A chord $ P Q $ of length $ 12 \mathrm{~cm} $ subtends an angle of $ 120^{\circ} $ at the centre of a circle. Find the area of the minor segment cut off by the chord $ P Q . $

Given:

A chord \( P Q \) of length \( 12 \mathrm{~cm} \) subtends an angle of \( 120^{\circ} \) at the centre of a circle.

To do: 

We have to find the area of the minor segment cut off by the chord \( P Q . \)

Solution:

Length of the chord $PQ = 12\ cm$

Angle at the centre $\theta = 120^o$

Draw $OD\ \perp\ DQ$ which bisects $PQ$ at $D$ and also bisects $\angle POQ$

This implies,

$\mathrm{PD}=\mathrm{DQ}=6 \mathrm{~cm}$

$\angle \mathrm{POD}=\frac{120^{\circ}}{2}$

$=60^{\circ}$

In right angled triangle $\Delta \mathrm{OPD}$,

$\sin \theta=\frac{\text { Base }}{\text { Hypotenuse }}$

$\Rightarrow \sin 60^{\circ}=\frac{\mathrm{PD}}{\mathrm{OP}}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{6}{r}$

$\Rightarrow r=\frac{6 \times 2}{\sqrt{3}}$

$=\frac{12}{\sqrt{3}}$

$=\frac{\sqrt{3} \times 12}{\sqrt{3} \times \sqrt{3}}$

$=\frac{12 \times \sqrt{3}}{3}$

$=4 \sqrt{3} \mathrm{~cm}$
Area of the minor segment $PRQ=(\frac{\pi \theta}{360^{\circ}}-\sin \theta \cos \theta) r^{2}$

$=(\frac{\pi \times 120^{\circ}}{360^{\circ}}-\sin \frac{120^{\circ}}{2} \cos \frac{120^{\circ}}{2})(4 \sqrt{3})^{2}$

$=(\frac{\pi}{3}-\sin 60^{\circ} \cos 60^{\circ}) 48$

$=48(\frac{\pi}{3}-\frac{\sqrt{3}}{2} \times \frac{1}{2})$

$=48(\frac{\pi}{3}-\frac{\sqrt{3}}{4})$

$=\frac{48 \times(4 \pi-3 \sqrt{3})}{12}$

$=4(4 \pi-3 \sqrt{3}) \mathrm{cm}^{2}$

The area of the minor segment cut off by the chord \( P Q \) is $4(4 \pi-3 \sqrt{3}) \mathrm{cm}^{2}$.

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Updated on: 10-Oct-2022

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