In the following figure, ACB is a right-angled triangle. AD is the altitude. Circles are inscribed within the triangle ACD and triangle BCD. P and Q are the centers of the circles. The distance PQ is

The length of AB is 15 m and AC is 20 m

By Pythagoras theorem we get BC = 25 . Let BD = x

Triangle ABD is similar to triangle CBA => AD/x = 20/15 

Also triangle ADC is similar to triangle ABC => AD/(25-x) = 15/20

From the 2 equations, we get x = 9, DC = 16 and AD = 12

We know that AREA = (semi perimeter ) * inradius

For triangle ABD, Area = 1/2 x BD X AD = 1/2 x 9 x 12 = 54 and semi perimeter = (15 + 9 + 12)/2 = 18. On using the above equation we get, inradius, r = 3.

Similarly for triangle ADC we get inradius R = 4 .

PQ = R + r = 7 cm