A chord of a circle of radius 12 cm subtends an angle of 120° at the center. Find the area of the corresponding segment of the circle. (Use π = 3.14 and $$\sqrt{3} = 1.73$$)

Asked by Pragya Singh | 1 year ago |  67

##### Solution :-

Now, draw a perpendicular OD on chord AB and it will bisect chord AB.

Now, the area of the minor sector = ($$\dfrac{θ}{360°}$$)×πr2

= ($$\dfrac{120}{360}$$)×($$\dfrac{22}{7}$$)×122

= 150.72 cm2

Consider the ΔAOB,

∠ OAB = 180°-(90°+60°) = 30°

Now, cos 30° = $$\dfrac{AD}{OA}$$

$$\dfrac{\sqrt{3}}{2}$$ = $$\dfrac{AD}{12}$$

Or, AD =$$6 \sqrt{3}$$ cm

We know OD bisects AB. So,

AB = 2×AD = $$12 \sqrt{3}$$ cm

Now, sin 30° = $$\dfrac{OD}{OA}$$

Or, $$\dfrac{1}{2}=$$ $$\dfrac{AD}{12}$$

OD = 6 cm

So, the area of ΔAOB = $$\dfrac{1}{2}$$ × base × height

Here, base = AB = $$12 \sqrt{3}$$ and

Height = OD = 6

So, area of ΔAOB = $$\dfrac{1}{2}$$×$$12 \sqrt{3}$$×6 = $$36 \sqrt{3}$$ cm = 62.28 cm2

Area of the corresponding Minor segment = Area of the Minor sector – Area of ΔAOB

= 150.72 cm2– 62.28 cm= 88.44 cm2

Answered by Abhisek | 1 year ago

### Related Questions

#### Prove that the area of a circular path of uniform width hsurrounding a circular region of radius r is πh(2r + h).

Prove that the area of a circular path of uniform width hsurrounding a circular region of radius r is πh(2r + h).

#### A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the area of the road.

A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the area of the road.

#### The outer circumference of a circular race-track is 528 m. The track is every­where 14 m wide.

The outer circumference of a circular race-track is 528 m. The track is every­where 14 m wide. Calculate the cost of levelling the track at the rate of 50 paise per square metre