Question

A racetrack is in the form of a ring whose inner circumference is $352m$ and outer circumference is $396m.$ Find the width and the area of the track.

Answer 1

Given: Inner circumference of racetrack (in the ring form) $=352m$

Outer circumference of racetrack $=396m$

Let the radius of inner track be $r$
$2\pi r=352$ [$\because$ Circumference of the circle $=2\pi r$]

$\Rightarrow 2\times \frac{22}{7}\times r=352$

$\Rightarrow r=\frac{352\times 7}{2\times 22}$

$\Rightarrow r=56m$
let the radius of outer track be $R$
$2\pi R=396$

$\Rightarrow 2\times \frac{22}{7}\times R=396$

$\Rightarrow R=\frac{396\times 7}{2\times 22}$

$\Rightarrow R=63m$

$\therefore$ Width of track $=$ radius of outer track $-$ radius of inner track
$=63-56$
$=7m$

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

$=\pi R^2-\pi r^2$ [$\because$ Area of the circle $=\pi (radius{)}^2$]

$=\pi (R^2-r^2)$

$=\pi (R-r)(R+r)$ [$\because (a^2-b^2)=(a-b)(a+b)$]

$=\frac{22}{7}\times (63-56)(63+56)$

$=\frac{22}{7}\times 7\times 119$

$=2618m^2$

Hence, the area of the track is $2618m^2$