A piece of wire of length 108 cm is bent to form a semicircular arc bounded by its diameter. Find its radius and area enclosed.

Given:

Total length of the wire = 108 cm

Let r be the radius of the circle.

Total length of wire = Circumference of the semicircle + diameter

As we know, the circumference of the semicircle = πr

$⇒ 108 = \dfrac{22}{7} \times r + 2r\\[1em] ⇒ 108 = \dfrac{22}{7}r + \dfrac{2}{1}r\\[1em] ⇒ 108 = \dfrac{22}{7}r + \dfrac{2 \times 7}{1 \times 7}r\\[1em] ⇒ 108 = \dfrac{22}{7}r + \dfrac{14}{7}r\\[1em] ⇒ 108 = \dfrac{22 + 14}{7}r\\[1em] ⇒ 108 = \dfrac{36}{7}r\\[1em] ⇒ r = \dfrac{7 \times 108}{36}\\[1em] ⇒ r = \dfrac{756}{36}\\[1em] ⇒ r = 21 \text{ cm}$

And, area of the semicircle = $\dfrac{1}{2}$πr²

$= \dfrac{1}{2} \times \dfrac{22}{7} \times 21^2\\[1em] = \dfrac{22}{14} \times 441\\[1em] = \dfrac{9,702}{14}\\[1em] = 693 \text{ cm}^2$

Hence, the radius of the circle is 21 cm and the area is 693 cm².