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

Let a, b and c be the sides of the triangle.

Let a = 18 cm, b = 10 cm.

Given,

Perimeter = 42 cm

∴ a + b + c = 42

⇒ 18 + 10 + c = 42

⇒ 28 + c = 42

⇒ c = 42 - 28 = 14 cm.

By formula,

Semi Perimeter (s) = $\dfrac{\text{Perimeter of triangle}}{2} = \dfrac{42}{2}$ = 21 cm.

By Heron's formula,

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

Substituting values we get :

$A = \sqrt{21(21 - 18)(21 - 10)(21 - 14)} \\[1em] = \sqrt{21 \times 3 \times 11 \times 7} \\[1em] = \sqrt{4851} \\[1em] = 21\sqrt{11} \text{ cm}^2$

Hence, area of triangle = $21\sqrt{11}$ cm².