Find the difference of the areas of two segments of a circle formed by a chord of length $5 \, cm$ subtending an angle of $90^{\circ}$ at the centre.

(D) Let the radius of the circle be $r$.
$OA = OB = r \, cm$.
Given that the length of the chord $AB = 5 \, cm$ and the central angle $\theta = 90^{\circ}$.
In $\triangle AOB$,by Pythagoras theorem,$AB^2 = OA^2 + OB^2 \implies 5^2 = r^2 + r^2 \implies 2r^2 = 25 \implies r^2 = 12.5$.
Area of $\triangle AOB = \frac{1}{2} \times OA \times OB \times \sin(90^{\circ}) = \frac{1}{2} \times r^2 \times 1 = \frac{12.5}{2} = 6.25 \, cm^2$.
Area of sector $OAB = \frac{\theta}{360^{\circ}} \times \pi r^2 = \frac{90^{\circ}}{360^{\circ}} \times \pi \times 12.5 = \frac{12.5 \pi}{4} = 3.125 \pi \, cm^2$.
Area of minor segment = Area of sector - Area of $\triangle AOB = (3.125 \pi - 6.25) \, cm^2$.
Area of circle = $\pi r^2 = 12.5 \pi \, cm^2$.
Area of major segment = Area of circle - Area of minor segment = $12.5 \pi - (3.125 \pi - 6.25) = (9.375 \pi + 6.25) \, cm^2$.
Difference of areas = $(9.375 \pi + 6.25) - (3.125 \pi - 6.25) = 6.25 \pi + 12.5 = (6.25 \pi + 12.5) \, cm^2$ or $(\frac{25 \pi}{4} + \frac{25}{2}) \, cm^2$.