The length of the hypotenuse of a right triangle is $h$ , and the radius of the inscribed circle is $r$. The ratio of the area of the circle to the area of the triangle is
$\textbf{(A) }\frac{\pi r}{h+2r}\qquad \textbf{(B) }\frac{\pi r}{h+r}\qquad \textbf{(C) }\frac{\pi}{2h+r}\qquad \textbf{(D) }\frac{\pi r^2}{r^2+h^2}\qquad \textbf{(E) }\text{none of these}$

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 $\boxed{\frac{\pi r}{h+r}}$.
\frac{\pir}{h+r}
Level 5
Geometry
Problem: The length of the hypotenuse of a right triangle is $h$ , and the radius of the inscribed circle is $r$. The ratio of the area of the circle to the area of the triangle is
$\textbf{(A) }\frac{\pi r}{h+2r}\qquad \textbf{(B) }\frac{\pi r}{h+r}\qquad \textbf{(C) }\frac{\pi}{2h+r}\qquad \textbf{(D) }\frac{\pi r^2}{r^2+h^2}\qquad \textbf{(E) }\text{none of these}$

Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the ratio of the area of the inscribed circle to the area of the triangle, we need to establish the formulas for both areas. The area of the inscribed circle is given by the formula A_circle = πr^2, where r is the radius of the circle. 

Next, we need to determine the area of the right triangle. The area A_triangle can be expressed in terms of the hypotenuse h and the radius r of the inscribed circle. For a right triangle, the area can also be calculated using the formula A_triangle = r * s, where s is the semi-perimeter of the triangle. The semi-perimeter s can be expressed as s = (a + b + h) / 2, where a and b are the lengths of the other two sides of the triangle.

Using the relationship between the radius of the inscribed circle and the area of the triangle, we can derive that A_triangle = r * s = r * (a + b + h) / 2. However, we can also express the area of the triangle in terms of the hypotenuse and the radius. 

For a right triangle, the relationship between the radius of the inscribed circle and the hypotenuse is given by r = (A_triangle / s). Substituting the area of the triangle into this equation allows us to express the area in terms of r and h. 

Now, we can set up the ratio of the areas: 

Ratio = A_circle / A_triangle = (πr^2) / (A_triangle). 

Substituting A_triangle with the expression derived from the relationship with r and h leads us to simplify the ratio. After simplification, we find that the ratio can be expressed as πr / (h + r). 

Thus, the final answer is confirmed to be πr / (h + r).<|END_OF_LATENT|> The final answer is $\frac{\pir}{h+r}$.