#### Question

A chord PQ of a circle with radius 15 cm subtends an angle of 60^{°} with the centre of the circle. Find the area of the minor as well as the major segment. ( \[\pi\] = 3.14, \[\sqrt{3}\] = 1.73)

#### Solution

Radius of the circle, *r* = 15 cm

Let O be the centre 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 cm

^{2}^{∴ Area of the major segment = Area of the circle − Area of the minor segment = 706.5 − 20.44 = 686.06 cm2Thus, the areas of the minor segment and major segment are 20.44 cm2 and 686.06 cm2, respectively.}

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Solution A Chord Pq of a Circle with Radius 15 Cm Subtends an Angle of 60° with the Centre of the Circle. Find the Area of the Minor as Well as the Major Segment. ( π = 3.14, √ 3 = 1.73) Concept: Areas of Sector and Segment of a Circle.