Question

A chord PQ of a circle with a radius of cm subtends an angle of 60^° with the center of the circle. Find the area of the minor as well as the major segment. ( \[\pi\] = 3.14,  \[\sqrt{3}\] = 1.73)

Answer 1

The radius of the circle, r = 15 cm
Let O be the center and PQ be the chord of the circle. 

 

∠POQ = θ = 60º
Area of the minor segment = Area of the shaded region

\[= r^2 \left( \frac{\pi\theta}{360°} - \frac{\sin\theta}{2} \right)\]
\[ = \left( 15 \right)^2 \times \left( \frac{3 . 14 \times 60° }{360° } - \frac{\sin60° }{2} \right)\]
\[ = 225 \times \frac{3 . 14}{6} - 225 \times \frac{\sqrt{3}}{4}\]
\[ = 117 . 75 - 97 . 31\]
\[ = 20 . 44 {cm}^2\]

Now,
Area of the circle = 

\[\pi r^2 = 3 . 14 \times \left( 15 \right)^2 = 3 . 14 \times 225\]  = 706.5 cm²

∴ Area of the major segment = Area of the circle − Area of the minor segment = 706.5 − 20.44 = 686.06 cm2
Thus, the areas of the minor segment and major segment are 20.44 cm2 and 686.06 cm2, respectively.