Math!!!

Assuming that 2√5  and 3√5  are the lengths of the legs......the hypotenuse is given by:

sqrt [ (2√5)^2  + (3√5)^2 ]  =  sqrt  [ 20 + 45]  = sqrt(65)

So....the perimeter is

2√5 + 3√5 + √65   =

5√5  + √65  =   about 19.24  units

I assume the 'bottom'  and the 'side' are not the hypotenuse

Then:   a^2 + b^2 = c^2  to find the hypotenuse

(3sqrt5)^2  +  (2sqrt5)^2  = c^2

45  +20  = c^2 = 65

c=sqrt65 =          so the perimeter is all of these sides added together

sqrt65 + 3sqrt5 + 2sqrt5  =  19.24

What is the perimeter to a right triangle with the bottom being 3√5 and the side 2√5 (use pythagorean theorem)

[3sqrt(5)]^2 + [2sqrt(5)]^2 =H^2

[9*5] + [4*5] =H^2

45 +20 =H^2

H =+or-sqrt(65)

The perimeter of the triangle=sqrt(45) + sqrt(20) + sqrt(65)=19.24 (rounded)