In the given figure, AB is the diameter of a circle with centre O and OA = 7 cm. Find the area of the shaded region.

In the figure,

AB and CD both are diameter of circle and they intersect at 90°.

OA = OB = OC = OD = radius of big circle (R) = 7 cm.

∠BOC = 90°

In triangle BOC,

By pythagoras theorem,

BC² = OB² + OC²

BC² = 7² + 7²

BC² = 49 + 49

BC² = 98

BC = $\sqrt{98} = 7\sqrt{2}$ cm.

In figure,

The smaller circle has diameter = OA = 7 cm, so radius (r) = $\dfrac{7}{2}$ cm.

Calculating,

$\text{Area of smaller circle} = πr^2 \\[1em] = \dfrac{22}{7} \times \Big(\dfrac{7}{2}\Big)^2 \\[1em] = \dfrac{22}{7} \times \dfrac{49}{4} \\[1em] = \dfrac{11 × 7}{2} \\[1em] = \dfrac{77}{2} \\[1em] = 38.5 \text{ cm}^2.$

Calculating area of semi-circle CBD,

$\text{Area of semi-circle CBD} = \dfrac{1}{2}πR^2 \\[1em] = \dfrac{1}{2} \times \dfrac{22}{7} \times 7^2 \\[1em] = \dfrac{11}{7} \times 49 \\[1em] = 11 \times 7 \\[1em] = 77 \text{ cm}^2.$

Calculating area of triangle DBC,

$\text{Area of △DBC} = \dfrac{1}{2} \times CD \times OB \\[1em] = \dfrac{1}{2} \times 14 \times 7 \\[1em] = 49 \text{ cm}^2.$

From figure,

Area of shaded area = Area of smaller circle + Area of semicircle - Area of triangle

= 38.5 + 77 - 49

= 115.5 - 49 = 66.5 cm².

Hence, area of shaded region = 66.5 cm².