The lengths of the two sides of a right triangle containing the right angle differ by 2 cm. If the area of the triangle is $24c{m}^2$, find the perimeter of the triangle.

Let x cm be the one of the sides, then $(x-2)cm$ be another side.
Area of triangle $=24c{m}^2$ (given)

We know, Area of triangle $=\frac{1}{2}(Base\times height)$
$\Rightarrow 24=\frac{1}{2}\times x\times (x-2)$
$\Rightarrow 48=x^2-2x$
or $x^2-2x-48=0$

Solving above equation, we have
$(x+6)(x-8)=0$
$x=-6$ or $x=8$
Since length measure cannot be negative, so neglect $x=-6$
One side $=8cm$
Another Side $=x-2=8-2=6cm$
Apply Pythagoras theorem:
Hypotenuse${}^2=$ Base${}^2+$ Perpendicular${}^2$
Hypotenuse${}^2=\sqrt{(8^2+6^2)}$
Hypotenuse $=\sqrt{100}=10$
Therefore, perimeter of triangle $=$ Sum of all the sides $=(6+8+10)cm=24cm$