Find the radius of a circle circumscribed about an isosceles triangle with base 96 and height 64 to it.

Let AB be the base of the isosceles triangle ABC, CD the height, O the center of the circumscribed circle (link to the figure below).
In an isosceles triangle, the height and median at the base coincide, so DB = AB / 2 = 48.
The center of the circumscribed circle lies at the intersection of the median perpendiculars. Since D divides the segment AB in half and the angle CDB is a straight line, O lies on the segment CD.
OB = OC = R is the radius of the circle.
OC + OD = CD = 64
OD = 64 – OC
OD ^ + DB ^ = OB ^ = OC ^
OC ^ – OD ^ = DB ^
(OC + OD) * (OC – OD) = DB ^ = 48 ^ = 2304
since OD + OC = 64, then
OC – OD = 2304/64 = 36
OC + OC – 64 = 36
OC = (64 + 36) / 2 = 50
Answer: 50