Find the area of an isosceles trapezoid, the base of which is 12 cm and 8 cm, the lateral edge

Find the area of an isosceles trapezoid, the base of which is 12 cm and 8 cm, the lateral edge is 10 cm, the acute angle at the lower base is 45 degrees.

From the top B of the trapezoid, we will build the height BH

Since the trapezoid is isosceles, the height drawn to the larger base divides it into two segments, the length of the smaller of which is equal to the half-difference of the dynes of its bases.

AH = (AD – BC) / 2 = (12 – 8) / 2 = 4/2 = 2 cm.

In a right-angled triangle ABH, the acute angle BAH, by condition, is equal to 45, then the triangle ABH is rectangular and isosceles, BH = AH = 2 cm.

The area of the trapezoid is:

Savsd = (BC + AD) * BH / 2 = (8 + 12) * 2/2 = 20 cm2.

Answer: The area of the trapezoid is 20 cm2.