In the figure (ii) given below, ABC is an isosceles right angled triangle with ∠ABC = 90°. A semicircle is drawn with AC as diameter. If AB = BC = 7 cm, find the area of the shaded region. Take π = $\dfrac{22}{7}$.

By formula,

Area of △ABC = $\dfrac{1}{2}$ × base × height

= $\dfrac{1}{2}$ × BC × AB

= $\dfrac{1}{2}$ × 7 × 7

= $\dfrac{49}{2}$ = 24.5 cm².

In right angle triangle,

Using pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = 7² + 7²

⇒ AC² = 49 + 49 = 98

⇒ AC = $\sqrt{98} = 7\sqrt{2}$ cm.

From figure,

Radius of semi-circle (r) = $\dfrac{AC}{2} = \dfrac{7\sqrt{2}}{2}$

By formula,

Area of semi-circle = $\dfrac{πr^2}{2}$

$= \dfrac{1}{2} \times \dfrac{22}{7} \times (\dfrac{7\sqrt{2}}{2})^2 \\[1em] = \dfrac{1}{2} \times \dfrac{22}{7} \times \dfrac{98}{4} \\[1em] = \dfrac{2156}{56} = 38.5 \text{ cm}^2.$

Area of the shaded region = Area of the semi-circle – Area of △ABC

= 38.5 - 24.5

= 14 cm².

Hence, area of the shaded region = 14 cm².