In the given figure, ABCP is a quadrant of a circle of radius 14 cm. With AC as diameter, a semi-circle is drawn. Find the area of shaded region.

Given,

ABCP is a quadrant of radius = 14 cm

AB = BC = 14 cm

AC is the diameter of semicircle.

Using Pythagoras theorem in △ABC,

⇒ AC² = AB² + BC²

⇒ AC² = 14² + 14²

⇒ AC² = 196 + 196

⇒ AC² = 392

⇒ AC = $\sqrt{392} = 14\sqrt{2}$ cm.

So, radius of semicircle = $\dfrac{14\sqrt{2}}{2} = 7\sqrt{2}$ cm.

Calculating area of semicircle :

$\text{Area of semicircle} = \dfrac{1}{2}πr^2 \\[1em] = \dfrac{1}{2} \times \dfrac{22}{7} \times (7\sqrt{2})^2 \\[1em] = \dfrac{11}{7} \times 49 \times 2 \\[1em] = 11 \times 7 \times 2 \\[1em] = 154 \text{ cm}^2.$

Calculating area of quadrant :

$\text{Area of quadrant} = \dfrac{1}{4}πr^2 \\[1em] = \dfrac{1}{4} \times \dfrac{22}{7} \times 14^2 \\[1em] = \dfrac{11}{14} \times 196 \\[1em] = 154 \text{ cm}^2.$

Calculating area of triangle :

Area of triangle ABC = $\dfrac{1}{2}$ × AB × BC

= $\dfrac{1}{2}$ × 14 × 14

= 7 × 14 = 98 cm².

Shaded area = (Area of semicircle AQC) - [(Area of quadrant APCB) - (Area of triangle ABC)]

Area of shaded region = 154 - [(154 - 98)]

= 154 - 56 = 98 cm².

Hence, area of shaded region = 98 cm².