Question

A chord of a circle of radius 5cm subtends an angle of 45 degree at the centre.find the area of minor segments of circle​

Answer 1

Answer:

From the attachment....

O is the centre of circle

AB is a chord in circle

AYB is a major arc

Radius = OA = OB = 5cm

AYB subtends an angle 45° at O

Now we have to find the area of sector:-

${ \boxed{area \: of \: sector = \frac{θ}{360} \pi {r}^{2} }}$

${ \boxed{\pi = 3.14}}$

${ \sf{area \: of \: sector \: aob = \frac{45}{360} \times 3.14 \times {5}^{2} }}$

${ \to{ \frac{1}{8} \times 78.5 }}$

${ \to{9.81\: cm}}$

Area of minor segment = Area of sector AOB− Area of △ AOB

From trigonometry:-

AC = 5 sin30°

OC = 5 Cos 30°

AB = 2AC

AB = 2× 5 Sin 30

${ \to{2 \times 5 \times \frac{1}{2} }}$

${ \to{5cm}}$

OC = 5cos 30°

${ \to{5 \times \frac{ \sqrt{3} }{2} }}$

${ \boxed{ \sqrt{3} = 1.73cm }}$

${ \to{ \frac{5 \times 1.73}{2} }}$

${ \to{4.325cm}}$

Area of △ AOB = 1/2×5×4.35 = 10.875cm

Area of minor segment =10.875- 9.81 = 1.06cm²

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

segment

${ \to{3.14 \times 5 \times 5 - 1.06}}$

${ \to{77.44 {cm}^{2} }}$