Question

1. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the area of the corresponding

minor segments of the circle.​

Answer 1

Answer :-

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° at O.

$area \: of \: sector \: aob \: = \frac{60}{360} \times \pi \times {r}^{2} \\ = \frac{60}{360} \times 3.14 \times {(15)}^{2} \\ = 117.75 \: {cm}^{2}$

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

− Area of △ AOBBy trigonometry,

AC=15sin30°

OC=15cos30°

And, AB=2AC

And, AB=2AC∴ AB=2 × 15sin30° =15 cm

$oc \: = 15cos30 = 15 \frac{ \sqrt{3} }{2} = 15 \times \frac{1.73}{2} = 12.975 \: cm \\ \\ area \: of \: aob \: = 0.5 \times 15 \times 12.975 = 97.3125 \: {cm}^{2}$

∴ 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

Area of major segment = Area of circle − Area of minor segment =(3.14×15×15)−20.4375

Area of major segment = Area of circle − Area of minor segment =(3.14×15×15)−20.4375

=686.0625cm^2

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