Question

A chord of a circle of radius 14cm subtends an angle of 120° at the centre. Find the area of the corresponding minor segment of the circle. Use $\pi =\frac{22}{7}$ and $\sqrt{3}=1.73$.

Answer 1

Given : Radius , r = 14 cm
Chord AB subtend angle ${120}^{\circ }$ at the centre
Area of circle $=\pi r^2$
$=\frac{22}{7}\times \left(14{)}^2\right\{\pi =\frac{22}{7}\}=616c{m}^2$
area of sector $OABO=\frac{O}{{360}^{\circ }}\times \pi r^2$
$=\frac{120}{{360}^{\circ }}\times \pi r^2\Rightarrow \frac{1}{3}\times 616=205.33c{m}^{2}AC=CRIn\Delta OCA\frac{OC}{OA}=cos60\circ \frac{OC}{14}=\frac{1}{2},OC=7cm\frac{AC}{AO}=sin{60}^{\circ }=\frac{\sqrt{3}}{2}\frac{AC}{14}=\frac{\sqrt{3}}{2},2\times 7\sqrt{3}AB=2AC=2\times 7\sqrt{3}=14\sqrt{3}$
Area of $\Delta OAB=\frac{1}{2}\times AB\times OC=\frac{1}{\text{/}2}\times \text{/}{14}^3\sqrt{3}\times 7=49\sqrt{3}=49\times 1.732$
Area of minor segment = Area of sector OABO - Area of $\Delta AOB$
$=205.33-84.87=-120.46c{m}^2$