Find the area of an isosceles trapezoid, the bases of which are 8 and 14 cm, and the lateral side is 5 cm.

The area of ​​the trapezoid is equal to the product of its midline by the height:

S = m * h.

The middle line is equal to the half-sum of the length of the bases of the trapezoid:

m = (BC + AD) / 2;

m = (8 + 14) / 2 = 22/2 = 11 cm.

The segment of the larger base located between the heights is equal to the length of the smaller base

HK = BC, and the other two segments AH and KD are equal, since the trapezoid is isosceles.

AH = KD = (AD – BC) / 2;

AH = KD = (14 – 8) / 2 = 6/2 = 3 cm.

Considering the triangle ABH, we can find the height BH by the Pythagorean theorem:

AB ^ 2 = BH ^ 2 + AH ^ 2;

BH ^ 2 = AB ^ 2 – AH ^ 2;

BH ^ 2 = 5 ^ 2 – 3 ^ 2 = 25 – 9 = 16;

BH = √16 = 4 cm.

S = 11 4 = 44 cm2.

Answer: the area of ​​the trapezoid is 44 cm2.