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.

Answer 1

Let O be the centre of the circle and AB be the chord.

Consider $∆$OAB.

$\mathrm{OA}=\mathrm{OB}=5\sqrt{2}\mathrm{cm}$

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

Now,
$\sqrt{100}=10\mathrm{cm}=AB$

Thus, $∆$OAB is a right isosceles triangle.

Thus, we have:
Area of $∆$OAB = $\frac{1}{2}\times 5\sqrt{2}\times 5\sqrt{2}=25{\mathrm{cm}}^2$

Area of the minor segment = Area of the sector $-$ Area of the triangle
$$\begin{gathered} =\left(\frac{90}{360}\times \mathrm{\pi }\times {\left(5\sqrt{2}\right)}^2\right)-25 \\ =14.25{\mathrm{cm}}^2 \end{gathered}$$

Area of the major segment = Area of the circle $-$ Area of the minor segment
$$\begin{gathered} =\mathrm{\pi }\times {\left(5\sqrt{2}\right)}^2-14.25 \\ =142.75{\mathrm{cm}}^2 \end{gathered}$$