Chapter 12

Area Related to Circles

Exercise 12.2

Question 1

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

Answer

angle of the sector is 60°

the area of sector = (θ/360°)×πr²

∴ Area of the sector with angle 60° = (60°/360°)×πr² cm²

= (36/6)π cm²

= 6×22/7 cm² = 132/7 cm²

Question 2

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

Answer

Circumference of the circle = 22 cm (given)

quadrant of a circle is a sector which is making an angle of 90°.

radius of the circle = r

here
2πr = 22,

r = 22/2π cm = 7/2 cm

∴ Area of the quadrant = (θ/360°) × πr²

Here, θ = 90°

So, A = (90°/360°) × π r² cm²

= (49/16) π cm²

= 77/8 cm² = 9.6 cm²

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.

Answer

Length of minute hand = radius of the clock

Radius of the circle r= 14 cm (given)

Angle swept by minute hand in 60 minutes = 360°

The angle swept by the minute hand in 1 minutes = 360°/60 = 6^o

The angle swept by the minute hand in 5 minutes =5 x 6^o =30^o

Area of a sector = (θ/360°) × πr²

= (30°/360°) × πr² cm²

= (1/12) × π14²

= (49/3)×(22/7) cm²

= 154/3 cm²

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)

Answer

Given radius of a circle AO=10cm

A perpendicular is drawn from the center of circle to the chord of the circle which bisect the chord

AD=DC
∠AOD=∠COD =45°
∠AOC =∠AOD +∠COD
=45° +45°=90

In ΔAOD

sin45°= AD/AO= 1/√2 =AD/10

AD= 5√2
and

cos45°= OD/AO= 1/√2 =OD/10

OD= 5√2
and

AC = 2 AD =2 x 5√2 =10√2

Also, area of ΔAOC = 1/2 × AC x OD

So, area of ΔAOB = 1/ × 10√2 × 5√2

= 50 cm²

(i) Area of minor sector = (90/360°)×πr²

= 90/360 x 3.14 x 10²

= 78.5 cm²

Now, area of minor segment = area of minor sector – area of ΔAOB

= 78.5 – 50

= 28.5 cm²

(ii) Area of major sector = Area of circle – Area of minor sector

= (3.14×10²)-78.5

= 235.5 cm²

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

Answer

.

Given,

Radius = 21 cm

θ = 60°

(i) Length of an arc = θ/360°×2πr

∴ = (60°/360°)×2×(22/7)×21

= (1/6)×2×(22/7)×21

Or Arc AB Length = 22cm

(ii) the angle subtend at center = 60°

= (60°/360°)×π r² cm²

= 441/6×22/7 cm²

= 231 cm²

In ΔOAB
∠OAB =∠OBA

∠OAB + ∠AOB + ∠OBA =180

2∠OAB +60 =180

∠OAB =60

So
ΔOAB is equileteral triangle

area will be √3/4×a² sq. Units.

√3/4×21² = (441×√3)/4 cm²

(iii)Area of segment APB = Area of sector OAPB – Area of ΔOAB

Or, Area of segment APB = [231-(441×√3)/4] cm²

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)

Answer

.

Given,

Radius = 15 cm

θ = 60°

So,

Area of sector OAPB = (60°/360°)×πr² cm²

= 225/6 πcm²

In ΔOAB
∠OAB =∠OBA

∠OAB + ∠AOB + ∠OBA =180

2∠OAB =120

∠OAB =60

So
ΔOAB is equileteral triangle

So, Area of ΔAOB = (√3/4) ×a²

Or, (√3/4) ×15²

∴ Area of ΔAOB = 97.31 cm²

Now, area of minor segment APB = Area of OAPB – Area of ΔAOB

Or, area of minor segment APB = ((225/6)π – 97.31) cm² = 20.43 cm²

And,

Area of major segment = Area of circle – Area of segment APB

Or, area of major segment = (π×15²) – 20.4 = 686.06 cm²

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)

Answer

.

cord AB substended an angle 120° to the center

∠ AOB = 120°
draw OD perpendicular to AB
AB=BD

∠ AOD =∠ BOD= 60°

here radius r=12 cm
IN ΔAOD
sin60°=AD/AO= √3/2=AD/12
AD =6√3 =10.38
cos60°=OD/AO= 1/2=OD/12
OD =6cm

length AB = 2AD =2 x 10.38 = 20.76

So, the area of ΔAOB = 1/2 × base × height

So, area of ΔAOB = 1/2 × 20.76 × 6 = 36√3 cm = 62.28 cm²

Now, the area of the minor sector = (θ/360°)×πr²

= (120/360)×(22/7)×12²

= 150.72 cm²

∴ Area of the corresponding Minor segment = Area of the Minor sector – Area of ΔAOB

= 150.72 cm²– 62.28 cm² = 88.44 cm²

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. 12.11). 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)

Answer

side of square =15 m

radius =5m

Here, the length of rope will be the radius of the circle i.e. r = 5 m

(i)

Now, the area of the part of the field where the horse can graze = (θ/360°) × πr²
(90/360°) x 3.14 x 5² = 19.625 m²

(ii)If the rope is increased to 10 m,

Area of circle will be = πr² =22/7×10² = 314 m²

Now, the area of the field graze by horse=(90/360°) x 3.14 x 10²

= 314/4 = 78.5 m²

∴ Increase in the grazing area = 78.5 m² – 19.625 m² = 58.875 m²

Question 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. 12.12. Find:

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

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

Answer

Diameter (D) = 35 mm

number of diameters = 5

Now, length of 5 diameters = 35×5 = 175

Circumference of the circle = 2πr

C = πD = 22/7×35 = 110

Area of the circle = πr²

Or, A = (22/7)×(35/2)² = 1925/2 mm²

(i)Total length of silver wire required = Circumference of the circle + Length of 5 diameter

= 110+175 = 185 mm

(ii)Number of sectors in the brooch = 10

angle of each sector = 360° /10 =36°

So, the area of each sector =(θ/360°) × πr²
(36/360°)× 22/7 x (35/2)²
= 22/70 x 1225/4
=26950/280

∴ = 385/4 mm²

Question 10

An umbrella has 8 ribs which are equally spaced (see Fig. 12.13). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Answer

The radius of the umbrella = 45 cm

Total number of ribs (n) = 8

The Angle between two rib = 360°/8 =45°

the area of one rib =(θ/360°) × πr²
45/360° × 22/7 x (45)²
22/7 x 8

∴ The area between the two consecutive ribs of the umbrella = A/n

6364.29/8 cm²

Or, The area between the two consecutive ribs of the umbrella = 795.5 cm²

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.

Answer

Given,

Radius (r) = 25 cm

Sector angle (θ) = 115°

Since there are 2 blades,

The total area of the sector made by wiper = 2×(θ/360°)×π r²

= 2×(115/360)×(22/7)×25²

= 2×158125/252 cm²

= 158125/126 = 1254.96 cm²

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.

(Use π = 3.14)

Answer

Let O bet the position of Lighthouse.

Given, radius (r) = 16.5 km

Sector angle (θ) = 80°

Or, Area of sector = (θ/360°)×πr²

= (80°/360°)×πr² km²

= 189.97 km²

Question 13

A round table cover has six equal designs as shown in Fig. 12.14. 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)

Answer

Total number of equal designs = 6

AOB= 360°/6 = 60°

Radius of the cover = 28 cm

Cost of making design = ₹ 0.35 per cm²

Since the two arms of the triangle are the radii of the circle and thus are equal, and one angle is 60°, ΔAOB is an equilateral triangle. So, its area will be (√3/4)×a² sq. units

Here, a = OA

∴ Area of equilateral ΔAOB = (√3/4)×28² = 333.2 cm²

Area of sector ACB = (60°/360°)×πr² cm²

= 410.66 cm²

So, area of a single design = area of sector ACB – area of ΔAOB

= 410.66 cm² – 333.2 cm² = 77.46 cm²

∴ Area of 6 designs = 6×77.46 cm² = 464.76 cm²

So, total cost of making design = 464.76 cm² ×Rs.0.35 per cm²

= Rs. 162.66

Question 14

Tick the correct solution 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

Answer

The area of a sector = (θ/360°)×πr²

Given, θ = p

So, area of sector = p/360×πR²

Multiplying and dividing by 2 simultaneously,

= (p/360)×2/2×πR²

= (2p/720)×2πR²

So, option (D) is correct.