Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm in length

Solution:

Let the sides of the triangle be ​a = 20 cm, b = 34 cm and c = 42 cm.
Let s be the semi-perimeter of the triangle.

$s=\frac{1}{2}(a+b+c)$

$s=\frac{1}{2}(20+34+42)$

$s=48 \mathrm{~cm}$

Area of the triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$\Rightarrow \sqrt{48(48-20)(48-34)(48-42)}$

$\Rightarrow \sqrt{48 \times 28 \times 14 \times 6}$

$\Rightarrow \sqrt{112896}$

$\Rightarrow 336 \mathrm{~cm}^{2}$

Length of the longest side is 42 cm.

Area of a triangle $=\frac{1}{2} \times b \times h$

$\Rightarrow 336=\frac{1}{2} \times 42 \times h$

$\Rightarrow 672=42 h$

$\Rightarrow \frac{672}{42}=h$

$\Rightarrow h=16 \mathrm{~cm}$

The height corresponding to the longest side is 16 cm.