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. - Noon Academy

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.

Areas Related to Circles | Exercise 11.4 | Maths | NCERT Exemplar

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.

Solution

From the given question,

We got Sides of the triangle are

$15$

m,

$16$

m and

$17$

m.

Then, perimeter of the triangle =

$(15+16+17)$

m =

$48$

m

Therefore, Semi-perimeter of the triangle = s =

$48/2$

=

$24$

m

By using Heron’s formula,

Where Area of the triangle =

$\sqrt{(s(s-a)(s-b)(s-c))}$

,here a, b and c are the sides of triangle

=

$\sqrt{(24(24-15)(24-16)(24-17))}$

=

$109.982$

${{m}^{2}}$

Let us consider B, C and H be the corners of the triangle on which buffalo, cow and horse are tied respectively with ropes of

$7$

m each.

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

∴ Radius of each sector = r =

$7$

m

Let us consider x, y, z be the angles at corners B, C, H respectively.

∴ Area of sector with central angle x will be,

=

$\frac{1}{2}(x/180)\times \pi {{r}^{2}}$

=

$(x/360)\times \pi {{(7)}^{2}}$

Area of sector with central angle y will be,

=

$\frac{1}{2}(y/180)\times \pi {{r}^{2}}$

=

$(y/360)\times \pi {{(7)}^{2}}$

Area of sector with central angle z will be,

=

$\frac{1}{2}(z/360)\times \pi {{r}^{2}}$

=

$(z/360)\times \pi {{(7)}^{2}}$

To get the Area of field not grazed by the animals = Area of triangle – (area of the three sectors)

=

$(109.982)-\left( \left( \frac{x}{360} \right)\times \pi {{(7)}^{2}}+\left( \frac{y}{360} \right)\times \pi {{(7)}^{2}}+\left( \frac{z}{360} \right)\times \pi {{(7)}^{2}} \right)$

=

$109.982-\left( \left( \frac{x+y+z}{360} \right)\times \pi {{(7)}^{2}} \right)$

=

$109.982-\left( \left( \frac{180}{360} \right)\times \left( \frac{22}{7} \right)\times {{(7)}^{2}} \right)$

=

$109.892-77$

=

$32.982$

$c{{m}^{2}}$

i

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