The height of a rectangular trapezoid with an area of 50 cm2 and a perimeter of 32 cm is 5 cm

The height of a rectangular trapezoid with an area of 50 cm2 and a perimeter of 32 cm is 5 cm. Find the sides of the trapezoid.

Since ABCD is a rectangular trapezoid, AB is the height of the trapezoid. Hence AB = 5 cm. The area of the trapezoid is equal to half the sum of the bases multiplied by the height: S = ((BC + AD) / 2) * AB, 50 = ((BC + AD) / 2) * 5, we express BC + AD. BC + AD = (50/5) * 2 = 20 (cm). The perimeter of a trapezoid is the sum of all its sides: P = AB + BC + CD + AD, 32 = 5 + BC + AD + CD. Since BC + AD = 20 cm, then CD = 32-5-20 = 7 (cm).
Answer: the sides are 5 cm and 7 cm.