Question

A chord of a circle subtends an angle of 60° at the centre of the circle. If the length of the chord is 10 cm, then the area of the major segment is
(a) 305 cm²
(b) 295 cm²
(c) 310 cm²
(d) 335 cm²

Answer 1

(a) 305 cm²
Let AB be the chord of a circle with centre O.
Now,
OA = OB = AB = 10 cm
Thus, we have:
$r=10\mathrm{cm}\mathrm{and}\theta =60°$
Area of the minor segment$=\left(\frac{\mathrm{\pi }{r}^2\theta }{360}-\frac{1}{2}r\right)$²sin θ cm²
$$\begin{gathered} =\left|\left(3.14\times 10\times 10\times \frac{60}{360}\right)-\left(\frac{1}{2}\times 10\times 10\times \sin 60°\right)\right|{\mathrm{cm}}^2 \\ =\left(\frac{157}{3}-50\times \frac{\sqrt{3}}{2}\right){\mathrm{cm}}^2 \\ =\left|\frac{157}{3}-\left(25\times 1.732\right)\right|{\mathrm{cm}}^2 \\ =\left(\frac{157}{3}-\frac{433}{10}\right){\mathrm{cm}}^2 \\ =\frac{271}{30}{\mathrm{cm}}^2 \\ =9{\mathrm{cm}}^2 \end{gathered}$$

Area of the major segment $=\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{circle}-\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{minor}\mathrm{segment}$

$$\begin{gathered} =\left|\left(\mathrm{\pi }\times 10\times 10\right)-9\right|{\mathrm{cm}}^2 \\ =\left(3.14\times 10\times 10-9\right){\mathrm{cm}}^2 \\ =\left(314-9\right){\mathrm{cm}}^2 \\ =305{\mathrm{cm}}^2 \end{gathered}$$