In the adjoining figure, ABCD is a square of side 21 cm. AC and BD are two diagonals of the square. Two semicircle are drawn with AD and BC as diameters. Find the area of the shaded region. Take π = $\dfrac{22}{7}$.

Area of square = (side)² = 21² = 441 cm².

We know that,

Diagonals of a square divide it into four triangles of equal area.

Area of a triangle = $\dfrac{\text{Area of square}}{4} = \dfrac{441}{4}$ = 110.25 cm².

Area of △AOD + Area of △BOC = 110.25 + 110.25 = 220.50 cm²

Diameter of each semicircle = 21 cm

∴ Radius = $\dfrac{21}{2}$ = 10.5 cm.

Area of each semicircle = $\dfrac{πr^2}{2}$

$= \dfrac{\dfrac{22}{7} \times 10.5 \times 10.5}{2} \\[1em] = \dfrac{22 \times 10.5 \times 10.5}{14} \\[1em] = \dfrac{2425.5}{14} \\[1em] = 173.25 \text{ cm}^2$

From figure,

Area of shaded region = Area of both semicircles + Area of △AOD + Area of △BOC

= (2 × 173.25) + 220.50

= 346.50 + 220.50

= 567 cm².

Hence, area of shaded region = 567 cm².