Question

A chord of circle $14cm$ makes an angle of ${60}^o$ at the centre of the circle. Find
(i) Area of minor sector
(ii) Area of the major segment
(iii)Area of the major sector
(iv)Area of the major segment

• A. $\left(i\right)=102.57c{m}^2,\left(ii\right)=17.80c{m}^2,\left(iii\right)=512.87c{m}^2,\left(iv\right)=597.64c{m}^2$
• B. $\left(i\right)=12.57c{m}^2,\left(ii\right)=17.80c{m}^2,\left(iii\right)=512.87c{m}^2,\left(iv\right)=597.64c{m}^2$
• C. $\left(i\right)=102.57c{m}^2,\left(ii\right)=1.80c{m}^2,\left(iii\right)=512.87c{m}^2,\left(iv\right)=597.64c{m}^2$
• D. None of these

Answer 1

The correct option is A $\left(i\right)=102.57c{m}^2,\left(ii\right)=17.80c{m}^2,\left(iii\right)=512.87c{m}^2,\left(iv\right)=597.64c{m}^2$

Given, $r=14cm,\theta ={60}^o$
(i)Area of minor sector $OAPB$$=\frac{\theta }{{360}^o}\pi r^2$
$=\frac{{60}^o}{{360}^o}\times 3.14\times 14\times 14$

$=102.57c{m}^2$
(ii)Area of minor segment APB$=\frac{\pi r^2\times \theta }{{360}^o}-\frac{r^2}{2}\sin \theta$
$=102.57-\frac{14\times 14}{2}\sin {60}^o$
$=102.57-98\times \frac{\sqrt{3}}{2}$
$=17.80c{m}^2$
(iii)Area of major sector $=$Area of circle-Area of minor sector $OAPB=$$\pi (14{)}^2-102.57$
$=615.44-102.57=512.87c{m}^2$
(iv)Area of major segment $AQB$
$=$Area of circle-area of minor segment $APB$
$=615.44-17.80=597.64c{m}^2$