Sides of a triangular field are $15 \, m$,$16 \, m$,and $17 \, m$. At 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.

(A) Given,a triangular field with sides $a = 15 \, m$,$b = 16 \, m$,and $c = 17 \, m$.
Each animal is tied at a corner with a rope of length $r = 7 \, m$. Each animal grazes a sector of the field at its respective corner.
The sum of the areas of the three sectors is given by:
Sum of areas $= \frac{\angle A}{360^{\circ}} \pi r^2 + \frac{\angle B}{360^{\circ}} \pi r^2 + \frac{\angle C}{360^{\circ}} \pi r^2 = \frac{(\angle A + \angle B + \angle C)}{360^{\circ}} \pi r^2$
Since the sum of angles in a triangle is $180^{\circ}$,the sum of areas $= \frac{180^{\circ}}{360^{\circ}} \times \pi \times (7)^2 = \frac{1}{2} \times \frac{22}{7} \times 49 = 77 \, m^2$.
Now,calculate the area of the triangular field using Heron's formula:
Semi-perimeter $s = \frac{15 + 16 + 17}{2} = \frac{48}{2} = 24 \, m$.
Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{24(24-15)(24-16)(24-17)} = \sqrt{24 \times 9 \times 8 \times 7} = \sqrt{12096} = 24\sqrt{21} \, m^2$.
The area which cannot be grazed is the difference between the total area and the grazed area:
Area not grazed $= (24\sqrt{21} - 77) \, m^2$.