Inscribe a circle and label its radius, r.

What is the ratio of the area of the circle to that of the triangle?

This is going be hard without a picture. Imagine a right triangle ABC where C is the right angle. Let the orientation be that C is at the bottom left and A is at the right and B is at the top. The sides are a, b, and h.

Draw the inscribed circle along with the radii to the sides of the triangle. The picture can be seen as a circle with 3 pairs of external tangents to the circle. And external tangents originating from the same point are congruent (proved by SSS for congruent triangles).

For the legs, the radii break them into (r, a-r) and (r, b-r). The hypotenuse is broken into two segments too. Because the external tangents are congruent, their lengths turn out to be a-r and b-r. So we have (a-r)+(b-r)=h or a+b=2r+h.

Squaring that we get a^2+b^2+2ab=4r^2+4rh+h^2. Since a^2+b^2=h^2. We end up with 2ab=4r^2+4rh.

At=ab/2=r^2+rh
Ac=pi*r^2

Ac/At=(pi*r^2)/(r^2+rh)
    =pi*r/(r+h)

I hope I didn't make any calculation errors.

Edited on November 20, 2004, 7:40 pm

Posted by np_rt on 2004-11-20 18:34:17