The perimeter of a triangular field is $540\ m$ and its sides are in the ratio $25 : 17 : 12$. Find the area of the triangle.

Given:

The perimeter of a triangular field is $540\ m$ and its sides are in the ratio $25 : 17 : 12$.

To do:

We have to find the area of the triangle.

Solution:

Let the sides of the triangular field be $25x, 17x$ and $12x$.

This implies,

$25x+17x+12x=540\ m$

$54x=540\ m$

$x=\frac{540}{54}$

$x=10\ m$

Therefore,

$a=25x=25(10)=250\ m$

$b=17x=17(10)=170\ m$

$c=12x=12(10)=120\ m$

$s=\frac{\text { Perimeter }}{2}$

$=\frac{540}{2}$

$=270$

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

$=\sqrt{270(270-250)(270-170)(270-120)}$

$=\sqrt{270 \times 20 \times 100 \times 150}$

$=\sqrt{90\times3\times20\times20\times5\times50\times3}$

$=\sqrt{9\times10\times3\times20\times20\times5\times5\times10\times3}$

$=3 \times 10 \times 3 \times 20 \times 5$

$=9000 \mathrm{~m}^{2}$

The area of the triangle is $9000\ m^2$.

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Updated on: 10-Oct-2022

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