NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle

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Written by Team Trustudies
Updated at 2021-02-23


NCERT solutions for class 10 Maths Chapter 12 Areas Related To Circle Exercise - 12.1

Q1 ) The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumference of the two circles.



NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle


Answer :


Let r be the radius of the circle whose circumference is equal to the sum of the circumference of two circles of radii 19 cm and 9 cm.

? 2?r = 2?(19) + 2?(9)

? r = 19 + 9 = 28

Hence, the radius of the new circle is 28 cm.

Q2 ) The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the area of the two circles.



NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle


Answer :


Let r be the radius of the circle whose area is equal to the sum of the area of the circles of radii 8 cm and 6 cm.

??r2 = ?(8)2 + ?(6)2

?r2 = 82 + 62

? r2 = 64 + 36 
? r = 10

Hence, the radius of the new circle is 10 cm.

Q3 ) Figure depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue , Black and White. The diameter of the region representing Gold score is 21 cm and reach of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.



NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle


Answer :


The area of each of the five scoring regions are as under :

For Gold : ?(10.5)2
= 227×110.25

= 2425.57 
= 346.5 cm2

For Red : ?[(21)2 ? (10.5)2] 
= 227(441 ? 110.25) 
= 227×330.75

= 7276.57 
= 1039.5 cm2

For Blue : ?[(31.5)2 ? (21)2] 
= 227(992.25 ? 441)

= 227×551.25 
= 12127.57 
= 1732.5 cm2

For Black : ?[(42)2 ? (31.5)2] 
= 227(1764 ? 992.25)

= 227×771.75 
= 16978.57 
= 2425.5 cm2

For White : ?[(52.5)2 ? (42)2] 
= 227(2756.25 ? 1764)

= 227×992.25 
= 21829.57 
= 3118.5 cm2

Q4 ) The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?



NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle


Answer :


Distance covered by the car in 10 minutes

= 66 × 1000 × 100 × 1060 = 66000006 cm

Circumference of the wheel of the car

= (2 × 227 × 40)cm

? Number of revolutions in 10 minutes when the car is travelling at a speed of 66 km/hr.

= 66000006 × 72 × 22 × 40 = 4375

Hence, the wheel of car makes 4375 revolutions in 10 minutes.

Q5 ) Tick the correct answer in the following and justify your choice : If the perimeter and the area of a circle are numerically equal, then the radius of the circle is :
(a) 2 units
(b) ? units
(c) 4 units
(d) 7 units



NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle


Answer :


Here, 2?r = ?r2, where r is the radius

? r2 ? 2r = 0

? r(r ? 2) = 0

? r = 0 or r = 2

But, r ? 0 ? r = 2 units

Hence the correct option is (a).

NCERT solutions for class 10 Maths Chapter 12 Areas Related To Circle Exercise - 12.2

Q1 ) Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°



NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle


Answer :


We know that the area A of a sector of angle ? in a circle of radius r is given by A = ?360 × ?r2.

Here r = 6 cm, ? = 60°

? A = 60360 × 227 × 36 
= (1327) cm2

Q2 ) Find the area of a quadrant of a circle whose circumference is 22 cm.



NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle


Answer :


Let r be the radius of the circle. Then,

Circumference = 22 cm

? 2?r = 22 
? 2 × 227 × r = 22
? r = 72cm

Area of the quadrant of a circle

= 14?r2 = (14 × 227 × 494)

= 53956 = 778 cm2

Q3 ) The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.



NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle


Answer :


Clearly, minute hand of a clock describes a circle of radius equal to its length i.e., 14 cm. Since the minutes hand rotates through 6° in one minute, therefore, area swept by the minute hand in one minute is the area of sector of angle 6° in a circle of radius 14 cm. Hence, the required area i.e., the area swept in 5 minutes.

= (?360 × ?r2 × 5) 
= 6360 × 227 × (14)2 × 5
= 160 × 227 × 196 × 5 
= 1543 cm2

Q4 ) A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:
(i) minor segment

(ii) major sector (Use ? = 3.14)



NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle


Answer :


Here, r = 10 cm, ? = 90°

(i) Area of the minor segment

= r2[??360 ? 12sin?]

= (10)2[3.14 × 90360 ? 12sin90°]

= 100(0.785 ? 0.5) 
= 100 × 0.285 = 28.5 cm2

(ii) Area of the major sector

= ?360 × ?r2,
where r = 10 and ? = 270°

= 270360 × 3.14 × 102
= (34 × 3.14 × 100) = 235.5 cm2

Q5 ) In a circle of a radius 21 cm, an arc substends an angle of 60º at the centre. Find :
(i) the length of the arc

(ii) area of the sector formed by the arc

(iii) area of the segment formed by the corresponding chord.



NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle


Answer :


Here r = 21cm ? = 60°
(i) Length of the arc, l = ?360 × 2?r

= 60180 × 227 × 21 = 22cm

(ii) Area of the sector, A = ?360 × ?r2

= 60360 × 227 × 21 × 21 
= 231 cm2

(iii) Area of the segment

= r2[??360 ? 12sin?]

= (21)2[227 × 60360 ? 12sin60°] 
= 441(1121 ? 12 × 32)

= 21 × 11 ? 44134 
= (231 ? 44134)cm2

Q6 ) A chord of a circle of radius 15 cm subtends an angle of 60º at the centre. Find the areas of the corresponding minor and major segment of the circle. (Use ? = 3.14and3 = 1.73)



NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle


Answer :


Here r = 15 cm and ? = 60°

? Area of the minor segment

= r2[??360 ? 12sin?]

= (15)2[3.14 × 60360 ? 12 × sin60]

= 225[3.146 ? 12 × 32] 
= 225(3.14 × 2 ? 3312)

= 225(6.28 ? 3 × 1.7312) 
= 225 × (6.28 ? 5.19)12

= 225 × 1.0912 
= 20.4375cm2

Area of the major segment

= Area of the circle - Area of the minor segment

= 3.14 × 225 ? 20.4375 
= 706.5 ? 20.4375

= 686.0625 cm2

Q7 ) A chord of a circle of radius 12 cm subtends an angle of 120º at the centre. Find the area of the corresponding segment of the circle.
(Use ? = 3.14 and 3 = 1.73)



NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle


Answer :


Herer = 12 cm and ? = 120°

Area of the corresponding segment of the circle

= Area of the minor segment

= r2[??360 ? 12sin?]

= (12)2[3.14 × 120360 ? 12 × sin120]

= 144[3.143 ? 12 × 32] 
= 225(3.14 × 2 ? 3312)
[ ?sin120° = sin(180° ? 60°) 
= sin60° = 32]

= 144 × 3.14 × 4 ? 3312 
= 12 × (12.56 ? 3 × 1.73)

= 12 × (12.56 ? 5.19) 
= 12 × 7.37 = 88.44cm2

Q8 ) A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see figure). Find


(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 ? = 3.14 )



NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle


Answer :


(i) The horse will graze over a quadrant of a circle with centre at the corner A of the field and radius AF = 5 m.

Then the area of the quadrant of this circle

= ? × 524 = 3.14 × 254 = 19.625m2

(ii) In the 2nd case, radius = 10 m.

The area of the quadrant of this circle

= ? × 1024 = 3.14 × 1004 = 78.5 m2

? Increase in the grazing area

= 78.5 ? 19.625 = 58.875m2

Q9 ) 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 sector as shown in figure. Find :
(i) The total length of the silver wire required.

(ii) The area of each sector of the brooch.




NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle


Answer :


(i) Length of silver wire used to make a brooch

= 2?r, where r = 352mm

= 2 × 227 × 352 = 110 mm

Wire used in 5 diameters =5 × 35 = 175 mm

? Total wire used =(110 + 175) = 285 mm

(ii) The area of each sector of the brooch

= 110 × Area of the circle

= 110 × ?r2, where r = 352mm

= 110 × 227 × (352)2 = 3854 mm2

Q10 ) An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.



NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle


Answer :


Area between two consecutive ribs

= 18th times the Total Area of the umbrella

= 18 ×?r2 ,where r = 45cm

= 18 × 227 × (45)2 = 2227528 cm2

Q11 ) A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115º. Find the total area cleaned at each sweep of the blades.



NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle


Answer :


Here r = 25 cm, ? = 115º

Total area cleaned at each sweep of the blades

= 2 × Area of the sector
(having r = 25 and ? = 115º )

= 2 × ?360 × ?r2

=2 × 115360 × 227 × (25)2

= 23 × 11 × 62518 × 7 = 158125126 cm2

Q12 ) To warn ships for underwater rocks a lighthouse spreads a red coloured light over a sector of angle 80º to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use ? = 3.14 )



NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle


Answer :


Area of the sea over which the ships are warned

= Area of the sector
(having r = 16.5 km and ? = 80°)

= ?360 × ?r2

= 80360 × 3.14 × (16.5)2

= 68389.2360

= 189.97 km2

Q13 ) A round table cover has six equal designs as shown in figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs. 0.35 per cm2 . (Use 3 = 1.7)



NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle


Answer :


Clearly from the figure,



Area of one design

= Area of the sector AOB - Area ( ? AOB )

= ?360 × ?r2 ?12 × (r)2 × sin60°, where r = 28 cm

= 60360 × 227 × (28)2 ? 12 × (28)2 × 32

= 12323 ? 333.2

= 1232 ? 999.63

= 232.43cm2

Area of 6 such design = 6 × 232.43 = 464.8cm2

Cost of making such designs @ Rs. 0.35 per cm2

= Rs 0.35 × 464.8

= Rs 162.68

Q14 ) Tick the correct answer in the following : Area of a sector of angle p (in degrees) of a circle with radius R is :
(a) p180 × 2?R
(b) p180 × ?R2
(c) p360 × 2?R
(d) p720 × 2?R2



NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle


Answer :


We know that area A of a sector of angle ? in a circle of radius r is given by

A = ?360 × ?r2

But here , r = R and ? = p

? A = p360 × ?R2 = p720 × 2?R2

?(D) is the correct answer.

NCERT solutions for class 10 Maths Chapter 12 Areas Related To Circle Exercise - 12.3

Q1 ) Find the area of the shaded region in figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.



NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circle


Answer :


? ROQ is a diameter

?? RPQ = 90°

In ? PRQ ,
RQ2 = RP2 + PQ2

? RQ2 = 72 + 242 
= 49 + 576 
?RQ2= 625

? RQ = 25 cm

? Radius r = 12RQ = 252cm

Area of the semi circle = 12?r2 
= 12 × 227 × (252)2 
= 687528cm2

and area of ? RPQ = 12 × RP × PQ

= 12 × 7 × 24 = 84 cm2

Area of the shaded region
= Area of the semi circle - Area (