Question

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.

Answer 1

In the mentioned figure,

O is the centre of circle,

AB is a chord

AXB is a major arc,

OA=OB= radius = 15 cm

Arc AXB subtends an angle 60

o

at O.

Area of sector AOB=

360

60

×π×r

2

=

360

60

×3.14×(15)

2

=117.75cm

2

Area of minor segment (Area of Shaded region) = Area of sector AOB− Area of △ AOB

By trigonometry,

AC=15sin30

OC=15cos30

And, AB=2AC

∴ AB=2×15sin30=15 cm

∴ OC=15cos30=15 √3/2 = 15 × 1.73/2 = 12.975cm

∴ Area of △AOB=0.5×15×12.975=97.3125cm ^2

∴ Area of minor segment (Area of Shaded region) =117.75−97.3125=20.4375 cm^2

Area of major segment = Area of circle − Area of minor segment

=(3.14×15×15)−20.4375

=686.0625cm^2

Answer 2

Required answer:–

✠Given:–

• Radius of a chord of a circle = 15cm

• Measure of an angle = 60⁰

✠To find:–

• Areas of the corresponding minor and major segments of circle

✠As we know:–

• Area of circle = πr²

where,

π = 3.14

r = radius

• Area of minor segment = Area of sector OAB - Area of triangle OAB.

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

✠ Step by step explaination:–

First we have to calculate area of minor segment..

That is,

⟹$\blue{ \dfrac{<}{360 {}^{0} } \: πr {}^{2} } - \blue{ \dfrac{ \sqrt{3} }{4} r {}^{2} }$

Evaluating values...

⟹ $\blue{ \dfrac{60}{360} \times 3.14 \times 15 {}^{2}} - \blue{ \dfrac{1.73}{4} \times 15 {}^{2} }$

⟹ $\blue{117.75 - 97.3125}$

⟹ $\blue{20.4375cm {}^{2}}$

Now calculating the area of major segment

That is,

⟹ $\blue{πr {}^{2} - 20.4375cm {}^{2} }$

Evaluating values...

⟹ $\blue{3.14 × 15 {}^{2} - 20.4375 }$

⟹ $\blue{686.0625 \: cm {}^{2} }$

Hence

Area of major segment is 686.0625 cm²

Area of minor segment is 20.4375 cm²