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Q5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.

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

Let us consider the length of one side of the triangle be $a = 12x$

Then, the remaining two sides are $b = 17x$ and $c = 25x.$

So, by the given perimeter, we can find the value of x:

$Perimeter = a+b+c = 12x+17x+25x = 540cm$

$\implies 54x = 540cm$

$\implies x = 10$

So, the sides of the triangle are:

$a = 12\times10 =120 cm$

$b = 17\times10 =170 cm$

$c = 25\times10 =250 cm$

So, the semi perimeter of the triangle is given by

$s = \frac{540cm}{2} = 270cm$

Therefore, using Heron's Formula, the area of the triangle is given by

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

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

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

$= \sqrt{81000000} = 9000cm^2$

Hence, the area of the triangle is $9000cm^2$ .

Posted by

Divya Prakash Singh

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