he length of the hypotenuse of a right triangle is hh , and the radius of the inscribed circle is rr. the ratio of the area of the circle to the area of the triangle is
Solution by e_power_pi_times_i
Since rs = A, where r is the inradius, s is the semiperimeter, and A is the area, we have that the ratio of the area of the circle to the area of the triangle is $\frac{\pi r^2}{rs} = \frac{\pi r}{s}$. Now we try to express s as h and r. Denote the points where the incircle meets the triangle as X,Y,Z, where O is the incenter, and denote AX = AY = z, BX = BZ = y, CY = CZ = x. Since XOZB is a square (tangents are perpendicular to radius), r = BX = BZ = y. The perimeter can be expressed as 2(x+y+z), so the semiperimeter is x+y+z. The hypotenuse is AY+CY = z+x. Thus we have s = x+y+z = (z+x)+y = h+r. The answer is (B) 3,14r/h+r