# An oval indoor running track with semicircular ends has the dimensions of 80' long, 34' wide, and...

## Question:

An oval indoor running track with semicircular ends has the dimensions of 80' long, 34' wide, and the individual lane width is 5' wide. If the track is to be resurfaced how many square feet of material will be required?

## Area:

The area of a rectangle with length l units and width w units is {eq}l{\times}w {/eq} square units.

The area of a semicircle with radius r unit is {eq}\frac{\pi r^2}{2} {/eq} square units.

Taking the value of pi as 3.14

Given: the dimensions of running track is 80 inches long, 34 inches wide. the individual lane width is 5 inches wide.

The area of the inner running track is the sum of the area of the rectangular part with dimension 80 by 34 inches and the area of two semicircles with radius 17 inches.

\begin{align} A_1&=l{\times}w+2{\times}(\frac{\pi r^2}{2}) \\ A_1&=80{\times}34+2{\times}(\frac{3.14{\times} 17^2}{2}) \\ A_1&=2720+3.14{\times} 289 \\ A_1&=2720+907.46 \\ A_1&=3627.46 \mbox{ square inches} \\ \end{align}

The area of the outer running track is the sum of the area of the rectangular part with dimension (80+10)=90 by (34+10)=44 inches and the area of two semicircles with radius (17+5)=23 inches.

\begin{align} A_2&=l{\times}w+2{\times}(\frac{\pi r^2}{2}) \\ A_2&=90{\times}44+2{\times}(\frac{3.14{\times} 23^2}{2}) \\ A_2&=3960+3.14{\times} 529 \\ A_2&=3960+1661.06 \\ A_2&=5621.06 \mbox{ square inches} \\ \end{align}

The area of the surface which needs to be resurfaced is

\begin{align} &=A_2-A_1 \\ &=5621.06 - 3627.46 \\ &=1993.6 \mbox{ square inches} \\ \end{align}