One of the sides of a triangle inscribed in a circle is equal to the diameter, the other two sides are equal to 9 and 12.

One of the sides of a triangle inscribed in a circle is equal to the diameter, the other two sides are equal to 9 and 12. Find the radius of the circle.

Let the inscribed triangle ABC and the side AC be equal to the diameter.

Lemma: if the chord in the circle is equal to the diameter, then it passes through the center of the circle.

Proof: Suppose there is a chord EF that is equal to the diameter and does not pass through the center of the circle. Then consider the triangle EOF. EO and FO are equal to the radius of the circle R. Since any side of a triangle is always less than the sum of the other two sides, we get that EF <EO + FO = R + R = 2 * R = D, where D is the diameter of the circle. We got a contradiction.

The lemma is proved.

It follows from the proved lemma that the side AC passes through the center of the circle.

Therefore, the angle ABC is based on an arc of 180 degrees and is equal to 90 degrees.

Hence it follows that triangle ABC is right-angled with right angle B.

Then, by the Pythagorean theorem:

AC ^ 2 = AB ^ 2 + BC ^ 2 = 9 ^ 2 + 12 ^ 2 = 225 = 15 ^ 2,

AC = 15.

Therefore, the radius of the circle is AC / 2 = 7.5.

Answer: 7.5.