Question

Please answer to this question........​

Answer 1

Answer:

hope it help you.

Answer 2

Solution

radius of the circle =10 cm

chord PQ subtends an angle = 60° at the centre

PQO is an equilateral triangle , O being the centre of the circle

area of mirror segment = 8/360×

$\pi \: r2 - 1 \2 \: r2 \: \sin \: 8$

= 60/360 × 22/7 × 10 × 10 - 1/2×10×10 3/2

= 9.03

area of circle =

$\pi \: r2$

= 22/7 ×10×10= 314 (approx.)

area of mirror segment = area of circle- area of mirror segment

= 314-9.03

= 304.97 cm2 ( approx.)