Question

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

Answer 1

Let the third side of the triangle be $x.$

Perimeter of the given triangle = 42 cm

$18cm+10cm+x=42$

$\therefore x=14cm$

$s=\frac{Perimeter}{2}=\frac{42cm}{2}=21cm$

By Heron’s formula,
Area of a triangle = $\sqrt{s(s-a)(s-b)(s-c)}$

Area of the given triangle = $\sqrt{s(s-a)(s-b)(s-c)}=\left(\sqrt{21(21-18)(21-10)(21-14)}\right)c{m}^2=\left(\sqrt{21\left(3\right)\left(11\right)\left(7\right)}\right)c{m}^2=21\sqrt{11}c{m}^2$