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,

\(\Rightarrow \) 18 + 10 + c = 42
\(\Rightarrow \) 28 + c = 42cm
\(\Rightarrow \) c = 42 - 28 cm
\(\Rightarrow \) 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
= \(\sqrt{21(21 - 18)(21 - 10)(21 - 14)}\)
(Since, Heron's formula [area = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

= \(\sqrt{21 × 3 × 11 × 7}\)
= \(\sqrt{7 × 3 × 3 × 11 × 7}\)
= \(21\sqrt{11} {cm}^2\)