The difference between the sides at right angle in a right-angled triangle is 7 cm.

Solution:

Given:

Area of the triangle $=60 \mathrm{~cm}^{2}$

Let the sides of the triangle be a, b and c, where a is the height, b is the base and c is hypotenuse of the triangle.

$a-b=7 \mathrm{~cm}$

$a=7+b \ldots \ldots .(1)$

Area of triangle $=\frac{1}{2} \times b \times h$

$\Rightarrow 60=\frac{1}{2} \times b \times(7+b)$

$\Rightarrow 120=7 b+b^{2}$

$\Rightarrow b^{2}+7 b-120=0$

$\Rightarrow(b+15)(b-8)=0$

$\Rightarrow b=-15$ or 8

Side of a triangle cannot be negative.
Therefore, b = 8 cm.

Substituting the value of b = 8 cm, in equation (1):
a = 7+8 = 15 cm

Now,  a = 15 cm, b = 8 cm

Now, in the given right triangle, we have to find third side.

$(\mathrm{Hyp})^{2}=(\text { First side })^{2}+(\text { Second side })^{2}$

$\Rightarrow \mathrm{Hyp}^{2}=8^{2}+15^{2}$

$\Rightarrow \mathrm{Hyp}^{2}=64+225$

$\Rightarrow \mathrm{Hyp}^{2}=289$

$\Rightarrow \mathrm{Hyp}=17 \mathrm{~cm}$

So, the  third side is 17 cm

Perimeter of a triangle $=a+b+c$.

Therefore, required perimeter of the triangle $=15+8+17=40 \mathrm{~cm}$.