Question

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 three animals.

Answer 1

Solution:-

Let ABC be the triangular field with sides Ac = 15 m, AB = 16 m and BC = 17 m
And,
Let the place where the cow, the buffalo and the horse are tied, are three sectors i.e. sector ADE, sector BFG and sector CHI
The area of triangular field by Heron's formula =

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

$s=\frac{a+b+c}{2}=\frac{15+16+17}{2}=\frac{48}{2}=24cm$

$\sqrt{24(24-15)(24-16)(24-17)}=\sqrt{24\times 9\times 8\times 7}=\sqrt{12096}$

Area of triangular field = 109.98 sq m
Area of the grazed part = Area of the triangular field = Area of the sector ADE + Area of sector BFG + Area of sector CHI

$=\pi \times 7^2\times \frac{\angle A}{360}+\pi \times 7^2\times \frac{\angle B}{360}+\pi \times 7^2\times \frac{\angle C}{360}$

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

$=\frac{22}{7}\times 7^2\times \frac{180}{360}=\frac{154}{2}$

Area of the grazed part = 77 sq m
Now, the area of the field which cannot be grazed by these animals
= 109.98 - 77
= 32.98 sq m