RBSE Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.2

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.2 Textbook Exercise Questions and Answers.

RBSE Class 10 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 a sector of a circle with a radius of 6 cm if the angle of the sector is 60°.
Solution:
According to the question,

The radius of the sector of the circle (R) = 6 cm.
Central angle (θ) = 60°
∵ Area of the sector = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)
= \(\frac{22}{7} \times \frac{6 \times 6 \times 60^{\circ}}{360^{\circ}}\)
= \(\frac{132}{7}\)
= 18.857 cm²
Hence area of the sector = \(\frac{132}{7}\) or 18.86 cm² (approximately)

Question 2.
Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
According to the question,
Circumference of the circle = 22 cm.
∴ 2πR = 22
⇒ R = \(\frac{22 \times 7}{2 \times 22}=\frac{7}{2}\) cm.
⇒ R = \(\frac{7}{2}\) cm.
Central angle (quadrant) (θ) = 90°
∴ Area of the quadrant = \(\frac{\pi \mathrm{R}^{2} \theta}{360}\)
= \(\frac{\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 90^{\circ}}{360^{\circ}}\)
= \(\frac{77}{8}\) 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:
According to the question,
length of the minute hand = radius of circle (R) = 14 cm.
∵ We know that
60' = 360°
⇒ 1' = \(\frac{360}{60}\) = 6°
⇒ 5' = 6° × 5 = 30°
Angle of sector (θ) = 30°
Hence area swept by the minute hand in 5 minutes.

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:

According to the question,
radius of circle (R) = 10 cm.
Central angle (θ) = 90°
∴ Area of minor sector = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)
= 3.14 × 10 × 10 × \(\frac{90^{\circ}}{360^{\circ}}\)
= 78.5 cm²
(i) Area of minor segment = Area of minor sector - Area of ΔAOB
= 78.5 - \(\frac{1}{2}\) × Base × Height
= (78.5 - \(\frac{1}{2}\) × 10 × 10) cm²
= (78.5 - 50) cm²
= 28.5 cm²
∴ Area of minor segment = 28.5 cm²
(ii) Area of corresponding major sector

∴ Area of major sector = 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.
Solution:
(i) According to the question,
Radius of circle (R) = 21 cm.
length of the arc (θ) = \(\frac{\theta}{360^{\circ}} \times 2 \pi \mathrm{R}\)
= \(\frac{60^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 21\)
∴ length of arc = 22 cm.
(ii) Area of the sector formed by the arc

(iii) ∵ ∆OAB is an equilateral triangle in which θ = 60°
∴ Area of the segment of circle = Area of sector - Area of ∆AOB

= 231 - 110.25 × 1.732
= 231 - 190.95 approx.
= 40.05 cm²
Hence area of segment 40.05 or (231 - \(\frac{441}{4} \sqrt{3}\)) 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)
Solution:
According to the question,

Radius of circle (R) = 15 cm.
Central angle (θ) = 60°
In ∆OAB,
Central angle θ = 60°
OA = OB = 15 cm
∴ ∠A = ∠B = 60°
∴ ∆OAB is an equilateral triangle.
Area of minor segment = Area of the minor sector - Area of an equilateral triangle
= \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}-\frac{\sqrt{3}}{4}(\text { side })^{2}\)

Area of major segment = Area of circle - Area of minor segment
= πR² - 20.43 
= 3.14 × 15 × 15 - 20.43
= 706.5 - 20.43
= 686.07 cm²
Area of major segment = 686.07 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)
Solution:
According to the question,
Radius of circle (R) = 12 cm.
Central angle (θ) = 120°
In ∆OAM, OM ⊥ AB
∴ AM = MB = \(\frac{1}{2}\) AB

∴ ∠OAM = 30° = ∠OBM
AB = 2AM

Area of segment = Area of sector - Area of of ∆OAB
Area of segment = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}-\frac{1}{2} \mathrm{AB} \times \mathrm{OM}\)

= 150.72 - 36 × 1.73
= (150.72 - 62.28) cm²
= 88.44 cm²
∴ Area of segment = 88.44 cm²

Question 8.
A horse is tied to a peg at one comer 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 π = 3.14)

Solution:
According to the question,
Side of square = 15 m
(i) Length of the rope of the peg = Radius of sector (R) = 5 m
Central angle (θ) = 90°

(ii) When the radius of the sector becomes 10 m.
The radius of the sector (R₁) = 10 m.
Central angle (θ) = 90°
∴ Area of sector = \(\frac{\pi \mathrm{R}_{1}^{2} \theta}{360^{\circ}}\)

∴ Increase in the grazing area = Area of larger sector - Area of smaller sector
= 78.5 - 19.625
= 58.875 m²
∴ Increase in the grazing area = 58.875 m²

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 Figure. Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.

Solution:
According to the question,
Diameter of circle (D) = 35 mm
Radius of circle (R) = \(\frac{35}{2}\) mm
Number of diameters = 5
Number of equal sectors = 10
(i) The total length of the silver wire required = Length of 5 diameters + Perimeter of circle (brooch)
= 5(35) + 2πR
= 175 + 2 × \(\frac{22}{7} \times \frac{35}{2}\)
= 175 + 110
= 285 mm
(ii) Angle of sector of brooch = \(\frac {360^{\circ}}{Number of sectors}\)
= \(\frac{360^{\circ}}{10}\)
= 36°
Area of each brooch (Area of the sector)

Hence area of each sector of the brooch = 96.25 mm²

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:
According to the question,
Radius of the circle is 45 cm.
Number of ribs = 8
Central angle (angle of sector) = \(\frac{360^{\circ}}{8}\) = 45°

Hence the area between the two consecutive ribs = \(\frac{22275}{28}\) cm²

Question 11.
A car has two wipers that do not overlap. Each wiper has a blade of the length of 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
Solution:
According to the question,
Length of blade (R) = 25 cm
The angle of the sector (θ) = 115°
The wiper revolves in the form of a sector.
Area of sector = Area swept by one blade

Question 12.
To warn ships of underwater rocks, a lighthouse spreads a red coloured light over a sector of an angle of 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warmed. (Use π = 3.14)
Solution:

Angle of sector (θ) = 80°
Radius of sector (R) = 16.5 km
Area of the sea over which the ships are warned,
i.e., Area of sector = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)
= \(\frac{3.14 \times 16.5 \times 16.5 \times 80^{\circ}}{360^{\circ}}\)
= 189.97 km²
Area of that portion of the sea over which the ships are warned = 189.97 km²

Question 13.
A round table cover has six equal designs as shown in the 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).

Solution:
According to the question,
Number of equal designs = 6
Radius of a design (R) = 28 cm
Each design is of the shape of a sector
Central angle (θ) = \(\frac{360^{\circ}}{6}\) = 60°
Since central angle is 60° and OA = OB,
therefore ∆OAB is an equilateral triangle whose side is 28 cm.

Area of the portion of a shaded design = Area of segment
= Area of sector OAB - Area of ∆OAB

= (410.66 - 333.2) cm²
= 77.46 cm²
Area of the portion of a shaded design = 77.46 cm²
Area of the portion of six designs = 6 [Area of the portion of a design]
= 6[77.46]
= 464.76 cm²
∵ Cost of making 1 cm² design = Rs. 0.35
∴ Cost of making 464.76 cm² design = Rs. 0.35 × 464.76 = Rs. 162.68
Hence the cost of making the designs = Rs. 162.68

Question 14.
Tick the correct answer in the following:
The area of a sector of angle p (in degrees) of a circle with radius R is-
(A) \(\frac{p}{180} \times 2 \pi \mathrm{R}\)
(B) \(\frac{p}{180} \times \pi \mathrm{R}^{2}\)
(C) \(\frac{p}{360} \times 2 \pi \mathrm{R}\)
(D) \(\frac{p}{720} \times 2 \pi \mathrm{R}^{2}\)
Solution:
Angle of sector (θ) = p°
Radius of circle = R
Area of sector = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)
= \(\frac{\pi \mathrm{R}^{2} \times p^{\circ}}{360^{\circ}}\)
= \(\frac{2 \pi \mathrm{R}^{2} \times p^{\circ}}{720^{\circ}}\)
∴ Correct option = (D)