$A$ chord of a circle of radius $10\, cm$ subtends a right angle at the centre. Find the area of the corresponding:
$(i)$ minor segment $\&$ $(ii)$ major sector. (Use $\pi=3.14$ )

(N/A) Let $AB$ be the chord of the circle subtending a $90^{\circ}$ angle at the centre $O$ of the circle.
Area of the major sector $OADB = \left(\frac{360^{\circ}-90^{\circ}}{360^{\circ}}\right) \times \pi r^{2} = \left(\frac{270^{\circ}}{360^{\circ}}\right) \times 3.14 \times 10^{2}$
$= \frac{3}{4} \times 3.14 \times 100 = 235.5\, cm^{2}$
Area of the minor sector $OACB = \frac{90^{\circ}}{360^{\circ}} \times \pi r^{2} = \frac{1}{4} \times 3.14 \times 100 = 78.5\, cm^{2}$
Area of $\triangle OAB = \frac{1}{2} \times OA \times OB = \frac{1}{2} \times 10 \times 10 = 50\, cm^{2}$
Area of the minor segment $ACB = \text{Area of minor sector } OACB - \text{Area of } \triangle OAB = 78.5 - 50 = 28.5\, cm^{2}$