A chord of a circle of radius 14 cm subtends an angle of 60 ° at the centre. Find the area of the corresponding minor segment. Use π= 22 7 and 3 =1.73

Given that

r = 14 cm, θ = 60^{\circ}

To find the area of the minor segment, use the following formula;
Area of the triangle

= \frac{1}{2} ab sin C

The area of the sector subtending an angle

θ

at the centre

= π r^{2} (\frac{θ}{360^{\circ}})

.
Substitute

a = b = 14 cm, C = 60^{\circ}

and

\frac{1}{2} ab sin C

to find the area of the triangle.

\Rightarrow \frac{1}{2} \times 14 \times 14 \times \frac{\sqrt{3}}{2}

= \frac{7 \times 14 \times 1.73}{2}

= 84.77 {cm}^{2}

Now subtract the area of the triangle from the area of the minor sector to find the area of the minor segment.
Area of minor segment =

π (14^{2}) (\frac{60}{360}) - 84.77

\Rightarrow Area of minor segment = \frac{22}{7} \times 14 \times 14 \times \frac{1}{6} - 84.77

\Rightarrow Area of minor segment = 102.67 - 84.77

\Rightarrow Area of minor segment = 17.89 {cm}^{2}

Hence, the area of minor segment is

17.89 {cm}^{2}

.
Therefore, option 1 is correct.

A chord of a circle of radius 14 cm subtends an angle of $60 °$ at the centre. Find the area of the corresponding minor segment.

$Use π = 227 and 3 = 1.73$