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

Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm in length. Hence, find the height corresponding to the longest side.

Solution:

Let :

$a=42 \mathrm{~cm}, b=34 \mathrm{~cm}$ and $c=20 \mathrm{~cm}$

$\therefore s=\frac{a+b+c}{2}=\frac{42+34+20}{2}=48 \mathrm{~cm}$

By Heron’s formula, we have :

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

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

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

$=\sqrt{4 \times 2 \times 6 \times 6 \times 7 \times 2 \times 7 \times 4}$

$=4 \times 2 \times 6 \times 7$

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

We know that the longest side is 42 cm.
Thus, we can find out the height of the triangle corresponding to 42 cm.
We have:

Area of triangle $=336 \mathrm{~cm}^{2}$

$\Rightarrow \frac{1}{2} \times$ Base $\times$ Height $=336$

$\Rightarrow$ Height $=\frac{336 \times 2}{42}=16 \mathrm{~cm}$