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# Sides of a triangle are in the ratio of 12:17:25 and its perimeter is $ 540 \mathrm{~cm} $. Find its area.

Given:

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

To find:

We have to find the area of the triangle.

Solution:

Let the common ratio between the sides of the triangle be $=\ a$

This implies,

The sides of the triangle as $12\ a, 17\ a$ and $25\ a$.

We know that,

Perimeter $P$ of a triangle with sides of length $a$ units, $b$ units and $c$ units

$P=(a+b+c)$units.

We have perimeter as $540\ cm$.

This implies,

$540\ cm=12\ a+17\ a+25\ a$

$540\ cm=54\ a$

This implies,

$a\ cm=\frac{540}{54}$

$a\ cm=10\ cm$

Therefore,

The sides of the triangle are $12\times10\ cm=120\ cm, 17\times\ 10\ cm=170\ cm$ and $25\times10\ cm=250\ cm$

By Heron's formula:

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

Since,

$S=\frac{a+b+c}{2}$

$S=\frac{120+170+250}{2}$

$S=\frac{540}{2}$

$S=270\ cm$

This implies,

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

$A=\sqrt{270(150)(100)(20)}$

$A=9000\ cm^2$

Therefore,

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