Question

a chord of a circle of radius10 cm subtends a right angle at the centre .Find the area of the corresponding 1)minor segment 2)major segment
plz answer this question

Answer 1

Answer:

Radius, r = 10 cm

Sector angle,x = 90

So area of minor segment =

$\frac{ {r}^{2} }{2} ( \frac{\pi \: x}{180} - sinx) \\ = \frac{22}{7} \times 100( \frac{22}{7} \times \frac{90}{180} - sin90) \\ = \frac{2200}{7} ( \frac{11}{7} - 1) \\ = \frac{2200}{7} \times \frac{4}{7} \\ = \frac{8800}{49} {cm}^{2}$

Area of major segment =

area of circle - area of minor segmant

=

$\frac{22}{7} \times 10 \times 10 - \frac{8800}{49} \\ \frac{2200}{7} - \frac{8800}{49} \\ = \frac{15400 - 8800}{49} \\ = \frac{6600}{49}$