The difference between the sides at right angle in a right-angled triangle is 7 cm. The area of the triangle is $60c{m}^2$. Find its perimeter.

Let x cm be the one of the sides, then $(x-7)$ cm be another side.
Area of triangle $=60c{m}^2$ (given)
We know, Area of triangle $=\frac{1}{2}(Base\times height)$
$\Rightarrow 60=\frac{1}{2}\times x\times (x-7)$
$\Rightarrow 120=x^2-7x$
or $x^2-7x-120=0$
Solving above equation, we have
$(x-15)(x+8)=0$
$x=15$ or $x=-8$
Since length measure cannot be negative, so neglect $x=-8$
One side $=15cm$
Another Side $=x-7=15-7=8cm$
Apply Pythagoras theorem:
Hypotenuse${}^2=$ Base${}^2+$ Perpendicular${}^2$
Hypotenuse${}^2=\sqrt{(15^2+8^2)}$
Hypotenuse $=\sqrt{289}=17$
Therefore,
perimeter of triangle $=$ Sum of all the sides $=(15+8+17)cm=40cm$