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

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

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

Question 1.
Find the area of sector of a circle with radius 6 cm, if angle of the sector is 60°.
Solution:
Radius of sector of circle (R) = 6 cm

Central angle (θ) = 60°
Area of sector = \(\frac{\pi \mathrm{R}^{2} \theta}{360}\)

= \(\frac{22}{7} \times \frac{6 \times 6 \times 60}{360}\)

= \(\frac{132}{7}\) cm²
∴ Area of sector = 18.86 cm².

Question 2.
Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
Circumference of circle = 22 cm
2πR = 22
R = \(\frac{22 \times 7}{2 \times 22}=\frac{7}{2}\) cm
R = \([latex]\frac{7}{2}\)[/latex] cm
Central angle [quadrant] (θ) = 90°
∴ Area of a quadrant = \(\frac{\pi \mathrm{R}^{2} \theta}{360}=\frac{\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 90}{360}\)

= \(\frac{77}{8}\)

Area of quadrant = 9.625 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:
Length of minute hand of clock = Radius of circle (R) = 14 cm

We know that
60’ = 360°
1’ = \(\frac{360^{\circ}}{60}\) = 6°
5′ = 60 × 5 = 30°
Angle of sector (θ) = 30°
∴ Area swept by minute hand in 5 minutes = \(\frac{\pi R^{2} \theta}{360}\)

= \(\frac{22}{7} \times 14 \times 14 \times \frac{30}{360}\)

= \(\frac{1}{12}\) × 22 × 28 = \(\frac{154}{3}\) cm².

Hence, area swept by minute hand in 5 minutes is 5133 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.
Solution:
Radius of circle (R) = 10 cm
Central angle (θ) = 90°

(i) Area of minor sector = \(\frac{\pi R^{2} \theta}{360}\)

= 3.14 × 10 × 10 × \(\frac{90}{360}\)

Area of minor sector = \(\frac{314}{4}\) = 78.5 cm²
Area of minor segment = Area of minor sector – Area of MOB
= 78.5 – \(\frac{1}{2}\) Base × Altitude
= 78.5 – \(\frac{1}{2}\) × 10 × 10
= 78.5 – 50 = 28.5
∴ Area of minor segment = 28.5 cm²

(ii) Area of major sector = \(\frac{(360-\theta)}{360}\)
= \(\frac{(360-90)}{360}\) × 3.14 × 10 × 10
= \(\frac{270}{360}\) × 3.14 × 100
= \(\frac{3 \times 314}{4}=\frac{3 \times 157}{2}\)
∴ 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:
(j) 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 of circle (R) 21 cm

Central angle(θ) = 60°
Length of arc = \(\frac{\theta}{360}\) × 2πR
= \(\frac{60}{360} \times 2 \times \frac{22}{7} \times 21\) = 22 cm
Hence, length of arc = 22 cm.

(ii) Area of sector formed by arc = \(\)
= \(\frac{22}{7} \times \frac{21 \times 21 \times 60}{360}\)
Area of sector = 231 cm²
∴ Since ∆OAB is equilateral triangle, θ = 60°.

(iii) Area of segment = Area of sector – Area of ∆AOB
= \(\frac{\pi R^{2} \theta}{360}-\frac{\sqrt{3}}{4}(\text { side })^{2}\)
= 231 – \(\frac{1.73}{4}\) × 21 × 21
= 231 – 0.4325 × 441
= 231 – 190.7325
Area of segment = 40.26 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:
Radius of circle = (R) = 15 cm
Central angle (θ) = 60°

In ∆OAB, central angle θ = 60°
OA = OB = 15 cm
∴ ∠A = ∠B
Now, ∠A + ∠B + ∠O = 180°
2∠A ÷ 600 = 180°
∠A = 60°
∴ ∠A = ∠B = 60
∴ ∠OAB is equilateral.
[ Area of minor segment] = [Area of minor sector] – [Area of equilateral triangle]

= \(\frac{\pi R^{2} \theta}{360}-\frac{\sqrt{3}}{4}(\text { side })^{2}\)

= 3.14 × \(\frac{15 \times 15 \times 60}{360}-\frac{1.73}{4} \times(15)^{2}\)

= 15 × 15 \(\left[\frac{3.14 \times 60}{360}-\frac{1.73}{4}\right]\)

= \(\frac{225}{100}\left[\frac{314}{6}-\frac{173}{4}\right]\)

= \(\frac{225}{100}\) [52.33 – 43.25]

= \(\frac{225}{100}\) × 9.08

= \(\frac{2043}{100}\) = 20.43 cm²

Area of minor segment = 20.43 cm².
Area of major segment = Area of circle – Area of major 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 it π = 3.14 and √3 = 1.73).
Solution:
Radius of circle (R) = 12 cm
Central angle (θ) = 120°

In ∆OAM, from O draw, angle bisector of ∠AOB as well as perpendicular bisector OM of AB.
∴ AM = MB = \(\frac{1}{2}\) AB
In ∆OMA,
∠AOM + ∠OMA + ∠OAM = 180°
LOAM = 30°
Similarly, ∠OAM = 30° = ∠OBM

AB = 2AM
= 2 \(\left(\frac{\mathrm{AM}}{\mathrm{OA}}\right)\) OA
= 2 (sin 60°) 12
AB = 2 × \(\frac{\sqrt{3}}{2}\) × 12 = 12√3 cm
OM = OA \(\left(\frac{O M}{O A}\right)\) = 12 (cos 60°)
= 12 × \(\frac{1}{2}\) = 6 cm
Area of the segment = Area of sector – Area of ∆OAB.
Area of segment = \(\frac{\pi R^{2} \theta}{360^{\circ}}-\frac{1}{2} A B \times O M\)

= \(\frac{3.14 \times 12 \times 12 \times 120}{360}\) – \(\frac{1}{2}\) × 12√3 × 6

= \(\frac{314}{100} \times \frac{144 \times 120}{360}\) – 36√3
= 150.72 – 36 × 1.73
= (150.72 – 62.28) cm²
= 88.44 cm²
∴ Area of the segment = 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.). 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:
Side of square = 15 m
(i) Length of Peg = Radius of rope (R) = 5 m
Centrral angle (θ) = 90° [Each angle of square]

Area of sector = \(\frac{\pi R^{2} \theta}{360}\)

= \(\frac{3.14 \times 5 \times 5 \times 90}{360}\)

Area of sector = \(\frac{3.14 \times 25}{4}=\frac{78.5}{4}\) m².
= 19.625 m²

(ii) When radius of sector increased to 10 m
Radius of sector OCD (R₁) = 10 m
Central angle (θ) = 90°
∴ Area of sector OCD = \(\frac{\pi R_{1}^{2} \theta}{360^{\circ}}\)

= \(\frac{3.14 \times 10 \times 10 \times 90}{360^{\circ}}\)

= \(\frac{314}{100} \times \frac{100 \times 90}{360^{\circ}}\)

= \(\frac{314}{4}\) = 78.5 m

∴ Increase in grazing area = Area of sector OCD – Area of sector OAB
= 78.5 – 19.625 = 58.875 m²
Increase of grazing area = 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. Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.

Solution:
Diameter of circle (D) = 35 mm
Radius of circle (R) = \(\frac{35}{2}\) mm
Number of diameter = 5
Number of equal sector = 10

(i) Wire used = Length of 5 diameters + circumference of circle (brooch) = 5(35) + 2πR
= 175 + 2 × \(\frac{22}{7}\) × \(\frac{35}{2}\)
= 175 + 110 = 285 mm

(ii) Sector angle of brooch = \(\frac{360^{\circ}}{\text { Number of sector }}\)
= \(\frac{360^{\circ}}{10}\) = 36°
Area of each brooch (Area of sector) = \(\frac{\pi R^{2} \theta}{360^{\circ}}\)

= \(\frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} \times \frac{36}{360}\)

= \(\frac{11 \times 35}{4}=\frac{385}{4}\) mm² = 96.25 mm²

∴ Area of each brooch = 96.25 m².

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

Solution:
Radius of circle = 45 cm
Number of ribs = 8
Central angle (sector angle) = \(\frac{360^{\circ}}{8}\) = 45°
Area of sector = \(\frac{\pi R^{2} \theta}{360^{\circ}}\)

= \(\frac{45}{360^{\circ}} \times \frac{22}{7} \times 45 \times 45\)

= \(\frac{1}{8} \times \frac{22}{7} \times 45 \times 45\)

= \(\frac{22275}{28}\) cm²

Area of sector = 795.53 cm²
∴ Area between two consecutive ribs of the umbrella = 795.53 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.
Solution:
Length of blade (R) = 25 cm
Sector angle (θ) = 115°
Wiper moves in form of sector.

Area of sector = Area çovered by one blade
= \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)

= \(\frac{22}{7} \times \frac{115}{360} \times 25 \times 25=\frac{316250}{504}\)

= 627.48 cm²
Area covered by two blades of wiper = 2 Area of sector
= 2 × 627.48 = 1254.96 cm².

Question 12.
To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 800 to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use x = 3.14)
Solution:
Sector angle (θ) = 80°
Radius of sector (R) = 16.5 km

Area of sea over which the ships are warned
Area of sector = \(\frac{\pi R^{2} \theta}{360^{\circ}}\)

= \(\frac{3.14 \times 16.5 \times 16.5 \times 80}{360}\)
= 189.97 km²
Area of sea over which the ships are warned = 189.97 km².

Question 13.
A round table cover has six equal designs as shown in fig. 1f 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:
Number of equal designs = 6
Radius of designs (R) = 28 cm
Each design is in the shape of sector central angle (θ) = 6 = 60°

Since central angle is 60° and OA = OB
∴ ∆OAB is equilateral triangle having side 28 cm.

Area of one shaded designed portion = area of segment = Area of the sector OAB – area of ∆OAB
= \(\frac{\pi R^{2} \theta}{360^{\circ}}-\frac{\sqrt{3}}{4}(\text { side })^{2}\)

= \(\frac{22}{7} \times \frac{28 \times 28}{360^{\circ}} \times 60^{\circ}-\frac{1.7}{4} \times 28 \times 28\)
= 410.66 – 333.2 = 77.46.

Area of one shaded designed portion = 77.46
Area of six designed portions = 6 [Area of one designed]
= 6 [77.46] = 464.76 cm²
Cost of making 1 cm² = ₹ 0.35
Cost of making of 464.76 cm² = 464.76 × 0.35 = ₹ 162.68.

Question 14.
Tick the correct answer in the following:
Area of a sector of angle p° of a circle with radius R is
(A) \(\frac{p}{180}\) × 2πR
(B) \(\frac{p}{180}\) × πR²
(C) \(\frac{p}{360}\) × 2πR
(D) \(\frac{p}{720}\) × 2πR²
Solution:
Angle of sector (θ) = p°
Radius of circle = R
Area of sector = \(\frac{\pi \mathrm{R}^{2} \theta}{360}\)

= \(\frac{\pi \mathrm{R}^{2} \times p^{\circ}}{360}\)

= \(\frac{p^{\circ}}{720}\) × 2πR²
∴ Correct option is (D).