Find the area of an isosceles trapezoid with bases of 5 cm and 9 cm and an angle of 45 degrees.

In an isosceles trapezoid, the sides are of equal length, and the angles at the bases are equal.

Segments АН and КD also have equal length, and the segment НК is equal to the base ВС:

AH = KD = (AD – BC) / 2;

AH = KD = (9 – 5) / 2 = 4/2 = 2 cm.

Consider a triangle ABH formed by the height BH. Since this triangle is rectangular, using the tangent of angle A, you can find the height of the BH:

tg A = BH / AH;

BH = AH · tg A;

tg 45 ° = 1;

BH = 2 1 = 2 cm.

The area of ​​the trapezoid is equal to the product of the half-sum of its bases by the height:

S = (BC + AD) / 2 · BH;

S = (5 + 9) / 2 2 = 14 cm2.

Answer: the area of ​​the trapezoid is 14 cm2.