In an isosceles trapezoid ABCD, the lateral side AB = 13 cm, and the bases are 7 and 31. Find the area of the trapezoid.

Let us construct from the vertices B and C heights BH and СK to the base of AD.

Triangles ABН and CDK are rectangular in which the hypotenuses AB and CD are equal since the trapezium ABCD is isosceles, the angle BAН = CDK as angles at the base of an isosceles trapezoid. Then the triangle ABН is equal to the triangle CDH in hypotenuse and acute angle, and then AH = DK.

Quadrangle НВСК is a rectangle, then КН = ВС = 7 cm.

Then АН = DH = (АD – НК) / 2 = (31 – 7) / 2 = 12 cm.

From the right-angled triangle ABН, according to the Pythagorean theorem, BH ^ 2 = AB ^ 2 – AH ^ 2 = 169 – 144 = 25.

BH = 5 cm.

Then Str = (ВС + АD) * ВН / 2 = (7 + 31) * 5/2 = 95 cm2.

Answer: The area of ​​the trapezoid is 95 cm2.