Question

please solve this ques

Answer 1

Required area= Area of circle-ar(ABC)-ar(OBD)-area of sector COD.

As CB is the diameter, therefore angle CAB=90°(as angle in semicircle is a right angle).

Therefore in triangle ABC,
${ab}^{2} + {ac}^{2} = {bc}^{2}$
${24}^{2} + {7}^{2} = {bc}^{2} \\ 625 = {bc}^{2} \\ bc = 25$
Therefore radius of the circle=25/2cm.
Area of triangle ABC=
$\frac{1}{2} \times 7 \times 24 = 84$
Area of triangle OBD=
$\frac{1}{2} \times r^{2} \times \sin(60) = \frac{1}{2} \times \frac{625}{4} \times \frac{ \sqrt{3} }{2}$
(Here angle is in degrees)

Area of sector OCD=
$\frac{120}{360} \times \pi {r}^{2} = \frac{1}{3} \times \frac{625\pi}{4}$
Area of circle=
$\pi {r}^{2} = \frac{625\pi}{4}$

Required area=
$\frac{625\pi}{4} - 84 - \frac{625 \sqrt{3} }{16} - \frac{625\pi}{12}$
which is approximately equal to
$175.59 {cm}^{2}$