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

Let the sides of a triangle a = 18 cm, b = 10 cm and c
We have, perimeter = 42 cm
So, a + b + c = 42
By substituting the values,
18 + 10 + c = 42
28 + c = 42cm
c = 42 - 28 cm
Thus, c = 14cm
Now, we know that, $s=\frac{a+b+c}{2}$
Therefore, $s=\frac{18+10+14}{2}=\frac{42}{2}=21cm$

Now, Area of triangle = $\sqrt{21\left(21?18\right)\left(21?10\right)\left(21?14\right)}$ ...(Since, Heron's formula [area = $\sqrt{s\left(s?a\right)\left(s?b\right)\left(s?c\right)}$])
= $\sqrt{21×3×11×7}$
= $\sqrt{7×3×3×11×7}$
= $21\sqrt{11}{cm}^{2}$