Question

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

Answer 1

In $△OAB,$

${OA}^2+{OB}^2=50+50=100$

${AB}^2={10}^2=100$

i.e., ${OA}^2+{OB}^2={AB}^2$ which satisfies the Pythagoras theorem. Therefor,e $\theta =90°$

Area of minor segment=area of sector-Area of $△AOB$

$=\frac{\theta }{360°}\times \pi r^2-\frac{1}{2}\times OA\times OB$

$=\frac{90°}{360°}\times \frac{22}{7}\times 5\sqrt{2}\times 5\sqrt{2}-\frac{1}{2}\times 5\sqrt{2}\times 5\sqrt{2}$

$=39.29-25=14.29c{m}^2$

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

$=\pi r^2-14.29=\frac{22}{7}\times 5\sqrt{2}\times 5\sqrt{2}-14.29$

$=142.85c{m}^2$