Question

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

• A. $686.57m^2$
• B. $656.57m^2$
• C. $636.57m^2$
• D. $616.57m^2$

Answer 1

The correct option is C $636.57m^2$

The length of the straight portion of the track is $90$ m.and

The inside perimeter $312$m

$\therefore \pi r+length+\pi r+length=312$

$\Rightarrow \pi r+90+\pi r+90=312$

$\Rightarrow 2\pi r=312-180$

$\Rightarrow 2\pi r=132$

$\Rightarrow r=\frac{132}{2\pi }$

$\Rightarrow r=\frac{132\times 7}{2\times 22}$

$\Rightarrow r=\frac{924}{44}$

$\Rightarrow r=21$

Width of the running track $=2m$

$\therefore$ Radius of outer perimeter=$21+2$$=23$

$\therefore$ Area of outer perimeter $=\frac{\pi (23{)}^2}{2}+(90\times 23\times 2)+\frac{\pi (23{)}^2}{2}$

$=\frac{22(23{)}^2}{7\times 2}+(90\times 23\times 2)+\frac{22(23{)}^2}{7\times 2}$

$=831.285+4140+831.285$$=5802.57$

$\therefore$ Area of inner perimeter $=\frac{\pi (21{)}^2}{2}+(90\times 21\times 2)+\frac{\pi (21{)}^2}{2}$

$=\frac{22(21{)}^2}{7\times 2}+(90\times 21\times 2)+\frac{22(21{)}^2}{7\times 2}$

$=693+3780+693$$=5166$

Area of the track$=$ Area of outer perimeter $-$ Area of inner perimeter

$=5802.57-5166$ $=636.57m^2$