A calf is tied with a rope of length $ 6 \mathrm{~m} $ at the corner of a square grassy lawn of side $ 20 \mathrm{~m} $. If the length of the rope is increased by $ 5.5 \mathrm{~m} $, find the increase in area of the grassy lawn in which the calf can graze.

Given: 

A calf is tied with a rope of length \( 6 \mathrm{~m} \) at the corner of a square grassy lawn of side \( 20 \mathrm{~m} \).

The length of the rope is increased by \( 5.5 \mathrm{~m} \).

To do: 

We have to find the increase in area of the grassy lawn in which the calf can graze.

Solution:

Let the calf be tied at the corner A of the square lawn.
The increase in area is equal to the difference of the two sectors of central angle $90^o$ each and radii $(6+ 5.5)\ m=11.5\ m$ and $6\ m$, which is the shaded region in the figure.

Area of a sector of radius $r$ and angle $\theta$ is $\frac{\theta}{360} \times \pi \times r^{2}$

Therefore,

Increase in area $=[\frac{90}{360} \times \pi \times 11.5^{2}-\frac{90}{360} \pi \times 6^{2}] \mathrm{m}^{2}$

$=\frac{\pi}{4} \times(11.5+6)(11.5-6) \mathrm{m}^{2}$

$=\frac{22}{7 \times 4} \times 17.5 \times 5.5 \mathrm{~m}^{2}$

$=75.625 \mathrm{~m}^{2}$

The increase in area of the grassy lawn in which the calf can graze is $75.625 \mathrm{~m}^{2}$.

Tutorialspoint

Updated on: 10-Oct-2022

52 Views

Kickstart Your Career

Get certified by completing the course

Get Started