HBSE 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2

Haryana State Board HBSE 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2

Unless stated otherwise, use π = \(\frac{22}{7}\).

Question 1.
Find the area of sector of a circle with radius 6 cm if angle of the sector is 60°.
Solution.
We have,
Radius = 6 cm
θ = 60°
Area of the sector = \(\frac{\pi r^2 \theta}{360^{\circ}}\)
= \(\frac{22}{7} \times \frac{6^2 \times 60^{\circ}}{360^{\circ}}\)
= \(\frac{22}{7} \times \frac{6 \times 6}{6}=\frac{132}{7}\)
Hence, area of sector = \(\frac{132}{7}\) cm2.

Haryana Board Solutions for 10th Class Maths Chapter 12 Areas Related to Circles Ex 12.2

Question 2.
Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
Let r be the radius of the circle whose circumference is 22 cm.
2πr = 22
r = \(\frac{22}{2 \pi}\)
r = \(\frac{22}{2 \times \frac{22}{-7}}\)
r = \(\frac{22 \times 7}{2 \times 22}\)
r = \(\frac{7}{2}\) cm
θ = 90°
[∵ sector is quadrant ∴ θ = \(\frac{360}{4}\) = 90°]
Therefore,area of the quadrant = \(\frac{\pi r^2 \theta}{360^{\circ}}\)
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{90^{\circ}}{360^{\circ}}\)
= \(\frac{22 \times 7}{4 \times 4}\)
= \(\frac{77}{8}\) cm2
Hence, area of the quadrant = \(\frac{77}{8}\) 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.
Solution.
Angle describe by the minute hand in 60 minutes = 360°
Angle, describe by the minute hand in 5 minutes = \(\frac{360^{\circ}}{60^{\circ}}\) × 5 = 30°
Now, θ = 30° and radius = 14 cm
Area swept by the minute hand in 5 minutes = Area of sector AOB

Haryana Board 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 1

= \(\frac{11 \times 14}{3}\)
= \(\frac{154}{3}\) cm2.
Hence, area swept by the minute hand = \(\frac{154}{3}\) cm2.

Haryana Board Solutions for 10th Class Maths Chapter 12 Areas Related to Circles Ex 12.2

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.
We have,
Radius = 10 cm
and θ = 90°
Area of the minor segment ACB = \(\frac{\pi r^2 \theta}{360}\) – Area of ∆AOB
= \(\frac{3.14 \times 10^2 \times 90^{\circ}}{360^{\circ}}\) – \(\frac{1}{2}\) × 10 × 10
= \(\frac{3.14 \times 100}{4}\) – 50
= 785 – 50 = 28.5 cm2
Area of major sector ADB

Haryana Board 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 2

= \(\frac{\pi r^2\left(360^{\circ}-90^{\circ}\right)}{360^{\circ}}\)
= \(\frac{3.14 \times 10^2 \times 270^{\circ}}{360^{\circ}}\)
= 314 × \(\frac{3}{4}\)
= 235.5 cm2
Hence, area of minor segment 285 cm2 and area of major sector = 235.5 cm2.

Haryana Board Solutions for 10th Class Maths Chapter 12 Areas Related to Circles Ex 12.2

Question 5.
In a circle of radius 21 cm, an arc subtends an angle of 600 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.
We have,
Radius (r) = 21 cm
Sector angle (θ) = 60°
(i) Length of the arc = \(\frac{\theta}{360^{\circ}}\) × 2πr

Haryana Board 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 3

= 22 cm
(ii) Area of the sector formed by the arc = \(\frac{\pi r^2 \theta}{360^{\circ}}\)
= \(\frac{22}{7} \times \frac{21^2 \times 60^{\circ}}{360^{\circ}}\)
= \(\frac{22}{7} \times \frac{21 \times 21 \times 60^{\circ}}{360^{\circ}}\)
= \(\frac{22 \times 3 \times 21 \times 1}{6}\)
= 11 × 21 = 231 cm2

(iii) In ∆OAB. OA = OB (equal radii)
⇒ ∠OAB = ∠OBA
Let ∠OAB = ∠OBA = x
∴ x + x + ∠AOB = 180°
⇒ 2x + 60° = 180°
⇒ 2x = 180° – 60° = 120°
⇒ x = \(\frac{120^{\circ}}{2}\) = 60°
Therefore, ∆OAB is an equilateral triangle of side 21 cm.
∴ Area of ∆OAB = \(\frac{\sqrt{3}}{4}\) (side)2
= \(\frac{\sqrt{3}}{4}\) × 212
= \(\frac{441 \sqrt{3}}{4}\) cm2

Area of the segment formed by the corresponding chord = Area of segment ACB.
= Area of the sector OACBO – Area of ∆OAB
= \(\frac{\pi r^2 \theta}{360^{\circ}}-\frac{441 \sqrt{3}}{4}\)
= \(=\frac{22}{7} \times \frac{21^2 \times 60^{\circ}}{360^{\circ}}-\frac{441 \sqrt{3}}{4}\)
= 11 × 21 – \(\frac{441 \sqrt{3}}{4}\)
= (231 – \(\frac{441 \sqrt{3}}{4}\)) cm2
Hence, length of the arc = 22 cm,
area of the sector = 231 cm2
and area of the segment (ACB) = (231 – \(\frac{441 \sqrt{3}}{4}\)) cm2.

Haryana Board Solutions for 10th Class Maths Chapter 12 Areas Related to Circles Ex 12.2

Question 6.
A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the area of the corresponding minor and major segments of the circle. [use π = 3.14 and √3 = 1.73]
Solution:
We heve,
Radius (r) = 15 cm
Sector angle (θ) = 60°
In ∆AOB
⇒ OA = OB (equal radii of the circle)
⇒ ∠OAB = ∠OBA
⇒ ∠OAB + ∠OBA + ∠AOB = 180°
⇒ 2∠OAB + 60°= 180°
⇒ 2∠OAB = 180° – 60° = 120°
⇒ ∠OAB = \(\frac{120^{\circ}}{2}\) = 60°
Therefore, ∆AOB is an equilateral triangle.
Its side = 15 cm
Area of ∆AOB = \(\frac{\sqrt{3}}{4}\) × 152
= \(\frac{225 \sqrt{3}}{4}\) cm2.

Area of minor segment (ACB) = Area of sector – area of ∆AOB

Haryana Board 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 4

= 117.75 – 97.31 = 2O.44 cm2.
Area of major segment = nr2 – Area of minor segment
= 3.14 × 152 – 20.44.
= 706.5 – 20.44 = 686.06 cm2
Hence, area of minor segment and major segment are 20.44 cm2 and 686.06 cm2 respectively.

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.
We have,
Radius (r) = 12 cm
∠AOB = 120°
Area of segment (ACB) = \(\frac{\pi r^2 \theta}{360^{\circ}}-r^2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\)
= \(\frac{3.14 \times 12^2 \times 120}{360}-12^2 \sin \frac{120^{\circ}}{2} \cos \frac{120^{\circ}}{2}\)
= 3.14 × 12 × 4 – 144 sin 60° cos 60°

Haryana Board 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 5

= 3.14 × 48 – 144 × \(\frac{\sqrt{3}}{2} \times \frac{1}{2}\)
= 3.14 × 48 – 36√3
= 150.72 – 36 × 173
= 150.72 – 62.28
= 88.44 cm2.
Hence, area of segment (ACB)= 88.44 cm2.

Haryana Board Solutions for 10th Class Maths Chapter 12 Areas Related to Circles Ex 12.2

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 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 10m long instead of 5 m. (use π = 3.14)

Haryana Board 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 6

Solution:
(i) We have,
Side of the square = 15 m
The area of the field in which the horse can graze represented in the figure by APR, a quadrant of the circle with radius 5 m.
The area which horse can graze = \(\frac{\pi r^2 \theta}{360^{\circ}}\)
= \(\frac{3.14 \times 5^2 \times 90^{\circ}}{360^{\circ}}\)
= \(\frac{3 \cdot 14 \times 25}{4}\)
= 19.63 cm2.

(ii) Increase in area when the rope is 10 m long = Area of quadrant ASQ – Area of quadrant APR
= \(\frac{\pi r^2 \theta}{360^{\circ}}\) – 19.63
= \(\frac{3.14 \times 10^2 \times 90^{\circ}}{360^{\circ}}\) – 19.63
= \(\frac{314}{4}\) – 19.63
= 78.5 – 19.63 = 58.87 cm2.

Haryana Board 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 7

Hence, area which horse can graze = 19.63 cm2
and increase in area = 58.87 cm2.

Question 9.
A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also usal in making 5 centimeters which divide the circle into 10 equal sectors as shown in figure. Find:
(i) The total length of the silver wire required.
(ii) The area of each sector of the brooch.

Haryana Board 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 8

Solution:
(i) We have,
Radius of the circle = \(\frac{35}{2}\) mm.
Circumference of the circle = 2πr
= 2 × \(\frac{22}{7}\) × \(\frac{35}{2}\)
= 110 mm
The total length of wire required = circumference of the circle + length of 5 diameters
= 110 + 5 × 35
= 110 + 175 = 285 mm.

(ii) Area of the circle = πr2
= \(\frac{22}{7} \times \frac{35}{2} \times \frac{35}{2}\)
= \(\frac{11 \times 5 \times 35}{2}\)
= 962.5 mm2

Since circle is divided into 10 equal sectors.
Therefore, area of each sector = \(\frac{\text { Area of the circle }}{10}\)
= \(\frac{962.5}{10}\)
= 96.25 mm2.
Hence, (i) total length of wire required = 285 mm.
(ii) Area of each sector = 96.25 mm2

Haryana Board Solutions for 10th Class Maths Chapter 12 Areas Related to Circles Ex 12.2

Question 10.
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.

Haryana Board 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 9

Solution:
We have,
Radius of circle = 45 cm
Area of the circle = πr2
= π × 452
= 2025π
= \(\frac{2025 \times 22}{7}\)
= 6364.29 cm2
Since the ribs divide the circle into 8 sectors of equal area
Area between the two consecutive ribs = Area of each sector of the circle
= \(\frac{\text { Area of the circle }}{8}\)
= \(\frac{6364 \cdot 29}{8}\)
= 795.54 cm2
Hence, area between the two consecutive ribs of the umbrella = 795.54 cm2

Haryana Board Solutions for 10th Class Maths Chapter 12 Areas Related to Circles Ex 12.2

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.
We have,
One blade of a wiper sweeps a sector area of circle of a radius 25 cm.
The sector angle (θ) = 115°

Haryana Board 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 10

Hence, total area cleaned at each sweep of the blades = \(\frac{158125}{126}\) cm2

Question 12.
To warn ships for underwater rocks, a light house spreads a red coloured 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)
Solution.
We have,
Radius = 16.5 km

Haryana Board 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 11

Sector angle (θ) = 80°
Area of the sea over which the ships are warned = \(\frac{\pi r^2 \theta}{360}\)
= \(\frac{3.14 \times 16.5 \times 16.5 \times 80}{360}\)
= \(\frac{3 \cdot 14 \times 272 \cdot 25 \times 2}{9}\)
= 189.97 km2
Hence, area of the sea over which the ships are warned = 189.97 km2.

Haryana Board Solutions for 10th Class Maths Chapter 12 Areas Related to Circles Ex 12.2

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)

Haryana Board 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 12

Solution:
We have,
r = 28 cm
and θ = \(\frac{360^{\circ}}{6}\) = 60°

Haryana Board 10th Class Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 13

In the figure OA = OB
∠OAB = ∠OBA
∴ ∠OAB + ∠OBA + ∠AOB = 180°
⇒ 2∠OAB + 60° = 180°
⇒ 2 ∠OAB = 180° – 60° = 120°
⇒ ∠OAB = \(\frac{120}{2}\) = 60°
∴ ∠AOB is an equilateral triangle having side 28 cm.
Area of shaded designed portion = Area of the sector OAB – Area of ∆OAB
= \(\frac{\pi r^2 \theta}{360^{\circ}}-\frac{\sqrt{3}}{4} \times 28^2\)
= \(\frac{22}{7} \times \frac{28^2 \times 60^{\circ}}{360^{\circ}}-\frac{\sqrt{3}}{4} \times 28 \times 28\)
= \(\frac{22 \times 4 \times 28 \times 1}{6}\) – √3 × 7 × 28
= \(\frac{44 \times 28}{3}\) – 1.7 × 196
= 410.67 – 333.2 = 77.47 cm2.
The total area of six shaded designed portion = 6 × 77.47 = 464.82 cm2.
Cost of making designs at the rate of 0.35 ₹/cm2 = 464.82 × 0.35 = ₹ 162.68
Hence, cost of the making designs = ₹ 162.68.

Haryana Board Solutions for 10th Class Maths Chapter 12 Areas Related to Circles Ex 12.2

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) \(\frac{p}{180^{\circ}}\) × 2πR
(B) \(\frac{p}{180^{\circ}}\) × πR
(C) \(\frac{p}{360^{\circ}}\) × 2πR
(D) \(\frac{p}{720^{\circ}}\) × 2πR2
Solution:
(D) \(\frac{p}{720^{\circ}}\) × 2πR2

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