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