Find the area of a triangle two sides of which are $ 18 \mathrm{~cm} $ and $ 10 \mathrm{~cm} $ and the perimeter is $ 42 \mathrm{~cm} $.

Given:

The sides of the triangle are $18\ cm$ and $10\ cm$ and the perimeter is $42\ cm$.

To do:

We have to find the area of the triangle.

Solution:

Let us assume the third side of the triangle as $x$

This implies,

The three sides of the triangle are $18\ cm, 10\ cm$ and $x\ cm$.

We have,

The perimeter of the triangle as $42\ cm$

We know that,

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

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

This implies,

$42\ cm=18\ cm+10\ cm+x\ cm$

$42\ cm=28\ cm+x\ cm$

This implies,

$x\ cm=42\ cm-28\ cm$

$x\ cm=14\ cm$

By Heron's formula:

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

Since,

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

$S=\frac{18+10+12}{2}$

$S=\frac{42}{2}$

$S=21\ cm$

This implies,

$A=\sqrt{21(21-18)(21-10)(21-14)}$

$A=\sqrt{21(3)(11)(7)}$

$A=21\sqrt{11}\ cm^2$

Therefore,

The area of the triangle is $21\sqrt{11}\ cm^2$.