Jharkhand Board JAC Class 10 Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2 Textbook Exercise Questions and Answers.

## JAC Board 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 radius 6 cm if angle of the sector is 60°.

Solution :

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

r = 6, θ = 60°

= \(\frac{22}{7} \times \frac{6 \times 6 \times 60}{360}=\frac{132}{7}\) = 18.85 cm².

Question 2.

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

Solution :

С = 2πr = 22

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 the minute hand = 14 cm = r.

In 15 minutes the minute hand sweeps an area equal to a quadrant.

Area of the quadrant = \(\frac{\pi r^2}{4}\)

Alternative Method:

One minute = 6°

5 minutes = 30°

Area swept by the minute hand

= πr² \(\frac{θ}{360°}\)

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

= 51.33 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)

Solution :

Radius of the circle = 10 cm

Major segment is making 360° – 90° = 270°

Area of the sector making angle 270° = \(\frac{270°}{360°}\) × πr² cm²

= \(\frac{1}{4}\) × 10²π = 25 π cm²

= 25 × 3.14 cm² = 235.5 cm²

∴ Area of the major segment = 235.5 cm²

Height of ΔAOB = OA = 10 cm

Base of ΔAOB = OB = 10 cm

Area of ΔAOB = \(\frac{1}{2}\) × OA × OB

= \(\frac{1}{2}\) × 10 × 10 = 50 cm²

Major segment is making 90°

Area of the sector making angle 90° = \(\frac{90°}{360°}\) × πr² cm²

= \(\frac{1}{4}\) × 10²

= 25 × 3.14 cm² = 78.5 cm²

Area of the minor segment = Area of the sector making angle 90° – Area of ΔAOB

= 78.5 cm² – 50 cm² = 28.5 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) Length of the arc AB = \(\frac{2 \pi r \theta}{360^{\circ}}\) θ = 60°, r = 21

= 2 × \(\frac{22}{7} \times \frac{21 \times 60}{360}\) = 22cm

(ii) Area of the sector formed by the arc = \(\frac{\pi r^2 \theta}{360^{\circ}}\)

= \(\frac{22}{7} \times \frac{21 \times 21 \times 60}{360^{\circ}}\)

= 11 × 21 = 231 cm².

(iii) Area of the segment formed APBLA = Area of the sector PAOB – Area of the ΔOAB

Area of ΔOAB = ?

From O, draw OL ⊥ AB. AL = ?, OL = ?

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 \(\sqrt{3}\) = 1.73).

Solution :

Area of the minor segment APB = Area of the sector OAPB – Area of ΔOAB

= \(\frac{\pi r^2 \theta}{360^{\circ}}\) – Area of ΔOAB

Area of the sector OAPB = \(\frac{3.14 \times 15 \times 15 \times 60}{360}\) = 1.57 × 75 cm²

OPB is an isosceles triangle. OA = OB ∴ \(\hat{A}\) = \(\hat{B}\) = 60°.

The triangle becomes an equilateral triangle.

Area of the sector OAPB = 1.57 × 75 = 117.75 cm²

∴ Area of the minor segment APB = 117.75 – 97.31 = 20.44 sq.cm.

Area of the major segment AQBA = Area of the circle – Area of the minor segment

= πr² – 20.44

= (3.14 × 15 × 15) – 20.44

= 706.50 – 20.44 = 686.06 sq.cm.

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 \(\sqrt{3}\) = 1.73).

Solution :

Area of the segment APB = Area of the sector PAOB – Area of ΔOAB

r = 12, θ = 120°, π = 3.14

∴ Area of the sector PAOB = \(\frac{\pi r^2 \theta}{360^{\circ}}\)

= \(\frac{3.14 \times 12 \times 12 \times 120}{360}\)

= 3.14 × 12 × 4

= 3.14 × 48

= 150.72 cm².

Area of ΔOAB = \(\frac{1}{2}\) × b × h b = AB, h = OC

Draw OC ⊥ AB.

AOB is an isosceles triangle. OC ⊥ AB.

ΔAOC ≅ BOC (RHS)

∴ AC = CB.

A\(\hat{O}\)C = 60°, O\(\hat{A}\)C = 30°, A\(\hat{C}\)O = 90°

In ΔOAC, O\(\hat{C}\)A = 90°. sin O\(\hat{A}\)C = \(\frac{OC}{OA}\)

sin 30° = \(\frac{OC}{12}\) sin 30° = \(\frac{1}{2}\)

\(\frac{1}{2}\) = \(\frac{OC}{12}\)

2OC = 12

OC = \(\frac{12}{2}\) = 6 cm.

In ΔOAC, O\(\hat{C}\)A = 90°

OC² + AC² = OA²

6² + AC² = 12²

AC² = 12² – 6²

= (12 + 6) (12 – 6)

= 18 × 6 = 108

AC = \(\sqrt{108}\) = \(\sqrt{36 \times 3}\) = 6\(\sqrt{3}\)

AC = CB = 6\(\sqrt{3}\)

∴ AB = 12\(\sqrt{3}\)

Area of ΔOAB = \(\frac{1}{2}\) × b × h

= \(\frac{1}{2}\) × AB × OC

= \(\frac{1}{2}\) × 12\(\sqrt{3}\) × 6 = 36\(\sqrt{3}\)

Area of the segment APB = Area of the sector PAOB – Area of triangle OAB

= 150.72 – 62.28

= 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. 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) When the rope is 5m long, area grazed = \(\frac{\pi r^2}{4}\) (quadrant)

= \(\frac{3.14 \times 5 \times 5}{4}=\frac{78.5}{4}\) = 19.625 m².

(ii) When the rope is 10 m long, area grazed = \(\frac{\pi r^2}{4}\)

= \(\frac{3.14 \times 10 \times 10}{4}=\frac{3.14 \times 100}{4}=\frac{314}{4}\)

= 78.5 cm².

Increased area available when the rope is 10 m long is

= 78.500 – 19.625 = 58.875 cm².

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 the figure. Find:

(i) the total length of the silver wire required

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

Solution :

A\(\hat{O}\)B = 180°

It is divided into five equal parts,

∴ Each part = θ = \(\frac{180°}{5}\) = 36°.

Total length of silver wire used = πd × 5 × 35

= \(\frac{22}{7}\) × 35 + 175

= 110 + 175

= 285 mm².

Area of each 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.

Question 10.

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

Solution :

Number of ribs in umbrella = 8

Radius of umbrella while flat = 45 cm.

Angle between two consecutive ribs of the umbrella = \(\frac{360}{8}\) = 45°

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 :

Given, length of wiper blade = 25 cm = r (say)

Angle made by the blade, θ = 115°

Question 12.

To warn ships for underwater rocks, a lighthouse 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 :

π = 3.14, r = 16.5, θ = 80°

Area warned = \(\frac{\theta \pi r^2}{360^{\circ}}\)

= \(\frac{3.14 \times 16.5 \times 16.5 \times 80}{360}\)

= 3.14 × 5.5 × 11

= 3.14 × 60.5 = 189.970

= 189.97 sq.kms.

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 Re. 0.35 per cm². (Use \(\sqrt{3}\) = 1.7).

Solution :

[Draw a circle with centre O using a convenient radius. Draw a diameter AOD. Keep the protractor on AD and make three equal angles of 60° each such that A\(\hat{O}\)B = B\(\hat{O}\)C = C\(\hat{O}\)D. Produce BO and CO to meet the circumference at E and F respectively. Join AB, BC, CD, DE, EF, EA. We get a regular hexagon and six segments which are shaded.

Let one segment be APB.

Find the area of one segment. Multiply it by 6. We get the area of the six segments formed. Find the cost of making these designs as follows:

Area of one design × 6 × Rate]

The design APB is nothing but a segment. We have to find the area of this segment using the following relation:

Area of the sector OAPB – Area of equilateral ΔAOB

Cost of making these six designs = Area × Rate

= 464.8 × 0.35

= Rs. 162.68.

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) \(\frac{P}{180}\) × 2πR²

(B) \(\frac{P}{180}\) × πR²

(C) \(\frac{P}{360}\) × 2πR

(D) \(\frac{P}{720}\) × 2πR²

Solution :

Area of sector = \(\frac{\theta \pi r^2}{360^{\circ}}\) (Given, θ = p, r = R)

= \(\frac{p \pi R^2}{360}=\frac{p 2 \pi R^2}{720}\)