In the adjoining figure, O is the center of a circular arc and AOB is a line segment. Find the perimeter and the area of the shaded region correct to one decimal place. (Take π = 3.142)

Angle in semi-circle = 90°

∴ ∠ACB = 90°

Using pythagoras theorem,

⇒ AB² = AC² + BC²

⇒ AB² = 12² + 16²

⇒ AB² = 144 + 256

⇒ AB² = 400

⇒ AB = $\sqrt{400}$ = 20 cm.

AB is the diameter of circle,

∴ Radius = $\dfrac{\text{AB}}{2}$ = $\dfrac{20}{2}$ = 10 cm.

Area of shaded region = Area of semi-circle - Area of triangle

$= \dfrac{πr^2}{2} - \dfrac{1}{2} \times \text{ base × height } \\[1em] = \dfrac{3.142 \times 10^2}{2} - \dfrac{1}{2} \times AC \times BC \\[1em] = \dfrac{3.142 \times 100}{2} - \dfrac{1}{2} \times 12 \times 16 \\[1em] = \dfrac{314.2}{2} - 96 \\[1em] = 157.1 - 96 \\[1em] = 61.1 \text{ cm}^2.$

Perimeter of shaded region = Circumference of semi-circle + AC + CB

= πr + 12 + 16

= 3.142 × 10 + 28

= 31.42 + 28

= 59.42 cm.

Hence, perimeter of shaded region = 59.42 cm and area of semi-circle = 61.1 cm².