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

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

Solution:

Whenever we are given the measurements of all sides of a triangle, we basically look for Heron’s formula to find out the area of the triangle.

If we denote area of the triangle by A, then the area of a triangle having sides a, b, c and s as semi-perimeter is given by:

$A=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$

Where, s = (a + b + c)/2

We are given:

a = 18 cm

b = 10 cm, and perimeter = 42 cm

We know that perimeter = 2s,

So, 2s = 42

Therefore, s = 21 cm

We know that, s = (a + b + c)/2

21 = (18 + 10 + c)/2

42 = 28 + c

c = 14 cm

So the area of the triangle is:

$A=\sqrt{21 \times(21-18) \times(21-10) \times(21-14)}$

$A=\sqrt{21 \times(3) \times(11) \times(7)}$

$A=\sqrt{4851}$

$A=21 \sqrt{11} \mathrm{~cm}^{2}$