Find the area of a triangle whose perimeter is 22 cm, one side is 9 cm and the difference of other two sides is 3 cm.

Perimeter of triangle = 22 cm

One side = 9 cm

Difference between other two sides = 3 cm

Let the unknown sides be a and b, where a > b.

⇒ a - b = 3 ………..(1)

⇒ a + b = 22 - 9 = 13 ………..(2)

Adding equations (1) and (2), we get

$\begin{matrix} & a & - & b & = & 3 \ & a & + & b & = & 13 \ \hline & & & 2a& = & 16 \ & & & a & = & 8 \ \end{matrix}$

Substituting a = 8 in equation (2):

8 + b = 13

b = 13 - 8 = 5

So, the three sides of the triangle are 9 cm, 8 cm, and 5 cm.

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

where, s = $\dfrac{a + b + c}{2}$

s = $\dfrac{8 + 5 + 9}{2}$

= $\dfrac{22}{2}$

= 11

Area of the triangle:

$= \sqrt{11(11 - 8)(11 - 5)(11 - 9)}\\[1em] = \sqrt{11 \times 3 \times 6 \times 2}\\[1em] = \sqrt{396}\\[1em] = 6 \sqrt{11}$

Hence, the area of triangle = 6 $\sqrt{11}$ cm².