Let the sides be 'a' and 'b' and the hypotenuse 'c'.

According to the question:

2(ab) = c^2

Now .5(ab) = 32 (Area). therefore, ab = 64

from the above 2 equation, c = 8sqrt(2)

Now, c^2 = a^2 + b^2
2ab = (a+b)^2 - 2ab
4ab = (a+b)^2
(a+b) = 16

Now a+b = 16 and ab=64, therefore, a = b = 8

perimeter = 16+8sqrt(2)

Is there a faster method of solving this?

could you explain in detail

how did u get .5(ab) = 32 ?

sunil_snath wrote:Let the sides be 'a' and 'b' and the hypotenuse 'c'.

According to the question:

2(ab) = c^2

Now .5(ab) = 32 (Area). therefore, ab = 64

from the above 2 equation, c = 8sqrt(2)

Now, c^2 = a^2 + b^2
2ab = (a+b)^2 - 2ab
4ab = (a+b)^2
(a+b) = 16


Now a+b = 16 and ab=64, therefore, a = b = 8

perimeter = 16+8sqrt(2)

Is there a faster method of solving this?

bhumika.k.shah wrote:could you explain in detail

how did u get .5(ab) = 32 ?

Area = 1/2 * base * height

For a right angle triangle, the base and height are simply the non-hypotenuse sides, in this case a and b.

So:

Area = 1/2(a)(b) = .5(ab)

To review the other calculations:

2ab = a^2 + b^2
0 = a^2 - 2ab + b^2

For the GMAT, it's good to train yourself to recognize the two special quadratics, perfect squares and differences of squares. Here we have a classic perfect square, which reduces to:

0 = (a-b)^2

giving us:

a = b

So, a=b and ab = 64; accordingly: a=b=8 (for geometry we can ignore negative solutions).