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

=\( \Large \pi r^{2}-\left(\frac{25\pi-50}{8}\right)=\frac{25\pi}{2}-\frac{(25\pi-50)}{8} \)

=\( \Large \frac{100\pi-25\pi+50}{8}=\frac{75\pi+50}{8}\)

=\( \Large \frac{25}{8}(3\pi+2)=\frac{25}{4}\left(\frac{3\pi}{2}+1\right)cm^{2} \)