Question

A chord 10 cm long is drwan in a circle whose radius is $5\sqrt{2}$ cm . Find the areas of both the segments .
[Take $\pi$ = 3.14]

Answer 1

Let OD br the perpendicular fromO on AB.

Now in triangle AOD,

$cos\theta =\frac{B}{H}$

$cos\theta =\frac{5}{5\sqrt{2}}$

$cos\theta =\frac{1}{\sqrt{2}}$

$\theta ={45}^o$

Then the angle made by the chord at centre$={90}^o$

Then area of the minor segment$=(\frac{\pi \theta }{{360}^o}-sin\frac{\theta }{2}cos\frac{\theta }{2})r^2$

$=(\frac{3.14}{4}-\frac{1}{2})(5\sqrt{2}{)}^2$

$=\left(\frac{1.14}{2}\right)50$

$=28.5c{m}^2$

Area of the major segment,

$=$ area of the circle$-$ area of the minor segment.

$=\pi ({5\sqrt{2})}^2-28.5$

$=128.5c{m}^2$