Question

Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

Answer 1

Let a, b and c be the sides of the given triangle. Then
a = 18 cm, b = 10 cm and a + b + c = 42 cm
Substituting a = 18, b = 10 in a + b + c = 42, we get
18 + 10 + c = 42 $\Rightarrow c=42-28=14\mathrm{cm}$
Now
$s=\frac{a+b+c}{2}=\frac{18+10+14}{2}=21\mathrm{cm}$
Using Heron's formula, we get
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{21\left(21-18\right)\left(21-10\right)\left(21-14\right)} \\ =\sqrt{21\left(3\right)\left(11\right)\left(7\right)} \\ =21\sqrt{11}{\mathrm{cm}}^2 \end{gathered}$$
Hence, the area of the triangle is $21\sqrt{11}{\mathrm{cm}}^2$.