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

Given:

Two sides of a triangle are \( 18 \mathrm{~cm} \) and \( 10 \mathrm{~cm} \).

The perimeter of the triangle is \( 42 \mathrm{~cm} \).

To find:

We have to find the area of the triangle.

Solution :

Let the length of the third side be $x$.

The perimeter of a triangle is the sum of all the sides of the triangle.

Therefore,

Perimeter$= 18+10+x$

$\Rightarrow x+28=42$

$\Rightarrow x=42-28$

$\Rightarrow x=14\ cm$

Here,

\( a=18 \mathrm{~cm}, b=10 \mathrm{~cm} \) and \( c=14 \mathrm{~cm} . \)
So, the semi perimeter of the triangle \( s=\frac{a+b+c}{2}=\frac{18+10+14}{2}=\frac{42}{2}=21 \mathrm{~cm} \)
Therefore, by Heron's formula,
\( \begin{aligned} A &=\sqrt{s(s-a)(s-b)(s-c)} \\ &=\sqrt{21(21-18)(21-10)(21-14)} \\ &=\sqrt{21(3)(11)(7)} \\ &=\sqrt{(7 \times 3)(3)(11)(7)} \\ &=7 \times 3 \sqrt{11}=21 \sqrt{11} \mathrm{~cm}^{2} \end{aligned} \)
Hence, the area of the triangle is \( 21 \sqrt{11} \mathrm{~cm}^{2} \).