Question

Sides of a triangular field are $15$ m, $16$ m and $17$ m. With the tree corners of the field a cow, a buffalo and horse are tide separately with ropes of length $7$ m each to graze in the field. Find the area of the field which cannot be grazed by the three animals.

Answer 1

Given radius of each sector$=7m$

Now area of sector with $\angle C=\frac{\angle C}{360}\times \pi r^2=\frac{\angle c}{360}\times \pi \left(7^2\right)m^2$

Area of the sector with $\angle B=\frac{\angle B}{360}\times \pi r^2=\frac{\angle B}{360}\times \pi \left(7^2\right)m^2$

And area of the sector with $\angle H=\frac{\angle H}{360}\times \pi r^2=\frac{\angle H}{360}\times \pi \left(7^2\right)m^2$

Therefore sum of the areas $=\frac{\angle C+\angle B+\angle H}{360}\times \pi \times 49=77m^2$

Semiperimeter=24 m

Therefore area of the triangular field $=\sqrt{s(s-a)(s-b)(s-c)}$

$=24\sqrt{21}{m}^2$

So area which can not be grazed $=(24\sqrt{21}-77)m^2$