In trapezoid ABCD, angle A = 60 degrees, angle D = 45 degrees, base BC is 3 cm, BF and CE are trapezoid heights, ED = 4 cm. Find the area of the trapezoid.
The heights BF and CE divide the trapezoid into two right-angled triangles ABF and CDE and a rectangle BCEF. In a right-angled triangle CDE, one of the acute angles is 45, then the triangle is isosceles, CE = DE = 4 cm.
In a right-angled triangle ABF BF = CE = 4 cm.
Then AF = BF / tg60 = 4 / √3 = 4 * √3 / 3 cm.
Since BCEF is a rectangle, then EF = BС = 3 cm.
Then AD = AF + EF + DE = 4 * √3 / 3 + 7 cm.
Determine the area of the trapezoid.
Str = (BC + AD) * CE / 2 = (3 + 4 * √3 / 3 + 7) * 4/2 = 20 + 8 * √3 / 3 cm2.
Answer: The area of the trapezoid is 20 + 8 * √3 / 3 cm2.
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