NCERT Solutions Class 10 – Chapter 11 Areas Related to Circles – Exercise 11.1

Question 1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Solution:

Given: r=66cm and ∅=60°

Area of sector=∅/(360°) *πr2

=60/360*22/7*6*6

=132/7

Question 2. Find the area of a quadrant of a circle whose circumference is 22 cm. Quadrant of circle is sector making 90°.

Solution:

Given: circumference of circle=22cm and ∅=90°

To find r=?

2πr=22

Radius =r=22/2πr cm=7/2cm

Area of quadrant =∅/(360°) *πr2

=90/360*22/7*7/2*7/2

=77/8cm2

Question 3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Solution:

Let assume, Minute hand of clock acts as radius of the circle.

Angle rotated by min⁡ (hand in 5minutes)=∅=360/60*5=30°

Radius=r=14cm

Area of swept middle hand=

=∅/(360°) *πr2

=30/360*22/7*14*14

=154/3cm2

Area swept by the minute hand in 5min=154/3cm2

Question 4. 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 π = 3.14)

Solution:

Radius =r=10cm

Major segment is =360°-90°=270°

(i) Area of minor segment =Area of sector-Area of triangle

=∅/(360°) *πr2-1/2*h*b

=90/306*3.14*10*10-1/2*10*10

=314/4-50

=78.5-50

Area of minor segment =28.5cm2

(ii) Area of major sector=∅/(360°) *πr2

=270/360*3.14*10*10

=3*314/4=235.5cm2

Area of major segment=235.5cm2

Question 5. In a circle of radius 21 cm, an arc subtends an angle of 60° 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

Solution:

(i) Radius=r=21cm

Length of arc=∅/(360°) *2πr

60/360*2*22/7*21

=22cm

(ii) Area of sector=∅/(360°) *πr2

60/360*22/7*21*21

11*21=231cm2

(iii) Area of segment =Area of sector -Area of triangle

=231-√3/4(side)2

=231-1.73/4*21*21

=231-762.93/4

=231-190.73

=40.27cm2

Question 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 segments of the circle. (Use π = 3.14 and √3 = 1.73)

Solution:

Radius of circle=15cm

∆AOB is isosceles

∴∠A = ∠B

=∠A+∠B+C=180°

=2∠A=180°-60°

=∠A=120°/2

=∠A=60°

Area of minor segment =Area of sector-Area of triangle

=∅/(360°) *πr2-√3/4(side)2

=(60°)/360*3.14*15*15-1.73/4

=706.5/6-389.25/4

=117.75-97.31

=20.44cm2

Area of major segment-Area of circle-Area of minor segment

=πr2-20.44

=3.14*15*15-20.44

=686.06cm2

Question 7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73)

Solution:

Radius=r=12cm

Area of triangle=1/2*base*height

Area of segment=Area of sector -area of triangle

=∅/(360°)*π*r²-1/2(side)²*sin∅

=120°/360°*3.14-1/2(side)²*sin120°∅

=150.72-6*12*sin60°

=150.72-6*12*√3/2

=150.72-36*1.73

=150.72-62.28

=88.44cm2

Question 8. 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 Fig.). 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 π = 3.14)

Solution:

(i) Horse with graze=∅/(360°) × π × r²

= 90°/360° × 3.14 × 5 × 5

= 78.5/4

=19.625cm²

Area of circle the length of rope is increased to 10m

=∅/(360°) × π × r²

=90°/360° × 3.14 × 10 × 10

=314/4

=78.5cm2

(ii) Increasing in grazing area=78.5m²-19.635m²=58.875m²

Question 9. A brooch is made with silver wire in the form of a circle with a diameter of 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. Find:

(i) the total length of the silver wire required.

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

Solution:

(i) Total length of silver wire required=circumference of broach+5diameter

=5 × 35 mm × πd

=175+22/7 × 35

=175+110

=285mm

(ii) Area of each sector=1/10×Area of circle

=1/10×π×r2

=1/10×22/7×35/2×35×2

=11×35/4

=385/4mm2

Question 10. An umbrella has 8 ribs that are equally spaced (see Fig.). Assuming the umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Solution:

Total ribs in umbrella=8

Radius of umbrella is =45cm

Area between the two consecutive ribs=1/8π×r2

=1/8×22/7×45×45cm2

=22275/28

Question 11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Solution:

Angle made by sector=∅=115°

r=25cm

Total area clean at each sweep of the blades=2×Area of sector

=2×∅/(360°)×π×r2

=2×(115°)/(360°)×22/7×25×25cm2

=23×11×25×25cm2

=158125/126cm2

=1254.96cm2

The total area clean at each sweep of blades=1254.96cm2

Question 12. To warn ships for underwater rocks, a lighthouse spreads a red colored light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned.

Solution:

(Use π = 3.14)

Distance over which light fall =r=16.5km

Angle made by the sector=∅=80°

Area of the sea over which the ships are warned=Area of sector

=∅/(360°)×π×r2

=(80°)/(360°)×3.14×16.5×16.5

=1709.73

=189.97km2

The area of ships is +warned =189.97km2

Question 13. 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 ₹ 0.35 per cm². (Use √3 = 1.7)

Solution:

Total equal designs==6

Radius =28cm

Cost for making design=RS.035 per cm2

∠O=360°/6=60°

Area of 1 design =Area of sector-Area of triangle

=∅/(360°)×π×r2×-

Question 14. Tick the correct answer in the following:

Area of a sector of angle p (in degrees) of a circle with radius R is

(A) p/180 × 2πR (B) p/180 × π R² (C) p/360 × 2πR (D) p/720 × 2πR²

Solution:

Area of sector angle p=p/360×2πr2

Therefore, option (D) is correct.