Question

A chord of a circle of radius $12$ cm subtends an angle of ${120}^o$ at the centre. Find the area of the corresponding segment of the circle.
(Use $\pi =3.14$ and $\sqrt{3}=1.73$)

• A. $88.44$
• B. $94.88$
• C. $43.88$
• D. $54.88$

Answer 1

The correct option is A $88.44$

Consider the above drawn diagram of a circle and draw a perpendicular $OV$ on chord $ST$ such that $SV=VT$.

In $△OVS$,

$\frac{OV}{OS}=\cos {60}^0$
$\Rightarrow \frac{OV}{12}=\frac{1}{2}$
$\Rightarrow OV=\frac{12}{2}$

$\Rightarrow OV=6$ cm

Also,

$\frac{SV}{SO}=\sin {60}^0$
$\Rightarrow \frac{SV}{12}=\frac{\sqrt{3}}{2}$
$\Rightarrow SV=\frac{12\sqrt{3}}{2}$

$\Rightarrow SV=6\sqrt{3}$ cm

Since $ST=2SV$, therefore, we have:

$ST=2SV=2\times 6\sqrt{3}=12\sqrt{3}$ cm

Now, we find the area of $△OST$ as follows:

$Ar\left(△OVS\right)=\frac{1}{2}\times ST\times OV=\frac{1}{2}\times 12\sqrt{3}\times 6=36\sqrt{3}=36\times 1.73=62.28$ cm${}^2$

Now, area of sector $OSUT$ is given by:

$\frac{{120}^0}{{360}^0}\times \pi {\left(12\right)}^2=\frac{1}{3}\times 3.14\times 144=150.72$cm${}^2$

Thus, the area of the segment $SUT$ will be derived by subtracting the area of $△OST$ from the area of sector $OSUT$ as shown below:

$150.72-62.28=88.44$cm${}^2$

Hence, the area of the corresponding segment of the circle is $88.44$cm${}^2$.