Question

A horse is tied to a pole fixed at one corner of a 30 m $\times$ 30 m square field of grass by a 10 m long rope. ($\mathrm{\pi }=3.14$)
(i) Find the area of that part of the field in which the horse can graze.


(ii) What will be the area of the field in which the horse can graze, if the pole was fixed at the middle of the side?

Answer 1

Given:
Side of the square field = 30 m
Length of the rope, r = 10 m
(i) We know:
Area of a sector = $\frac{\mathrm{\theta }}{360}\times \mathrm{\pi }{r}^2$
∴ Area of the sector = $\frac{90}{360}\times 3.14\times 10\times 10$ m²

= 78.5 m²

Thus, the area of the part of the field where the horse can graze is 78.5 m².

(ii) If the pole is fixed at the middle of the side, then the horse can graze in a semicircular region.
∴ Area of the field where the horse can graze = $\frac{1}{2}\mathrm{\pi }{r}^2\mathit{=}\frac{1}{2}\times 3.14\times 10\times 10=157{\mathrm{m}}^2$