Question

$\mathrm{B},\mathrm{C},\mathrm{H}$Sides of a triangular field are $15m,16m$and $17m.$ With the three corners of the field a cow, a buffalo and a horse are tied separately with ropes of length $7m$each to graze in the field. Find the area of the field which cannot be grazed by the three animals.

Answer 1

Step 1. The area of the field :

Sides are$15\mathrm{m},16\mathrm{m},$ and$17\mathrm{m}$.

Length of the rope to which cow, horse, and buffalo are tied$=7\mathrm{cm}$

Now, the perimeter of the triangle:

$\begin{array}{rcl}\mathrm{Perimeter}& =& (15+16+17)\mathrm{m}\\ & =& 48\mathrm{m}\end{array}$

Semi-perimeter of the triangle:

$\begin{array}{rcl}s& =& \frac{48}{2}\\ & =& 24m\end{array}$

Step 2. Concept used formula :

By Heron’s formula,

$\mathrm{Area}\mathrm{of}\mathrm{a}\mathrm{triangle}=\sqrt{\left(\mathrm{s}\right(\mathrm{s}–\mathrm{a}\left)\right(\mathrm{s}–\mathrm{b}\left)\right(\mathrm{s}–\mathrm{c})}$

where $a,b\mathrm{and}c$ are the sides of the triangle.

$\begin{array}{rcl}\mathrm{Area}(△\mathrm{ABC})& =& \sqrt{\left(24\right(24–15\left)\right(24–16\left)\right(24–17)}\\ & =& 109.982{\mathrm{m}}^2\end{array}$

Let $\mathrm{B},\mathrm{C}\mathrm{and}\mathrm{H}$be the corners of the triangle on which buffalo, cow and horse are tied respectively with ropes of $7\mathrm{m}$ each.

So, the area grazed by each animal will be in the form of a sector.

Radius of each sector is $r=7m$.

Step 3. Area of the Sectors :

Let $x,y,z$ be the angles at corners respectively.

Area of the sector with central angle x,

$$\begin{gathered} {\mathrm{Area}}_x=\frac{1}{2}\frac{x}{180}\times {\mathrm{\pi r}}^2 \\ =\frac{1}{2}\frac{x}{180}\mathrm{\pi }{\left(7\right)}^2 \end{gathered}$$

Area of sector with central angle y,

$$\begin{gathered} {\mathrm{Area}}_y=\frac{1}{2}\frac{y}{180}\times {\mathrm{\pi r}}^2 \\ =\frac{y}{360}\times \mathrm{\pi }{\left(7\right)}^2 \end{gathered}$$

Area of sector with central angle z,

$$\begin{gathered} {\mathrm{Area}}_z=\frac{1}{2}\frac{z}{180}\times {\mathrm{\pi r}}^2 \\ =\frac{z}{360}\times \mathrm{\pi }{\left(7\right)}^2 \end{gathered}$$

The area of field not grazed by the animals is an area of the three sectors subtracted from the Area of a triangle :

$\begin{array}{rcl}\mathrm{Area}& =& {\mathrm{Area}}_{\mathrm{x}}+{\mathrm{Area}}_{\mathrm{y}}+{\mathrm{Area}}_{\mathrm{z}}\\ \mathrm{Area}& =& (109.982)-\left[\right(\frac{x}{360}\times \pi 7^2)+(\frac{y}{360}\times \pi 7^2)+(\frac{z}{360}\times \pi 7^2\left)\right]\\ \mathrm{Area}& =& (109.982)-\left[\right(\frac{x+y+z}{360}\times \pi 7^2\left)\right]\\ \mathrm{Area}& =& (109.982)-\left[\right(\frac{180}{360}\times \frac{22}{7}49\left)\right]\\ \mathrm{Area}& =& 109.892–77\\ \mathrm{Area}& =& 32.982{\mathrm{cm}}^2\end{array}$

Hence, the Area of field not grazed by the animals $32.982{\mathrm{cm}}^2$