In trapezoid ABCD (AD and BC-bases), the diagonals intersect at point O, AD = 12 cm, BC = 4 Find SABCD if SAOD = 45.

Triangles AOD and BOC are similar in two angles, since the angle BOC is equal to the angle AOD as vertical angles, the angle OBC is equal to the angle ODA as criss-crossing angles.

The coefficient of similarity of triangles is: K = BC / AD = 4/12 = 1/3.

Then Svos / Sаod = K2 = 1/9.

Swax = 45/9 = 5 cm2.

Let us determine the heights OK and OH of the triangles BOS and AOD.

Swax = BC * OK / 2.

OK = 2 * Svos / BC = 2 * 5/4 = 2.5 cm.

OH = 2 * Saod / AD = 2 * 45/12 = 7.5 cm.

KH = 2.5 + 7.5 = 10 cm.

Then Savsd = (BC + AD) * KН / 2 = (4 + 12) * 10/2 = 80 cm2.

Answer: The area of the trapezoid is 80 cm2.