Question

Question 3
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 7m each to graze in the field.
Find the area of the field which cannot be grazed by the three animals.

Answer 1

Given that, a triangular field with the three corners of the filed a cow, a buffalo an a horse are tied separately with ropes. So, each animal grazed the filed in each corner of triangular field as a sectorial form.

Radius of each sector = (r)=7m



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 $\angle H$
$=\frac{\angle H}{{360}^{\circ }}\times \pi r^2=\frac{\angle H}{{360}^{\circ }}\times \pi \times (7{)}^2{m}^2$

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

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

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

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

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

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

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

$\therefore Areaoftrianglularfield=\sqrt{s(s-a)(s-b)(s-c)}$ [by Heron's formula]

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

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

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

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

=Area of triangular field - Area of each sectorial field

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

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