In the figure (i) given below, ABC is a right angled triangle, ∠B = 90°, AB = 28 cm and BC = 21 cm. With AC as diameter a semi-circle is drawn and with BC as radius a quarter circle is drawn. Find the area of the shaded region correct to two decimal places.

In right angle △ABC,

Using Pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = 28² + 21²

⇒ AC² = 784 + 441

⇒ AC² = 1225

⇒ AC = $\sqrt{1225}$ = 35 cm.

Radius of semi-circle (R) = $\dfrac{AC}{2} = \dfrac{35}{2}$ = 17.5 cm.

Radius of quadrant (r) = BC = 21 cm.

From figure,

Area of shaded region = Area of △ABC + Area of semi-circle – Area of quadrant

$= \dfrac{1}{2} \times BC \times AB + \dfrac{πR^2}{2} - \dfrac{πr^2}{4}\\[1em] = \dfrac{1}{2} \times 21 \times 28 + \dfrac{\dfrac{22}{7} \times (17.5)^2}{2} - \dfrac{\dfrac{22}{7} \times (21)^2}{4} \\[1em] = 21 \times 14 + \dfrac{22 \times 306.25}{7 \times 2} - \dfrac{22 \times 441}{7 \times 4} \\[1em] = 294 + \dfrac{6737.5}{14} - \dfrac{9702}{28} \\[1em] = 294 + 481.25 - 346.5 \\[1em] = 428.75 \text{ cm}^2.$

Hence, area of shaded region = 428.75 cm².