2019/013) The length of perimeter of a right $\triangle$ is 60 inches and the length of altitude perpendicular to the hypotenuse is 12 inches. find the sides of the $\triangle$

Let the sides be a,b,c where c is the hypotenuse. without loss of generality we assume $a>=b$

We have
$a+b+c= 60$
Or $60-c = a+b\cdots(1)$
$a^2 + b^2 = c^2\cdots(2)$
Area of the triangle $\frac{1}{2}ab = \frac{1}{2}12c$ or
$ab= 12c\cdots(3)$
From (1)
$(60-c)^2 = ( a + b)^2 = a^2 + 2ab + b^2 = (a^2+b^2) + 2 * 12c= c^2 + 24c$ (using (2) and (3))
Or $3600 - 120c + c^2  = c^2 + 24c$
Or $144c = 3600$ or $c=25$
From (1) $a+b= 35\cdots(4)$
From (3) $ab= 300\cdots(5)$
So we have $(a-b)^2 = (a+b)^2 - 4ab = 1225 - 1200 = 25$ (using (4) and (5))
So $a-b = 5\cdots(6)$
From (4) and (6) we have a = 20, b = 15,
So sides of triangle $15,20,25$