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In the given figure O is the centre of the circle with radius 28 cm. Find the area of the

minor segment $ASB .$

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30.756 {cm}^{2}

30.856 {cm}^{2}

20.856 {cm}^{2}

30.816 {cm}^{2}

answer is B.

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Detailed Solution

Given,
Centre of the circle = O
Radius r = 28 cm
Angle

θ = 45^{\circ}

The minor segment area =

ASB

We have calculated
The sector of area

△ OAB

△ OAB = \frac{π r^{2}}{360} \times θ

△ OAB = (\frac{22}{7} \times \frac{1}{360} \times (28)^{2} \times 45) {cm}^{2}

△ OAB = 308 {cm}^{2}

Now calculate the area of

ΔOAB

Area of

ΔOAB = \frac{1}{2} \times OA \times OB \times \sin 45^{\circ} [since, Area of △ ABC = \frac{1}{2} AB \times AC \times \sin A]

= \frac{1}{2} \times 28 \times 28 \times \frac{1}{\sqrt{2}} {cm}^{2}

= 14 \times 14 \times \sqrt{2} {cm}^{2}

= 196 \times 1.414 {cm}^{2}

= 277.144 {cm}^{2}

Now, we have to find the area of minor segment using following steps
Area of minor segment

= Area of sector △ OAB - Area of △ OAB

= (308 - 277.144) {cm}^{2}

= 30.856 {cm}^{2}

Correct option is 2.

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