Question

A chord 10 cm long is drawn in a circle whose radius is $5\sqrt{2}$ cm. Find the areas of both the segments.

$[Take\pi =\mathrm{3.14.}]$

Answer 1

3

Given, a chord AB of length 10 cm and radius = OA = OB = $5\sqrt{2}$ cm.

Construction: Draw OC perpendicular to AB.

Now, AC = BC = $\frac{10}{2}$ = 5 cm [The perpendicular drawn from the centre of a circle to a chord always bisect the chord.]

In $△$ OAC,

$sinx=\frac{AC}{OA}sinx=\frac{5}{5\sqrt{2}}sinx=\frac{1}{\sqrt{2}}sinx=sin{45}^{o}x={45}^o$

Similarly, $\angle BOC={45}^o\Rightarrow \angle AOB=\angle AOC+\angle BOC={45}^o+{45}^o={90}^o$

We know that, area of sector $=\frac{\theta }{360}\times \pi r^2$

Therefore, area of sector OAB$=\frac{\theta }{360}\times \pi r^2=\frac{90}{360}\times \pi (5\sqrt{2}{)}^2=\frac{50\pi }{4}=\frac{25}{2}\times \frac{22}{7}=\frac{275}{7}=39.28c{m}^2$

Again, in $△$ OAC,

$cosx=\frac{OC}{OA}cos{45}^o=\frac{OC}{5\sqrt{2}}\frac{1}{\sqrt{2}}=\frac{OC}{5\sqrt{2}}OC=5cm$

Therefore, area of $△$OAB

$=\frac{1}{2}\times OC\times AB=\frac{1}{2}\times 5\times 10=25c{m}^2$

Now, area of minor segment = Area of sector OAB - Area of $△$ OAB

= 39.28 $c{m}^2$ - 25 $c{m}^2$​​​​​​​

= 14.28 $c{m}^2$​​​​​​​

Similarly, area of major segment = Area of circle - Area of minor segment

⇒ Area of major segment

$=\pi r^2-14.28=\frac{22}{7}\times (5\sqrt{2}{)}^2-14.28=157.1428-14.28=142.86c{m}^2$