A) \( \Large \frac{25}{4}\left(\frac{\pi}{2}+1\right)cm^{2} \) |
B) \( \Large \frac{25}{4}\left(\frac{\pi}{2}-1\right)cm^{2} \) |
C) \( \Large \frac{25}{4}\left(\frac{3\pi}{2}+1\right)cm^{2} \) |
D) None of the above |
C) \( \Large \frac{25}{4}\left(\frac{3\pi}{2}+1\right)cm^{2} \) |
In \( \Large \triangle AOB \),AO=OB=r[radius of circle]
By Pythagoras theorem,
\( \Large AB^{2}=OA^{2}+OB^{2} => (5)^{2}=r^{2}+r^{2} \)
\( \Large r^{2}=\frac{25}{2} \)cm
Now, area of sector AOB = \( \Large \frac{\theta}{360 degree}\times \pi r^{2} \)
=\( \Large \frac{90 degree}{360 degree}\times \pi\times 25/2=\frac{25\pi}{8}cm^{2} \)
Now, area of minor segment = area of sector - area of triangle
=\( \Large \frac{25\pi}{8}-\frac{r^{2}}{2}=\frac{25\pi}{8}-\frac{25}{4}=\left(\frac{25\pi-50}{8}\right) \)
Area of major segment = Area of circle - Area of minor segment
=\( \Large \pi r^{2}-\left(\frac{25\pi-50}{8}\right)=\frac{25\pi}{2}-\frac{(25\pi-50)}{8} \)
=\( \Large \frac{100\pi-25\pi+50}{8}=\frac{75\pi+50}{8}\)
=\( \Large \frac{25}{8}(3\pi+2)=\frac{25}{4}\left(\frac{3\pi}{2}+1\right)cm^{2} \)