The smaller main and lateral sides of the isosceles trapezoid are 24 and 12 cm

| June 23, 2021 | education

The smaller main and lateral sides of the isosceles trapezoid are 24 and 12 cm, respectively. Find the area of a trapezoid if its acute angle is 60 degrees.

Let’s denote the trapezoid ABCD, BC = 24 cm – smaller base, AB = CD = 12 cm – lateral sides, angle A = 60 °.
From the top B to the bottom base, let us lower the BH height. Consider a right-angled triangle ABH, in which the angle B = 90 ° – 60 ° = 30 °. The AH leg is located opposite an angle of 30 ° and is equal to:
AH = 1/2 * AB = 6 (cm).
We find the VN leg, it is also the height of the trapezoid:
BH = √ (AB² – AH²) = √ (144 – 36) = √108 = 6√3 (cm).
Find a larger base AD:
AD = BC + 2 * AH = 24 + 2 * 6 = 36 (cm).
We find the area of the trapezoid:
S = 1/2 * (a + b) * h = 1/2 * (BC + AD) * BH = 1/2 * (24 + 36) * 6√3 = 180√3 (cm²).
Answer: the area of the trapezoid is 180√3 cm².



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