In the figure (ii) given below, the inside perimeter of a practice running track with semi-circular ends and straight parallel sides is 312 m. The length of the straight portion of the track is 90 m. If the track has a uniform width of 2 m throughout, find its area.

Given,

Perimeter of inside semi-circular track = 312 m.

⇒ 90 + πr + 90 + πr = 312

⇒ 2πr + 180 = 312

⇒ 2πr = 312 - 180

⇒ 2πr = 132

⇒ πr = $\dfrac{132}{2}$

⇒ πr = 66

⇒ r = $\dfrac{66}{π} = \dfrac{66}{\dfrac{22}{7}} = \dfrac{66 \times 7}{22} = 21$ m.

So, length of AB = 2r = 2 × 21 = 42 m.

Since, width of track = 2 m.

So, HE = GF = 42 + 2 + 2 = 46 m.

Radius of outer semi-circle (R) = $\dfrac{46}{2}$ = 23 m.

From figure,

Area of track = Area of outer semi-circle (with diametre HE) + Area of outer semi-circle( with diametre GF) + Area of outer rectangle (EFGH) - [Area of inner semi-circle (with diameter AB) + Area of inner semi-circle (with diameter DC) + Area of inner rectangle ABCD]

$= \dfrac{πR^2}{2} + \dfrac{πR^2}{2} + HG \times HE - (\dfrac{πr^2}{2} + \dfrac{πr^2}{2} + AB \times BC) \\[1em] = πR^2 + 90 \times 46 - (πr^2 + 42 \times 90) \\[1em] = πR^2 - πr^2 + 90 \times 46 - 90 \times 42 \\[1em] = π(R^2 - r^2) + 90 \times (46 - 42) \\[1em] = \dfrac{22}{7} \times (23^2 - 21^2) + 90 \times 4 \\[1em] = \dfrac{22}{7} \times (529 - 441) + 360 \\[1em] = \dfrac{22 \times 88}{7} + 360 \\[1em] = \dfrac{1936}{7} + 360 \\[1em] = \dfrac{1936 + 2520}{7} \\[1em] = \dfrac{4456}{7} \\[1em] = 636\dfrac{4}{7} \text{ m}^2.$

Hence, area of semi-circular track = $636\dfrac{4}{7}$ m².