# Each of the lateral sides and the smaller base of the trapezoid is 5 cm

**Each of the lateral sides and the smaller base of the trapezoid is 5 cm, and one of the corners is 60 degrees. Find the radius of the circle around it.**

Let’s build the height BH. In a right-angled triangle ABH, the angle ABH = 90 – 60 = 30, then the leg AH lies opposite the angle 30, then AH = AB / 2 = 5/2 = 2.5 cm.

The length of the segment AH = (AD – BC) / 2.

AD = 2 * AH + BC = 5 + 5 = 10 cm.

In the triangle ABD, according to the cosine theorem, BD ^ 2 = AB ^ 2 + AD ^ 2 – 2 * AB * BD * Cos60 = 25 + 100 – 2 * 5 * 10 * (1/2) = 125 – 50 = 75.

AD ^ 2 – AB ^ 2 = 100 – 25 = 75.

Then the triangle ABD is rectangular, and then the radius of the circumscribed circle is equal to half of the side AD.

R = АD / 2 = 10/2 = 5 cm.

Answer: The radius of the circumscribed circle is 5 cm.