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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 $9000\ cm^2$.