Question

A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and \(\sqrt3\) = 1.73)

Answer 1

Let us draw a perpendicular OV on chord ST. It will bisect the chord ST.
SV = VT

In ΔOVS,

\(\frac{OV}{OS} = cos 60^{\degree}\)

\(\frac{ OV}{ 12} = \frac{1}2 \)

\( OV = 6 \,cm \)

\(\frac{SV}{ SO}  = sin 60^{\degree} = \frac{\sqrt3} 2\)

\(\frac{SV}{ 12} = \frac{\sqrt3}{2}\)

\( SV = 6 \sqrt3\,cm\)

\( ST = 2SV = 2 \times 6 \sqrt3 = 12 \sqrt3 \,cm\)

Area of ΔOST = \(\frac{1}2 \times ST \times OV\)

= \(\frac{1}2 \times 12 \sqrt3 \times 6\)

= \(36\sqrt3 = 36 \times 1.73 = 62.28 \,cm^2\)

Area of sector OSUT = \(\frac{120^{\degree} }{ 360^{\degree}} \times \pi (12)^2\)

 = \(\frac{1}3 \times 3.14 \times 144 = 150.72 \,cm^2\)

Area of segment SUT = Area of sector OSUT - Area of ΔOST 
                                   = 150.72 - 62.28 
                                   = 88.44 cm\(^2\)