 3.14). Solution : Given sector is OAPB (see Fig. 11.5). - a \
Area of the sector = o x mr? a v
~ 360
. A p B
=o 3 14x 4x 4 om? P
360 . . Fig. 11.5
: 12.56 |
C) = 3 cm? = 4.19cm?


3.14). Solution : Given sector is OAPB (see Fig. 11.5). - a \ Area of the sector = o x mr? a v ~ 360 . A p B =o 3 14x 4x 4 om? P 360 . . Fig. 11.5 : 12.56 | C) = 3 cm? = 4.19cm? (approx.)
\ = XQ
Area of the corresponding major sector
= mr - area of sector OAPB
8 = (3.14 x 16 - 4.19) cm?
4, "= 46.05 cm? = 46.1 cm?' (approx.)
oe 360 - 6
Alternatively, area of the major sector = st x Tr?
: = [P05 |x 3.14% 16 cm?
360
= 0 3.14 16cm? = 46.05 cm?
360
= 46.1 cm? (approx.)
Reprint 2025-26


AREAS RELATED TO CIRCLES 157
Example 2 : Find the area of the segment AYB x
shown in Fig. 11.6, if radius of the circle is 21 cm and A J B
22 120°. Z AOB = 120°. (Use t= =) Cee
Fig. 11.6 '
Solution : Area of the segment AYB
= Area of sector OAYB - Area of AOQAB (1)
120. 22 | . Now, area of the sector OAYB = 360 x = x 21x 21cm? = 462 cm'. (2)
For finding the area of A OAB, draw OM _ AB as shown in Fig. 11.7. Note that OA = OB. Therefore, by RHS congruence, A AMO = A BMO.
. » . : ; 1 ° oO
So, M is the mid-point of AB and Z AOM = Z BOM = 5 x 120° = 60°,
L = Ls
et OM =x cm A B
"om . So, fromAOMA, \ -- =cos 60° a oy
> > NS -_ OA . 7 <> a
. 1 1 °
x
-=- | cos 60° =- i
Or, 172 ( ;] Fig. 11.7
- . 21
or, . XG
. . 21
So, OM = - cm
2
.


So, fromAOMA, \ -- =cos 60° a oy > > NS -_ OA . 7 <> a . 1 1 ° x -=- | cos 60° =- i Or, 172 ( ;] Fig. 11.7 - . 21 or, . XG . . 21 So, OM = - cm 2 . AM J3
Al . -- =si ° = -
SO, OA sin 60 5
21V3
So, AM = aN cm
2x21
Therefore, AB = 2 AM = DONS cm = 21V3cm
Reprint 2025-26


158 MATHEMATICS
1 1 21
So, area of A OAB = 5 ABXOM = > x 21V3 x > em'
441
= -3cm? (3)
4
441 2
Therefore, area of the segment AYB = [460 - 3 cm" [From (1), (2) and (3)]
21
= 7 (88 - 21V3)em? :
EXERCISE 11.1
22 . Unless stated otherwise, use 7 = 7: .
1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
2. Find the area of a quadrant of a circle whose circumference is 22 cm.
3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute
hand in 5 minutes.
4. Achord of acircle of radius 10 cm subtends a right angle at the centre. Find the area of
the corresponding: (i) minorsegment (ii) major sector. (Use m=3.14)
5. Inacircle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
A \ . (i) the length of the arc (ii) area of the sector formed by the arc
~M " 5
(iii) area of the segment formed by the corresponding chord
6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre.


Find: A \ . (i) the length of the arc (ii) area of the sector formed by the arc ~M " 5 (iii) area of the segment formed by the corresponding chord 6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find
the areas of the corresponding minor and major
i <«¢ *.¢ ~
segments of the circle. vs 0 "3 vie ?, ' mn
(Use m=3.14 and ./3 = 1.73) Sfet ti fe% 208 23 +:
So ere Pose tos uy et
7. Acchord of a circle of radius 12 cm subtends an SESS St Je ye
angle of 120° at the centre. Find the area of the 3-240 ce 23 233 Pas
corresponding segment of the circle. +o BX wa + ot 3
+e ATR o
(Usen=3.14 and 3 = 1.73) 93S" : Ds
i. + d/ete ' +
8. A horse is tied to a peg at one corner of a square * "+, oe% pawn
shaped grass field of side 15 m by means of a5 m 7 247
long rope (see Fig. 11.8). Find Fig. 11.8
Reprint 2025-26


AREAS RELATED TO CIRCLES 159
(i) the area of that part of the field in which the
horse can graze. (ii) the increase in the grazing area if the rope
were 10 m long instead of 5 m. (Use x =3.14) lok\ [se
9. A brooch is made with silver wire in the form
of a circle with diameter 35 mm. The wire is also
used in making 5 diameters which divide the
circle into 10 equal sectors as shown in
Fig. 11.9. Find: °
. Fig. 11.9
(i) the total length of the silver wire required. (ii) the area of each sector of the brooch. . 10. Anumbrella has 8 ribs which are equally spaced
(see Fig. 11.10). Assuming umbrella to be a flat
circle of radius 45 cm, find the area between the
two consecutive ribs of the umbrella. _ oo | 11. Acar has two wipers which do not overlap. Each : 1
wiper has a blade of length 25 cm sweeping - WY
through an angle of 115°. Find the total area Fig. 11.10
cleaned at each sweep of the blades. 12. To warn ships for underwater rocks, a lighthouse" ' gr ve ON
spreads a red coloured light over a sector of I SA
angle 80° to a distance of 16.5 km.


through an angle of 115°. Find the total area Fig. 11.10 cleaned at each sweep of the blades. 12. To warn ships for underwater rocks, a lighthouse" ' gr ve ON spreads a red coloured light over a sector of I SA angle 80° to a distance of 16.5 km. Find the area A h
of the sea over which the ships are warned. SJ a
(Use n= 3.14) $ Ps ' s 13. A round table cover has six equal designs as Ye 4
shown in Fig. 11.11. If the radius of the cover is an LE
28 cm, find the cost of making the designs at the -
rate of 0.35 per cm'. (Use ./3 = 1.7) Fig. 11.11 14. Tick the correct answer in the following : Area of a sector of angle p (in degrees) of a circle
with radius R is
P P 2 P P 2
--xX2nR --xmR --x2nR --x2nR
() 180 (8) 180 © 360 ©) 720
Reprint 2025-26


160 MATHEMATICS
11.2 Summary
In this chapter, you have studied the following points :
1. Length of an arc of a sector of a circle with radius r and angle with degree measure 0 is
8 x 2mr.
360
2. Area of a sector of a circle with radius r and angle with degree measure 9 is a x Tr,
3. Area of segment of a circle "
= Area of the corresponding sector - Area of the corresponding triangle. OO. Reprint 2025-26


it a A any _ (a) thr 1062CH13
12.1 Introduction . From Class IX, you are familiar with some of the solids like cuboid, cone, cylinder, and
sphere (see Fig. 12.1). You have also learnt how to find their surface areas and volumes. (i) Cc) (ii) . (iii) (iv)
. Fig, 12.1
In our day-to-day life, we come across a number of solids made up of combinations
of two or more of the basic solids as shown above. You must have seen a truck with a
container fitted on its back (see Fig. 12.2), 8)
carrying oil or water from one place to RY
another. Is it in the shape of any of the four |.
basic solids mentioned above? You may NK WK A
guess that it is made of a cylinder withtwo [J - TT Fr
hemispheres as its ends. \e) - 0
Fig. 12.2
Reprint 2025-26


162 MATHEMATICS Again, you may have seen an object like the
one in Fig. 12.3. Can you name it? A test tube, right! 3
You would have used one in your science laboratory. _0°
This tube is also a combination of a cylinder and a
hemisphere. Similarly, while travelling, you may have
seen some big and beautiful buildings or monuments
made up of a combination of solids mentioned above. If for some reason you wanted to find the
surface areas, or volumes, or capacities of such ;
objects, how would you do it? We cannot classify .
these under any of the solids you have already studied. Fig. 12.3 In this chapter, you will see how to find surface areas and volumes of such
objects. . 12.2 Surface Area of a Combination of Solids a . Let us consider the container seen in Fig. 12.2. How do we find the surface area of
such a solid? Now, whenever we come across a new problem, we first try to see, if
we can break it down into smaller problems, we have earlier solved. We can see that
this solid is made up of a cylinder with two hemispheres stuck at either end. It would
look like what we have in Fig. 12.4, after we put the pieces all together.


see, if we can break it down into smaller problems, we have earlier solved. We can see that this solid is made up of a cylinder with two hemispheres stuck at either end. It would look like what we have in Fig. 12.4, after we put the pieces all together. SX
--_>(f N ) --
@0-- CO--
ML .
: Fig. 12.4 If we consider the surface of the newly formed object, we would be able to see
only the curved surfaces of the two hemispheres and the curved surface of the cylinder. So, the total surface area of the new solid is the sum of the curved surface
areas of each of the individual parts. This gives, TSA of new solid = CSA of one hemisphere + CSA of cylinder : + CSA of other hemisphere
where TSA; CSA stand for "Total Surface Area' and 'Curved Surface Area'
respectively. Let us now consider another situation. Suppose we are making a toy by putting
together a hemisphere and a cone. Let us see the steps that we would be going
through. Reprint 2025-26


SURFACE AREAS AND VOLUMES 163
First, we would take a cone and a hemisphere and bring their flat faces together. Here, of course, we would take the base radius of the cone equal to the radius of the
hemisphere, for the toy is to have a smooth surface. So, the steps would be as shown
in Fig. 12.5.
fa
oa " & -_ OS
Fig. 12.5 - At the end of our trial, we have got ourselves a nice round-bottomed toy. Now if
we want to find how much paint we would require to colour the surface of this toy,
what would we need to know? We would need to know the surface area of the toy,
which consists of the CSA of the hemisphere and the CSA of the cone. So, we can say: . oO _
Total surface area of the toy = CSA of hemisphere + CSA of cone Now, let us consider some examples. : Example I : Rasheed got a playing top (lattu) as his (seen \
birthday present, which surprisingly had no colour on k \
it. He wanted to colour it with his crayons. The top is 5cem
shaped like a cone surmounted by a hemisphere
(see Fig 12.6).


some examples. : Example I : Rasheed got a playing top (lattu) as his (seen \ birthday present, which surprisingly had no colour on k \ it. He wanted to colour it with his crayons. The top is 5cem shaped like a cone surmounted by a hemisphere (see Fig 12.6). The entire top is 5 cm in height and
the diameter of the top is 3.5 cm. Find the area he io.
22 ;
has to colour, (Take m = 7 ) Fig. 12.6
Solution : This top is exactly like the object we have discussed in Fig. 12.5. So, we
can conveniently use the result we have arrived at there. That is :
. _ TSA of the toy = CSA of hemisphere + CSA of cone
1
Now, the curved surface area of the hemisphere = 3 Anr °) = 2nr?
_|2~x 22 35 33 cm?
7 2 2
Reprint 2025-26


164 MATHEMATICS
Also, the height of the cone = height of the top - height (radius) of the hemispherical part
3.5
= |5-- |em =3.25 cm
2
3.5)
So, the slant height of the cone (J) = yr? +h? = 2 + (3.25)? cm =3.7 cm (approx.)
22 3.5 2
Therefore, CSA of cone = mrl = 7 x > x 3.7 | cm "
This gives the surface area of the top as ,
22.3.5. 3. 22. 3. |
a2 2 4 32 39) om? +f 2? x 32 3.7 Jem?
7 2 2 7 2° yO
22 3. ll , co
=> x "2 (3.5 + 3.7) cm' = 3X (3.5 + 3.7) cm? = 39.6 cm" (approx.)
You may note that 'total surface area of the top' is not the sum of the total
surface areas of the cone and hemisphere. ° 4.2 em
Example 2 : The decorative block shown : (....y >
in Fig. 12.7 is made of two solids - a cube . Pa
and a hemisphere. The base of the block isa .
cube with edge 5 cm, and the hemisphere 5 om,
fixed on the top has a diameter of 4.2 cm.


The decorative block shown : (....y > in Fig. 12.7 is made of two solids - a cube . Pa and a hemisphere. The base of the block isa . cube with edge 5 cm, and the hemisphere 5 om, fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block.
22
(Take n= -) Sem
7 oo Fig. 12.7
Solution : The total surface area of the cube = 6 x (edge) =6 x5 x5 cm' = 150 cm'. Note that the part of the cube where the hemisphere is attached is not included in the
surface area. So, ~ the surface area of the block = TSA of cube - base area of hemisphere
° + CSA of hemisphere
= 150--nr +2 ar = (150 + ar') cm
= 150 cm? + 22 42 42 cm?
7 2 2
= (150 + 13.86) cm? = 163.86 cm? Reprint 2025-26


SURFACE AREAS AND VOLUMES 165
Example 3 : A wooden toy rocket is in the
shape of a cone mounted on a cylinder, as
shown in Fig. 12.8. The height of the entire 6m
rocket is 26 cm, while the height of the conical
: . . 26 cm
part is 6 cm. The base of the conical portion
has a diameter of 5 cm, while the base
diameter of the cylindrical portion is 3 cm. If
the conical portion is to be painted orange
and the cylindrical portion yellow, find the Scm
area of the rocket painted with each of these
colours. (Take m = 3.14) .
base of cylinder
Solution : Denote radius of cone by r, slant
. . . base of cone
height of cone by /, height of cone by A, radius >
of cylinder by 7' and height of cylinder by hr'. - oo
Then r = 2.5 cm, h = 6 cm, r = 1.5 cm, - ow
h' = 26 -6 = 20 cmand oe Fig. 12.8
l= Jr th? = J2.57 +6? om =6.5 cm
Here, the conical portion has its circular base resting on the base of the cylinder, but
the base of the cone is larger than the base of the cylinder.


- ow h' = 26 -6 = 20 cmand oe Fig. 12.8 l= Jr th? = J2.57 +6? om =6.5 cm Here, the conical portion has its circular base resting on the base of the cylinder, but the base of the cone is larger than the base of the cylinder. So, a part of the base of the
cone (a ring) is to be painted. ,
: \ . So, the area to be painted orange = CSA of the cone + base area of the cone
So - base area of the cylinder
= mrl+ar-n(ry
= m[(2.5 x 6.5) + (2.5 - (1.5)] cm? SO = 7[20.25] cm? = 3.14 x 20.25 cm? SO = 63.585 cm? Now, _ the area to be painted yellow = CSA of the cylinder
+ area of one base of the cylinder
a = 2nrh' + n(r'y
= nr (2h' +r)
= (3.14 x 1.5) (2 x 20 + 1.5) cm?
= 4.71 x 41.5 cm?
= 195.465 cm? Reprint 2025-26


166 MATHEMATICS
Example 4 : Mayank made a bird-bath for his garden 30cm
in the shape of a cylinder with a hemispherical ee
depression at one end (see Fig. 12.9). The height of SEES)
the cylinder is 1.45 m and its radius is 30 cm. Find the 145m
22
total surface area of the bird-bath. (Take 1 => )
Solution : Let h be height of the cylinder, and r the | | |
common radius of the cylinder and hemisphere. Then, Fig. 12.9 ;
the total surface area of the bird-bath = CSA of cylinder + CSA of hemisphere
= 2arh + 2nr = 2nr(h+r)
22 '
= 2x a 30(145 +30) cm?
= 33000cm? =3.3m?"
EXERCISE 12.1.
22 . Unless stated otherwise, take m = 7 1. 2 cubes each of volume 64 cm? are joined end to end. Find the surface area of the resulting cuboid. - _
sa™ . 2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The
diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the
inner surface area of the vessel. 3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius.


a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel. 3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy. 4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest
diameter the hemisphere can have? Find the surface area of the solid. 5. A hemispherical depression is cut out from one face of a cubical wooden block such
that the diameter / of the hemisphere is equal to the edge of the cube. Determine the
surface area of the remaining solid. 6. A medicine capsule is in the shape of a
cylinder with two hemispheres stuck to each 5mm
of its ends (see Fig. 12.10). The length of <--_-_-_
the entire capsule is 14 mm and the diameter 14mm
of the capsule is 5 mm. Find its surface area. Fig. 12.10 Reprint 2025-26


SURFACE AREAS AND VOLUMES 167
7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and
diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the
top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of
the canvas of the tent at the rate of 500 per m2. (Note that the base of the tent will not be covered with canvas.) 8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the
same height and same diameter is hollowed out. Find the total surface area of the
remaining solid to the nearest cm', 9. A wooden article was made by scooping «Kg '
out a hemisphere from each end of a solid AI
cylinder, as shown in Fig. 12.11. If the
height of the cylinder is 10 cm, and its :
base is of radius 3.5 cm, find the total . MU
surface area of the article. . QUT 12.3. Volume of a Combination of Solids oO Fig. 12.11
In the previous section, we have discussed how to find the surface area of solids made
up of a combination of two basic solids.


of radius 3.5 cm, find the total . MU surface area of the article. . QUT 12.3. Volume of a Combination of Solids oO Fig. 12.11 In the previous section, we have discussed how to find the surface area of solids made up of a combination of two basic solids. Here, we shall see how to calculate their
volumes. It may be noted that in calculating the surface area, we have not added the
surface areas of the two constituents, because some part of the surface area disappeared
in the process of joining them. However, this will not be the case when we calculate
the volume. The volume of the solid formed by joining two basic solids will actually be
the sum of the volumes of the constituents, as we see in the examples below. Se" . Example 5 : Shanta runs an industry in
a shed which is in the shape of a cuboid
surmounted by a half cylinder (see Fig.
12.12). If the base of the shed is of
dimension 7 m x 15 m, and the height of (oa,
the cuboidal portion is 8 m, find the volume A
of air that the shed can hold.


in the shape of a cuboid surmounted by a half cylinder (see Fig. 12.12). If the base of the shed is of dimension 7 m x 15 m, and the height of (oa, the cuboidal portion is 8 m, find the volume A of air that the shed can hold. Further, ;
suppose the machinery in the shed 8m:
occupies a total space of 300 m°, and '
there are 20 workers, each of whom
occupy about 0.08 m? space on an v
average. Then, how much air is in the Fig. 12.12
22
shed? (Take m = 7? Reprint 2025-26


168 MATHEMATICS
Solution : The volume of air inside the shed (when there are no people or machinery)
is given by the volume of air inside the cuboid and inside the half cylinder, taken
together. Now, the length, breadth and height of the cuboid are 15 m, 7 m and 8 m, respectively. Also, the diameter of the half cylinder is 7 m and its height is 15 m.
; ., 1 ,
So, the required volume = volume of the cuboid + 3 volume of the cylinder
1 22
= 115K 7x84 bx ec 15 |? = 1128.75 m?-
2 7 2 2 ;
Next, the total space occupied by the machinery = 300 m? °
And the total space occupied by the workers = 20 x 0.08 m3 = 1.6 m?-
Therefore, the volume of the air, when there are machinery and workers
= 1128.75 - (300.00 + 1.60) = 827.15 m3
Example 6 : A juice seller was serving his ; :
customers using glasses as shown in Fig. 12.13. ,
The inner diameter of the cylindrical glass was
5 cm, but the bottom of the glass had a'.
hemispherical raised portion which reduced the
capacity of the glass.


827.15 m3 Example 6 : A juice seller was serving his ; : customers using glasses as shown in Fig. 12.13. , The inner diameter of the cylindrical glass was 5 cm, but the bottom of the glass had a'. hemispherical raised portion which reduced the capacity of the glass. If the height of a glass eects,
was 10 cm, find the apparent capacity of the Dates'
glass and its actual capacity. (Use a = 3.14.) Fig. 12.13
Solution : Since the inner diameter of the glass = 5 cm and height = 10 cm,
the apparent capacity of the glass = urh
. . = 3.14x2.5 x 2.5 x 10 cm? = 196.25 cm? But the actual capacity of the glass is less by the volume of the hemisphere at the
base of the glass.
. i 2 2 3
i.e., * itis less by 3 tr = 3 x 3.14 2.52.5 x 2.5cm™ = 32.71 cm
So, the actual capacity of the glass = apparent capacity of glass - volume of the
hemisphere
= (196.25 - 32.71) cm'
= 163.54 cm? Reprint 2025-26


SURFACE AREAS AND VOLUMES 169
: i is i oo OAS
Example 7: A solid toy is in the form of a EC - FF
hemisphere surmounted by a right circular cone. The ZA
height of the cone is 2 cm and the diameter of the pao Nc
; , oo
base is 4 cm. Determine the volume of the toy. Ifa
right circular cylinder circumscribes the toy, find the \ /
difference of the volumes of the cylinder and the toy. HYP
(Take x = 3.14) Fig. 12.14
Solution : Let BPC be the hemisphere and ABC be the cone standing on the base
of the hemisphere (see Fig. 12.14). The radius BO of the hemisphere (as well as
1 ;
of the cone) = > x4cm=2cm.
2 3, 1_, "
So, volume of the toy = 3 + 3 h ; .
2 3,1 2 - 3 3
= 3 <8 l4x@) +3 X34 (Q2)° x2} cm = 25.12 cm
Now, let the right circular cylinder EFGH circumscribe the given solid.


the cone) = > x4cm=2cm. 2 3, 1_, " So, volume of the toy = 3 + 3 h ; . 2 3,1 2 - 3 3 = 3 <8 l4x@) +3 X34 (Q2)° x2} cm = 25.12 cm Now, let the right circular cylinder EFGH circumscribe the given solid. The radius of
the base of the right circular cylinder = HP = BO = 2 cm, and its height is
EH = AO + OP=(24+2)cm=4cm
So, the volume required = volume of the right circular cylinder - volume of the toy
fe (3.14 x 2? x 4 ~ 25.12) cm?
. = 25.12 cm? :
NOY 5
Hence, the required difference of the two volumes = 25.12 cm*. EXERCISE 12.2
s 22
Unless stated otherwise, take T= 7 . 1. A solid is in the shape of a cone standing on a hemisphere with both their radii being
equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid
in terms of 7. 2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with
two cones attached at its two ends by using a thin aluminium sheet. The diameter of the
model is 3 cm and its length is 12 cm.


of the solid in terms of 7. 2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume
of air contained in the model that Rachel made. (Assume the outer and inner dimensions
of the model to be nearly the same.) Reprint 2025-26


170 MATHEMATICS 3. A gulab jamun, contains sugar syrup up to about KD
30% of its volume. Find approximately how much
syrup would be found in 45 gulab jamuns, each
shaped like a cylinder with two hemispherical ends
with length 5 cm and diameter 2.8 cm (see Fig. 12.15). eee fee 4. A pen stand made of wood is in the shape of a Fig. 12.15
cuboid with four conical depressions to hold pens. -
The dimensions of the cuboid are 15 cm by 10 cm by
3.5 cm. The radius of each of the depressions is 0.5
cm and the depth is 1.4 cm. Find the volume of
wood in the entire stand (see Fig. 12.16). | 5. A vessel is in the form of an inverted cone. Its .
height is 8 cm and the radius of its top, which is
open, is 5 cm. It is filled with water up to the brim. . ME VV VV
When lead shots, each of which is a sphere of radius ce
0.5 cm are dropped into the vessel, one-fourth of | |
the water flows out. Find the number of lead shots . . d di . Fig. 12.16
ropped in the vessel. 6.


brim. . ME VV VV When lead shots, each of which is a sphere of radius ce 0.5 cm are dropped into the vessel, one-fourth of | | the water flows out. Find the number of lead shots . . d di . Fig. 12.16 ropped in the vessel. 6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which
is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the
pole, given that 1 cm? of iron has approximately 8g mass. (Use m = 3.14) fs.' 7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on
a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water
such that it touches the bottom. Find the volume of water left in the cylinder, if the radius
of the cylinder is 60 cm and its height is 180 cm. 8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter
of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its
volume to be 345 cm'.


60 cm and its height is 180 cm. 8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm'. Check whether she is correct, taking the above as the inside
measurements, and 7 = 3.14. 12.4 Summary
In this chapter, you have studied the following points: 1. To determine the surface area of an object formed by combining any two of the basic
solids, namely, cuboid, cone, cylinder, sphere and hemisphere. 2. To find the volume of objects formed by combining any two of a cuboid, cone, cylinder,
sphere and hemisphere. Reprint 2025-26


[a]
eee
PY ae
[a] eer,
1062CH14
13.1 Introduction °
In Class IX, you have studied the classification of given data into ungrouped as well as
grouped frequency distributions. You have also learnt to represent the data pictorially
in the form of various graphs such as bar graphs, histograms (including those of varying
widths) and frequency polygons. In fact, you went a step further by studying certain
numerical representatives of the ungrouped data, also called measures of central
tendency, namely, mean, median and mode. In this chapter, we shall extend the study
of these three measures, i.e., mean, median and mode from ungrouped data to that of
grouped data. We shall also discuss the concept of cumulative frequency, the
cumulative frequency distribution and how to draw cumulative frequency curves, called
ogives. ( ) °
ML .
13.2 Mean of Grouped Data
The mean (or average) of observations, as we know, is the sum of the values of all the
observations divided by the total number of observations. From Class IX, recall that if
Xp Xye oo X, are observations with respective frequencies fof, ..6f, then this
means observation x, occurs f, times, x, occurs f, times, and so on. Now, the sum of the values of all the observations = Sx, +fx,+...+fx,, and
the number of observations =f, +f,+...+f..


IX, recall that if Xp Xye oo X, are observations with respective frequencies fof, ..6f, then this means observation x, occurs f, times, x, occurs f, times, and so on. Now, the sum of the values of all the observations = Sx, +fx,+...+fx,, and the number of observations =f, +f,+...+f.. So, the mean x of the data is given by
. t= AX t+ fary tot A %n
fithteoth,
Recall that we can write this in short form by using the Greek letter 2 (capital
sigma) which means summation. That is,
Reprint 2025-26


172 MATHEMATICS
DA
x=
Ls
i=l
; ; oo - 2x 2... vo
which, more briefly, is written as x = Ta , if it is understood that i varies from
1 ton. Let us apply this formula to find the mean in the following example. Example 1 : The marks obtained by 30 students of Class X of a certain school in a
Mathematics paper consisting of 100 marks are presented in table below. Find the
mean of the marks obtained by the students. Marks obtained | 10 | 20 | 36 50 | 56 70 | 72 88 | 92 | 95
(x,)
Number of 1) 1) 374 73]2)4)] 47172727371
students (/,)
Solution: Recall that to find the mean marks, we require the product of each x, with
the corresponding frequency f,. So, let us put them in a column as shown in Table 13.1.
- Table 13.1 ©
Marks obtained (x,) Number of students (f,)
10 1 10
20 1 20
36 3 108
40 4 160
50 3 150
56 2 112
60 4 240
70 4 280
72 1 72
80 1 80
88 2 176
92 3 276
95 1 95
Reprint 2025-26


STATISTICS 173
_ Lfx, 1779
Now, x= = - = 59.3
xf, 30
Therefore, the mean marks obtained is 59.3. In most of our real life situations, data is usually so large that to make a meaningful
study it needs to be condensed as grouped data. So, we need to convert given ungrouped
data into grouped data and devise some method to find its mean. Let us convert the ungrouped data of Example 1 into grouped data by forming
class-intervals of width, say 15. Remember that, while allocating frequencies to each
class-interval, students falling in any upper class-limit would be considered in the next
class, ¢.g., 4 students who have obtained 40 marks would be considered in the classinterval 40-55 and not in 25-40. With this convention in our mind, let us form a grouped
frequency distribution table (see Table 13.2). . - Table 13.2. o
Class interval 10-25 | 25-40] 40-55 | 55-70 | 70-85] 85-100
wmwrormon] 2 > 7 «lols Now, for each class-interval, we require a point which would serve as the 1 a
representative of the whole class. /t is assumed that the frequency of each classinterval is centred around its mid-point. So the mid-point (or class mark) of each
class can be chosen to represent the observations falling in the class.


we require a point which would serve as the 1 a representative of the whole class. /t is assumed that the frequency of each classinterval is centred around its mid-point. So the mid-point (or class mark) of each class can be chosen to represent the observations falling in the class. Recall that we
find the mid-point of a class (or its class mark) by finding the average of its upper and
lower limits. That is, : . Class saark = Upper class limit + Lower class limit
2
7 ° _ 10425 . With reference to Table 13.2, for the class 10-25, the class mark is > 1.€.,
17.5. Similarly, we can find the class marks of the remaining class intervals. We put
them in Table 13.3. These class marks serve as our x,'s. Now, in general, for the ith
class interval, we have the frequency f, corresponding to the class mark x, We can
now proceed to compute the mean in the same manner as in Example 1. Reprint 2025-26


174 MATHEMATICS
Table 13.3
Class interval | Number of students (/,) Class mark fe
10 - 25 2 17.5 35.0
25 - 40 3 32.5 97.5
40 - 55 7 47.5 332.5
55 - 70 6 62.5 375.0
70 - 85 6 715 465.0
85 - 100 6 92.5 555.0
rhe | Pini 18600
The sum of the values in the last column gives us X f.x,. So, the mean x of the
given data is given by Co
_ "fx, . ee
xr TAX, = 1860.0 = 62 . Xf, . 30 . ; - This new method of finding the mean is known as the Direct Method. We observe that Tables 13.1 and 13.3 are using the same data and employing the
same formula for the calculation of the mean but the results obtained are different. Can you think why this is so, and which one is more accurate? The difference in the
two values is because of the mid-point assumption in Table 13.3, 59.3 being the exact
mean, while 62 an approximate mean. - y Sometimes when the numerical values of x, and f, are large, finding the product
of x, and f, becomes tedious and time consuming. So, for such situations, let us think of
a method of reducing these calculations.


13.3, 59.3 being the exact mean, while 62 an approximate mean. - y Sometimes when the numerical values of x, and f, are large, finding the product of x, and f, becomes tedious and time consuming. So, for such situations, let us think of a method of reducing these calculations. We can do nothing with the f's, but we can change each x, to a smaller number
so that our calculations become easy. How do we do this? What about subtracting a
fixed number from each of these x,'s? Let us try this method. The first step is to choose one among the x; s as the assumed mean, and denote
it by 'a'. Also, to further reduce our calculation work, we may take 'a' to be that x,
which lies in the centre of x,, x,,...,x,. So, we can choose a = 47.5 or a = 62.5. Let
us choose a = 47.5. The next step is to find the difference d, between a and each of the x,'s, that is,
the deviation of 'a' from each of the x,'s.
1e., d,=x,-a=x,-475 Reprint 2025-26


STATISTICS 175
The third step is to find the product of d, with the corresponding f,, and take the sum
of all the f,d,'s. The calculations are shown in Table 13.4. Table 13.4
Class interval Number of Class mark d,= x,- 47.5 Sd,
students (f,) (x,)
10 - 25 2 17.5 -30 -60
25 - 40 3 32.5 -15 45
40 - 55 7 47.5 0 0
55 - 70 6 62.5 15 90
70 - 85 6 7715 30 180
85 - 100 6 92.5 45 270
ye | |=
dy
So, from Table 13.4, the mean of the deviations, d = yf
Now, let us find the relation between d and x. Since in obtaining d,, we subtracted 'a' from each x,, so, in order to get the mean
¥ , we need to add 'a' to d . This can be explained mathematically as:
f- ) ;
y - Lfd,
M f - oe - d =: ii
ean of deviations : ye. f
> Uf (x, -a
So, d= BA-@
- Lf,
yy 2Fix; - Ufa
SO Of OE
yf,
a a acti
. Xf;
. = xX-a
So, X=at+d
"fd,
i.e., X= at Hid
Xi
Reprint 2025-26


176 MATHEMATICS Substituting the values of a, Lfd, and Lf, from Table 13.4, we get xX = 475 4 9° L475 +14.5= 62,
30
Therefore, the mean of the marks obtained by the students is 62. The method discussed above is called the Assumed Mean Method. Activity 1 : From the Table 13.3 find the mean by taking each of x, (i.e., 17.5, 32.5,
and so on) as 'a'. What do you observe? You will find that the mean determined in
each case is the same, i.e., 62. (Why 7?) So, we can say that the value of the mean obtained does not depend on the
choice of 'a'. . Observe that in Table 13.4, the values in Column 4 are all multiples of 15. So, if
we divide the values in the entire Column 4 by 15, we would get smaller numbers to
multiply with f. (Here, 15 is the class size of each class interval.) . x; - 4a . - . : _ : So, let u, = hl where a is the assumed mean and s is the class size. Now, we calculate u, in this way and continue as before (i.e., find f, u,and
then Yfu.). Taking h = 15, let us form Table 13.5. -~ "Table 13.5
x; -a


_ : So, let u, = hl where a is the assumed mean and s is the class size. Now, we calculate u, in this way and continue as before (i.e., find f, u,and then Yfu.). Taking h = 15, let us form Table 13.5. -~ "Table 13.5 x; -a 10-25 2 17.5 -30 -2 -4 25 - 40 3 32.5 -15 -1 -3 40-55 7 47.5 0 0 0 55 - 70 6 62.5 15 1 6 70 - 85 6 715 30 2 12 85 - 100 6 92.5 45 3 18 - <Lfu,
L uo
et x f
Here, again let us find the relation between yw and x. Reprint 2025-26


STATISTICS 177
xX; --a
We have, "= ---
i h
(x; - a)
xf, > _
Therefore, ee 1 2X; - O2fi
Xf; h Lf;
12a Bh
hl Sf Xf, °
] _ .
= 7/* -a :
"[¥ - a] |
So, hu=xX-a : . .
ie., Foathi 9° |
So X= at h Phils
oi
Now, substituting the values of a, h, Sf, and Xf, from Table 14.5, we get
- 5
- X= 475+4+15x (*]
~) / 30
NL = 47.5+ 14.5 =62
So, the mean marks obtained by a student is 62. The method discussed above is called the Step-deviation method. We note that: - >
@ the step-deviation method will be convenient to apply if all the ds have a
common factor.
e The mean obtained by all the three methods is the same.
e@ The assumed mean method and step-deviation method are just simplified
forms of the direct method,
@ The formula ¥ =a+ Au still holds if a and / are not as given above, but are
any non-zero numbers such that u, = a
Let us apply these methods in another example. Reprint 2025-26


178 MATHEMATICS
Example 2 : The table below gives the percentage distribution of female teachers in
the primary schools of rural areas of various states and union territories (U.T.) of
India. Find the mean percentage of female teachers by all the three methods discussed
in this section. Percentage of | 15-25] 25-35 | 35-45] 45-55 | 55-65] 65-75] 75-85
female teachers
Number of 11 7 4 4 2 1
States/U.T. Source : Seventh All India School Education Survey conducted by NCERT
Solution : Let us find the class marks, x,, of each class, and put them in a column
(see Table 13.6): . oe
Table 13.6 9 -
Percentage of female Number of Xx,
teachers States /U.T. (f,)
15 - 25 6 20
25 - 35 11 30
35 - 45 7 40
45-55 4 50
55 - 65 4 60
65 - 75 2 70
75 - 85 1 80
x, -50
Here we take a = 50, h = 10, then d, = x,- 50 and 4, =o
We now find d, and u, and put them in Table 13.7. Reprint 2025-26


STATISTICS 179
Table 13.7
Percentage of | Number of | x, | d,=x,-50 yet Ix, | fd, fu,
female states/U.T. 10
teachers (f,)
15 - 25 6 20 -30 -3 120 | -180 | -18
25 - 35 ll 30 -20 -2 330 | -220 | -22
35 - 45 7 40 -10 -1 280} -70 | -7
45-55 4 50 0 0 200 0 0
55 - 65 4 60 10 1 240 40 4
65 - 75 2 70 20 2 140 40 4
75 - 85 1 80 30 3 80 30 3
Pw | ss | | | _[nm
From the table above, we obtain Xf = 35, Lfx,= 1390, ~ |
Sid. =- 360, Sfu, = -36. Using the direct method, x = 2% - 139 _ 39 74
of 35s
Using the assumed mean method, - . NK ) Yfd. - _
x= a + id = 50 + 2) _ 39.71
Lf 35
Using the step-deviation method,
_ fy .
ge at| MH |p = 504/22) 10=30.71
" Xf; 35
Therefore, the mean percentage of female teachers in the primary schools of
rural areas is 39.71. Remark : The result obtained by all the three methods is the same.


_ 39.71 Lf 35 Using the step-deviation method, _ fy . ge at| MH |p = 504/22) 10=30.71 " Xf; 35 Therefore, the mean percentage of female teachers in the primary schools of rural areas is 39.71. Remark : The result obtained by all the three methods is the same. So the choice of
method to be used depends on the numerical values of x, and f. If x, and f, are
sufficiently small, then the direct method is an appropriate choice. If x, and f, are
numerically large numbers, then we can go for the assumed mean method or
step-deviation method. If the class sizes are unequal, and x, are large numerically, we
can still apply the step-deviation method by taking / to be a suitable divisor of all the d,'s. Reprint 2025-26


180 MATHEMATICS
Example 3 : The distribution below shows the number of wickets taken by bowlers in
one-day cricket matches. Find the mean number of wickets by choosing a suitable
method. What does the mean signify? Number of 20-60 | 60-100] 100-150) 150-250] 250 - 350 | 350 - 450
wickets
Number of 7 5 16 12 2 3
bowlers
Solution : Here, the class size varies, and the x,s are large. Let us still apply the stepdeviation method with a = 200 and h = 20. Then, we obtain the data as in Table 13.8. Table 13.8 . oo
d,
Number of Number of x. d=x.-- 200 u; => wf.
. i i i 20 ivi
wickets bowlers
taken (f)
20 - 60 7 40 -160 -8 -56
60 - 100 5 80 -120 -6 -30
100 - 150 16 125 -75 -3.75 -60
150 - 250 12 200 0 0 0
250 - 350 2 300 100 5 10
350 - 450 3 400 200 10 30
es
_ -106
So, 7 = 106 | Therefore, ¥ = 200+ 20( I") = 200 - 47.11 = 152.89.
45 45
This tells us that, on an average, the number of wickets taken by these 45 bowlers
in one-day cricket is 152.89.


- 450 3 400 200 10 30 es _ -106 So, 7 = 106 | Therefore, ¥ = 200+ 20( I") = 200 - 47.11 = 152.89. 45 45 This tells us that, on an average, the number of wickets taken by these 45 bowlers in one-day cricket is 152.89. Now, let us see how well you can apply the concepts discussed in this section! Reprint 2025-26


STATISTICS 181
Activity 2 : Divide the students of your class into three groups and ask each group to do one of the
following activities. 1. Collect the marks obtained by all the students of your class in Mathematics in the
latest examination conducted by your school. Form a grouped frequency distribution
of the data obtained. 2. Collect the daily maximum temperatures recorded for a period of 30 days in your
city. Present this data as a grouped frequency table. 3, Measure the heights of all the students of your class (in cm) and form a grouped
frequency distribution table of this data. After all the groups have collected the data and formed grouped frequency distribution tables, the groups should find the mean in each case by the method which
they find appropriate. , me
EXERCISE 13.1. _
1. Asurvey was conducted by a group of students as a part of their environment awareness
programme, in which they collected the following data regarding the number of plants in
20 houses in a locality. Find the mean number of plants per house.
sO
Which method did you use for finding the mean, and why? 2. Consider the following distribution of daily wages of 50 workers of a factory.


they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house. sO Which method did you use for finding the mean, and why? 2. Consider the following distribution of daily wages of 50 workers of a factory. Daily wages (in 2) 500-520 | 520-540 | 540-560 | 560-580 | 580-600
Find the mean daily wages of the workers of the factory by using an appropriate method. 3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency /. Daily pocket 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25 allowance (in 2) sonar | 7 Le | Lets p ete Reprint 2025-26


182 MATHEMATICS
4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per
minute were recorded and summarised as follows. Find the mean heartbeats per minute
for these women, choosing a suitable method. Number of heartbeats | 65-68 | 68-71 | 71-74 | 74-77 | 77-80 | 80-83 | 83-86
per minute
ornsome [fe pe Pe pepe p
5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These
boxes contained varying number of mangoes. The following was the distribution of
mangoes according to the number of boxes. Number of mangoes | 50 - 52 | 53-55 | 56-58 | 59 - 61 62 - 64
Find the mean number of mangoes kept in a packing box. Which method of finding
the mean did you choose? . ' " _
6. The table below shows the daily expenditure on food of 25 households in a locality. Daily expenditure | 100-150 | 150-200 | 200-250 | 250-300 | 300-350
(in %)
Number of 4 5 2 2 2
households
Find the mean daily expenditure on food by a suitable method.
7.


6. The table below shows the daily expenditure on food of 25 households in a locality. Daily expenditure | 100-150 | 150-200 | 200-250 | 250-300 | 300-350 (in %) Number of 4 5 2 2 2 households Find the mean daily expenditure on food by a suitable method. 7. To find out the concentration of SO, in the air (in parts per million, i.e., ppm), the data
was collected for 30 localities in a certain city and is presented below:
Concentration of SO, (in ppm)
0.00 - 0.04 4
0.04 - 0.08 9
0.08 - 0.12 9
0.12 - 0.16 2
0.16 - 0.20 4
0.20 - 0.24 2
Find the mean concentration of SO, in the air. Reprint 2025-26