The extensions of the lateral sides AB and CD of the trapezoid ABCD meet at point P. The area of the triangle APD

The extensions of the lateral sides AB and CD of the trapezoid ABCD meet at point P. The area of the triangle APD is 80. Find the area of the trapezoid if it is known that BC: AD = 3: 4.

Triangles ABD and BPC are similar in two angles, since the angle PBC = PAD, and PCB = PDA, as the corresponding angles at the intersection of parallel AD and BC.

The ratio of the areas of similar triangles is equal to the square of the similarity coefficient.

Svrs / Sard = (3/4) 2.

Svrs / 80 = 9/16.

Svrs = 80 * 9/16 = 45 cm2.

Then the area of the trapezoid is equal to: Savsd = Sapd – S bpc = 80 – 45 = 35 cm2.

Answer: The area of the trapezoid is 35 cm2.