Question

. In a circle of radius 7 cm, a chord makes an angle of 60° at the
centre of the circle. Find: (a) area of the circle (b) area of sector
AOB (C) area of minor segment APB​

Answer 1

a)The Area of the circle is 154 sq. cm.

b)The Area of the sector AOB is 25.66 sq. cm.

c)The Area of minor segment APB​ is 4.45 sq. cm.

Step-by-step explanation:

We have drawn the circle for your reference.

Given;

AB is a chord of circle with centre O. And OA and OB are the two radii of the circle.

Length of OA = Length of OB = radius(r) =  7 cm.

Solving for part a.

The Area of the circle =$\pi r^2$

The Area of the circle =$\frac{22}{7}\times7\times7=154\ cm^2$

Hence The Area of the circle is 154 sq. cm.

Solving for part b.

Here the Arc AB forms an angle of 60° at the centre of the circle.

So, Area of the sector = $\frac{\pi r^2\theta }{360}$

On putting the values, we get;

Area of the sector =$\frac{\frac{22}{7}7\times7\times60}{360} =\frac{22\times7\times7\times60}{7\times360}=25.66\ cm^2$

Hence The Area of the sector AOB is 25.66 sq. cm.

Solving for part c.

Area of minor segment APB can be calculated by subtracting area of triangle AOB from area of sector AOB.

For ΔAOB

∠AOB = 60°

AO and OB are the radii, hence ∠OAB = ∠OBA

And by angle sum property,

Sum of all the angles of a triangle is equal to 180°.

So, ∠AOB+∠OAB+∠OBA=180°

$60+2\angle OAB=180\°\\\\2\angle OAB=180-60=120\°\\\\\angle OAB=\frac{120}{2}=60\°$

∠OAB = ∠OBA = ∠AOB = 60°

That means the triangle is equilateral.

So, we use the formula of area of equilateral triangle.

Area = $\frac{\sqrt3}{4}\times a^2$

Where 'a' is the side which is equal to 7 cm.

$Area = \frac{\sqrt3}{4}\times7\times7= 0.433\times49=21.21\ cm^2$

Area of minor segment APB=Area of the sector AOB-Area of ΔAOB

Area of minor segment APB=$25.66-21.21=4.45\ cm^2$

Hence The Area of minor segment APB​ is 4.45 sq. cm.

Answer 2

area of the triangle

area of the sector of angle 60