Question

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

Answer 1

The area grazed by the cow, buffalo and the horse are in the form of sectors of the circle with radius 7 m.
Let the angle formed in the three sectors be ${\theta }_1,{\theta }_2,{\theta }_3$
Now the area of these sectors will be
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{sector}\mathrm{with}\mathrm{angle}{\theta }_1=\frac{\mathrm{\pi }{\mathrm{\theta }}_1{\left(7\right)}^2}{360} \\ \mathrm{Area}\mathrm{of}\mathrm{sector}\mathrm{with}\mathrm{angle}{\theta }_2=\frac{\mathrm{\pi }{\mathrm{\theta }}_2{\left(7\right)}^2}{360} \\ \mathrm{Area}\mathrm{of}\mathrm{sector}\mathrm{with}\mathrm{angle}{\theta }_3=\frac{\mathrm{\pi }{\mathrm{\theta }}_3{\left(7\right)}^2}{360} \end{gathered}$$
Area of the triangle ABC wil be
$$\begin{gathered} s=\frac{a+b+c}{2}=\frac{15+16+17}{2}=\frac{48}{2}=24 \\ Area=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{24\left(24-15\right)\left(24-16\right)\left(24-17\right)} \\ =24\sqrt{21}{\mathrm{m}}^2 \end{gathered}$$
Area of the field not grazed by the animals = Area of the triangle ABC − Area of the three sectors
$$\begin{gathered} =24\sqrt{21}-\left[\left(\frac{{\mathrm{\pi \theta }}_1{\left(7\right)}^2}{360}\right)+\left(\frac{{\mathrm{\pi \theta }}_2{\left(7\right)}^2}{360}\right)+\left(\frac{{\mathrm{\pi \theta }}_3{\left(7\right)}^2}{360}\right)\right] \\ =24\sqrt{21}-\left(\frac{\pi {\left(7\right)}^2}{360}\right)\left({\theta }_1+{\theta }_2+{\theta }_3\right) \\ =24\sqrt{21}-\left(\frac{\pi {\left(7\right)}^2}{360}\right)\left(180\right)\left(\mathrm{Angle}\mathrm{sum}\mathrm{property}\right) \\ =\left(24\sqrt{21}-77\right){\mathrm{m}}^2 \end{gathered}$$