An arc subtends an angle of 90° at the centre of

We have given an angle subtended by an arc at the centre of the circle and radius of the circle.

$r=14 \mathrm{~cm}$

$\theta=90^{\circ}$

Now we will find the area of the minor sector.

Area of the minor sector $=\frac{\theta}{360} \times \pi r^{2}$

Substituting the values we get,

Area of the minor sector $=\frac{90}{360} \times \pi \times 14^{2}$.....(1)

Now we will simplify the equation (1) as below,

Area of the minor sector $=\frac{1}{4} \times \pi \times 14^{2}$

Area of the minor sector $=\frac{1}{4} \times \pi \times 14 \times 14$

Area of the minor sector = π × 7 × 7

Area of the minor sector = 49π

Therefore, area of the minor sector is $49 \pi \mathrm{cm}^{2}$