Question

A chord of a circle of radius 14 cm makes a right angle at the centre. Find the areas of the minor and major segments of the circle.

Answer 1

Answer:

The area of the minor segment is 56 cm² and  Area of major segment is 560 cm².

Step-by-step explanation:

Given :  

Given :

Radius of a circle,r = 14 cm  

Angle at the centre of a circle, θ = 90°

Area of the minor segment ,A = {πθ/360 - sin θ /2 cos θ/2 }r²

A = {90°π/360° - sin 90°/2 cos 90°/2 }× 14²

A = {π/4  - sin 45°cos 45°} × 196

A = {π/4  - 1/√2 × 1/√2} × 196

A = {π/4  - 1/2 } × 196

A = {196π/4 - 196/2}

A = {196 × 22/7× 1/4 - 98}

A = 154 - 98

A = 56 cm²

Area of the minor segment = 56 cm²

Area of circle = πr²  

= 22/7 × 14²

= 22/7 × 14 × 14  

= 22 × 2 × 14  

Area of circle = 616 cm²

Area of major segment = Area of circle - Area minor segment

Area of major segment = 616 - 56 = 560 cm²

Hence, the area of the minor segment is 56 cm² and  Area of major segment is 560 cm².

HOPE THIS ANSWER WILL HELP YOU….

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Answer 2

Hey mate..

Answer :

Radius of the circle = 14cm

Angle subtend at center = 90°

By Pythagoras theorem = AB2 = OA2 + OB2

= 14^2 +14^2

$ab = 14 \sqrt{2}$

Area of sector OAB =

$\frac{90}{360} \ \times \pi {r}^{2} \\ \\ \frac{1}{4 } \pi {r}^{2} \\ \\ = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = 154c {m}^{2}$

Area of triangle AOB =

$\frac{1}{2} \times 14 \times 14 = 98c {m}^{2}$

So area of minor segment – OACB =area of sector – area of triangle

= 154 – 98 = 56cm2

Area of major segment = area of circle - area of minor segment

$\frac{22}{7} \times 14 \times 14 - 56$

= 44×14 – 56 = 560cm2