# The base of an isosceles acute-angled triangle is 48, and the radius of the circumscribed circle around it is 25.

**The base of an isosceles acute-angled triangle is 48, and the radius of the circumscribed circle around it is 25. Find the distance between the centers of the inscribed and circumscribed circles of the triangle.**

The radius of the circumscribed circle is: R = AC / 2 * SinB.

SinB = AC / 2 * R = 48/50 = 24/25.

Let’s define CosB.

Cos2B = 1 – Sin2B = 1 – 576/625 = 49/625.

CosB = 7/25.

Let’s apply the cosine theorem.

AC ^ 2 = AB ^ 2 + BC ^ 2 – 2 * AB * BC * CosB = 2 * AB ^ 2 – 2 * AB ^ 2 * CosB = 2 * AB ^ 2 * (1 – CosB).

2304 = 2 * AB ^ 2 * (1 – 7/25).

AB ^ 2 = 2304/2 * (18/25) = 1600.

AB = 40 cm.

Determine the area of the triangle ABC.

Savs = AB * BC * SinB / 2 = (40 * 40 * 24/25) / 2 = 768 cm2.

Then the radius of the inscribed circle is equal to: O1K = O1H = 2 * S / P = 2 * 768/128 = 12 cm.

In a rectangular triangle OCH: OH ^ 2 = OC ^ 2 – CH ^ 2 = 625 – 576 = 49. OH = 7 cm.

Then OO1 = O1H – OH = 12 – 7 = 5 cm.

Answer: Between the centers of the circles is 5 cm.