Subject: General Aptitude · Chapter: Area · Exam: 2006 · Marks: · Difficulty:
An athletic track 14 m wide consists of two straight sections 120 m long joining semi-circular ends whose inner radius is 35 m. The area of the track is
(a)$7026 \mathrm{m}^{2}$
(b)$7036 \mathrm{m}^{2}$
(c)$7046 \mathrm{m}^{2}$
(d)$7056 \mathrm{m}^{2}$
Answer
Answer (as printed): D
Explanation
Area of the track = Area of the two rectangles + Area of the two semi-circular ring ends $$\begin{aligned} & =\left[(2 \times 120 \times 14)+2 \times \frac{\pi}{2} \times\left\{(49)^{2}-(35)^{2}\right\}\right] \mathrm{m}^{2} \\ & =\left[3360+\frac{22}{7} \times(49+35)(49-35)\right] \mathrm{m}^{2} \\ & =\left(3360+\frac{22}{7} \times 84 \times 14\right) \mathrm{m}^{2}=(3360+3696) \mathrm{m}^{2}=7056 \mathrm{m}^{2} . \end{aligned}$$