Question

In the given figure, if A is the centre of the circle. \[\angle\] PAR = 30^°, AP = 7.5, find the area of the segment PQR. (\[\pi\] = 3.14)

Answer 1

Radius of the circle, r = 7.5 cm

∠​PAR = θ = 30º 

∴ Area of segment PQR

\[= r^2 \left[ \frac{\pi\theta}{360^{\circ}} - \frac{\sin\theta}{2} \right]\]

= \[(7.5)^{2}\left[\frac{3.14\times30}{360^{\circ}}-\frac{\sin30^{\circ}}{2}\right]\]

= \[56.25\left[\frac{3.14}{12}-\frac{1}{2}\times\frac{1}{2}\right]\]

= \[56.25\left[\frac{3.14}{12}-\frac{1\times3}{4\times3}\right]\]

= \[56.25\left[\frac{3.14}{12}-\frac{3}{12}\right]\]

= \[56.25\left(\frac{3.14-3}{12}\right)\]

= \[56.25\left(\frac{0.14}{12}\right)\]

= \[\frac{7.875}{12}\]

= 0.65625 sq. units

Thus, the area of the segment PQR is 0.65625 sq. units.