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 the sides of the triangle be a = 18 cm, b = 10 cm and c = ?
∴ Perimeter (2s) = 42 cm
$\Rightarrow s=\frac{42}{2}=21$ cm
$c=42-(18+10)$ cm
∴ Area of the given triangle
$=\sqrt{21(21-18)(21-10(21-14\left)\right)}c{m}^2$
$=\sqrt{21\times 3\times 11\times 7}c{m}^2$
$21\sqrt{11}c{m}^2$
Thus, the required area of the triangle $=21\sqrt{11}c{m}^2$