#### Question

The inside perimeter of a running track (shown in Fig. 20.24) is 400 m. The length of each of the straight portion is 90 m and the ends are semi-circles. If track is everywhere 14 m wide, find the area of the track. Also, find the length of the outer running track.

#### Solution

It is given that the inside perimeter of the running track is 400 m . It means the length of the inner track is 400 m .

Let r be the radius of the inner semicircles .

Observe: Perimeter of the inner track = Length of two straight portions of 90 m + Length of two semicircles

∴ 400 = (2 x 90) + (2 x Perimiter of a semicircle)

\[400 = 180 + (2 \times \frac{22}{7} \times r)\]

\[400 - 180 = (\frac{44}{7} \times r)\]

\[\frac{44}{7} \times r = 220\]

\[r = \frac{220 \times 7}{44} = 35 m\]

∴ Width of the inner track = 2r = 2 x 35 = 70 m

Since the track is 14 m wide at all places, so the width of the outer track: 70 + (2 x 14) = 98 m

∴ Radius of the outer track semicircles \[= \frac{98}{2} = 49 m\]

Area of the outer track = (Area of the rectangular portion with sides 90 m and 98 m) + (2 x Area of two semicircles with radius 49 m)

\[ = (98 \times 90) + (2 \times \frac{1}{2} \times \frac{22}{7} \times {49}^2 )\]

\[ = (8820) + (7546)\]

\[ = 16366 m^2 \]

And, area of the inner track = (Area of the rectangular portion with sides 90 m and 70 m) + (2 x Area of the semicircle with radius 35 m)

\[ = (70 \times 90) + (2 \times \frac{1}{2} \times \frac{22}{7} \times {35}^2 )\]

\[ = (6300) + (3850)\]

\[ = 10150 m^2 \]

∴ Area of the running track = Area of the outer track - Area of the inner track

\[ = 16366 - 10150\]

\[ = 6216 m^2 \]

And, length of the outer track = (2 x length of the straight portion) + (2 \times perimeter of the semicircles with radius 49 m)

\[ = (2 \times 90) + (2 \times \frac{22}{7} \times 49)\]

\[ = 180 + 308\]

\[ = 488 m\]