A chord 10 cm long is drawn in a circle whose radius is

Solution:

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

Consider $\Delta \mathrm{OAB}$.

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

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

Now,

$\sqrt{100}=10 \mathrm{~cm}=A B$

Thus, $\triangle O A B$ is a right isosceles triangle.

Thus, we have:

Area of $\Delta O A B=\frac{1}{2} \times 5 \sqrt{2} \times 5 \sqrt{2}=25 \mathrm{~cm}^{2}$

Area of the minor segment = Area of the sector -">−Area of the triangle

$=\left(\frac{90}{360} \times \pi \times(5 \sqrt{2})^{2}\right)-25$

$=14.25 \mathrm{~cm}^{2}$

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

$=\pi \times(5 \sqrt{2})^{2}-14.25$

$=142.75 \mathrm{~cm}^{2}$