A circle inscribed in a rectangular trapezoid divides the large lateral side by the point of contact into 3cm

A circle inscribed in a rectangular trapezoid divides the large lateral side by the point of contact into 3cm and 12cm segments. Find the radius of the inscribed circle if P = 54cm.

Since a circle is inscribed in the trapezoid ABCD, then for such a circle the sum of the lengths of the bases of the trapezoid is equal to the sum of the lengths of its lateral sides.

Side length CD = CM + DM = 3 + 12 = 15 cm.

Then (AB + CD) = (BC + AD).

2 * (AB + CD) = Ravsd = 54 cm.

2 * (AB + 15) = 54.

2 * AB = 54 – 30 = 24.

AB = 24/2 = 12 cm.

Since trapezoid ABCD is rectangular, its height is equal to the diameter of the inscribed circle.

Then R = D / 2 = AB / 2 = 12/2 = 6 cm.

Answer: The radius of the inscribed circle is 6 cm.