Areas Related to Circles | Exercise 11.1

Question 7

A chord of a circle of radius 12 cm12\text{ cm} subtends an angle of 120∘120^\circ at the centre. Find the area of the corresponding segment of the circle.

(Use π=3.14\pi = 3.14 and 3=1.73\sqrt{3} = 1.73)

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Solution

We find the area of the segment by subtracting the area of the triangle from the area of the sector.

Step 1 — Find triangle area

Let's draw a perpendicular from the center OO to the chord STST. Let this point be VV. This perpendicular bisects the angle at the center. So, ∠SOV=120∘/2=60∘\angle SOV = 120^\circ / 2 = \mathbf{60^\circ}. In the right triangle △OVS\triangle OVS, we use trigonometry. The radius OSOS is 12 cm\mathbf{12 \text{ cm}}.

OV=OS×cos⁡60∘OV = OS \times \cos 60^\circ

=12×12= 12 \times \frac{1}{2}

=6 cm= 6 \text{ cm}

SV=OS×sin⁡60∘SV = OS \times \sin 60^\circ

=12×32= 12 \times \frac{\sqrt{3}}{2}

=63 cm= 6\sqrt{3} \text{ cm}

The length of the chord STST is twice SVSV.

ST=2×SVST = 2 \times SV

=2×63= 2 \times 6\sqrt{3}

=123 cm= 12\sqrt{3} \text{ cm}

Now, we can find the area of △OST\triangle OST.

Area of △OST=12×base×height\text{Area of } \triangle OST = \frac{1}{2} \times \text{base} \times \text{height}

=12×ST×OV= \frac{1}{2} \times ST \times OV

=12×123×6= \frac{1}{2} \times 12\sqrt{3} \times 6

=363 cm2= 36\sqrt{3} \text{ cm}^2

We use the given value 3=1.73\sqrt{3} = \mathbf{1.73}.

=36×1.73= 36 \times 1.73

62.28 cm2\boxed{62.28 \text{ cm}^2}

Diagram 1

Step 2 — Find sector area

The angle subtended by the chord at the center is 120∘\mathbf{120^\circ}. The radius rr is 12 cm\mathbf{12 \text{ cm}}. We use the formula for the area of a sector.

Area of sector OSUT=θ360∘×πr2\text{Area of sector OSUT} = \frac{\theta}{360^\circ} \times \pi r^2

=120∘360∘×π(12)2= \frac{120^\circ}{360^\circ} \times \pi (12)^2

=13×3.14×144= \frac{1}{3} \times 3.14 \times 144

=3.14×48= 3.14 \times 48

150.72 cm2\boxed{150.72 \text{ cm}^2}

Step 3 — Find segment area

The area of the segment is the difference. We subtract the area of the triangle from the area of the sector.

Area of segment=Area of sector OSUT−Area of △OST\text{Area of segment} = \text{Area of sector OSUT} - \text{Area of } \triangle OST

=150.72−62.28= 150.72 - 62.28

88.44 cm2\boxed{88.44 \text{ cm}^2}

Answer

The area of the corresponding segment of the circle is 88.44 cm2\mathbf{88.44 \text{ cm}^2}.

More questions in Exercise 11.1

Q1

Unless stated otherwise, use π=227\pi = \frac{22}{7}.

Find the area of a sector of a circle with radius 6 cm6\text{ cm} if angle of the sector is 60∘60^\circ.

Q2

Find the area of a quadrant of a circle whose circumference is 22 cm22\text{ cm}.

Q3

The length of the minute hand of a clock is 14 cm14\text{ cm}. Find the area swept by the minute hand in 55 minutes.

Q4

A chord of a circle of radius 10 cm10\text{ cm} subtends a right angle at the centre. Find the area of the corresponding :

(i) minor segment

(ii) major sector. (Use π=3.14\pi = 3.14)

Q5

In a circle of radius 21 cm21\text{ cm}, an arc subtends an angle of 60∘60^\circ 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

Q6

A chord of a circle of radius 15 cm15\text{ cm} subtends an angle of 60∘60^\circ at the centre. Find the areas of the corresponding minor and major segments of the circle.

(Use π=3.14\pi = 3.14 and 3=1.73\sqrt{3} = 1.73)

Q7

A chord of a circle of radius 12 cm12\text{ cm} subtends an angle of 120∘120^\circ at the centre. Find the area of the corresponding segment of the circle.

(Use π=3.14\pi = 3.14 and 3=1.73\sqrt{3} = 1.73)

Q8

A horse is tied to a peg at one corner of a square shaped grass field of side 15 m15\text{ m} by means of a 5 m5\text{ m} long rope (see Fig. 11.8). 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 m10\text{ m} long instead of 5 m5\text{ m}. (Use π=3.14\pi = 3.14)

Q9
  1. 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. 11.9. Find :

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

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

Q10
  1. An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
Q11
  1. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115∘115^\circ. Find the total area cleaned at each sweep of the blades.
Q12
  1. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80∘80^\circ to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π=3.14\pi = 3.14)
Q13
  1. A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm2\text{cm}^2. (Use 3=1.7\sqrt{3} = 1.7)
Q14
  1. Tick the correct answer in the following :

Area of a sector of angle pp (in degrees) of a circle with radius R is

(A) p180×2πR\frac{p}{180} \times 2\pi\text{R} (B) p180×πR2\frac{p}{180} \times \pi\text{R}^2 (C) p360×2πR\frac{p}{360} \times 2\pi\text{R} (D) p720×2πR2\frac{p}{720} \times 2\pi\text{R}^2

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