Solution: For a right triangle with legs $ a $ and $ b $, and hypotenuse $ z $, the inradius is given by $ c = raca + b - z2 $. The area $ A $ of the triangle is $ rac12ab $. The area of the incircle is $ \pi c^2 $. Using the identity $ a^2 + b^2 = z^2 $, and expressing $ a + b $ in terms of $ z $ and $ c $, we find $ a + b = z + 2c $. Also, the semiperimeter $ s = raca + b + z2 = rac2z + 2c2 = z + c $. The area can also be written as $ A = r \cdot s = c(z + c) $. Therefore, the ratio becomes: - Abu Waleed Tea