Sides of a triangular field are $15\, m , 16 \,m$ and $17\, m$. With the 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 the three animals.

Given that, a triangular field with the three corners of the field a cow, a buffalo and a horse are tied separately with ropes.

So, each animal grazed the field in each corner of triangular field as a sectorial form.

Given, radius of each sector $( r )=7\, m$

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

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

and area of the sector with $\angle H=\frac{\angle H}{360^{\circ}} \times \pi^{2}=\frac{\angle H}{360^{\circ}} \times \pi \times(7)^{2} \,m ^{2}$

Therefore, sum of the areas (in $cm ^{2}$ ) of the three sectors

$=\frac{\angle C}{360^{\circ}} \times \pi \times(7)^{2}+\frac{\angle B}{360^{\circ}} \times \pi \times(7)^{2}+\frac{\angle H}{360^{\circ}} \times \pi \times(7)^{2}$

$=\frac{(\angle C+\angle B+\angle H)}{360^{\circ}} \times \pi \times 49$

$=\frac{180^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 49=11 \times 7=77 \,cm ^{2}$

Given that, sides of triangle are $a=15, b=16$ and $c=17$

Now, semi-perimeter of triangle, $s=\frac{a+b+c}{2}$

$=\frac{15+16+17}{2}=\frac{48}{2}=24$

$\therefore \quad$ Area of triangular field $=\sqrt{s(s-a)(s-b)(s-c)} \quad$ [by Heron's formula]

$=\sqrt{24 \cdot 9 \cdot 8 \cdot 7}$

$=\sqrt{64 \cdot 9 \cdot 21}$

$=8 \times 3 \sqrt{21}=24 \sqrt{21} \,m ^{2}$

So, area of the field which cannot be grazed by the three animals

$=$ Area of triangular field $-$ Area of each sectorial field

$=24 \sqrt{21}-77\, m ^{2}$

Hence, the required area of the field which can not be grazed by the three animals. is $(24 \sqrt{21}-77) \, m^{2}$