In trapezoid ABCD, angle A = 60, D = 45? base BC = 3 cm, BF and CF – trapezoid heights. ED = 4 cm.

In trapezoid ABCD, angle A = 60, D = 45? base BC = 3 cm, BF and CF – trapezoid heights. ED = 4 cm. Find the S of the trapezoid.

Let’s draw the heights BF and CE. In a right-angled triangle CDE, the angle CDE = 45, then the triangle CDE is isosceles, CE = DE = 4 cm, then BF = 4 cm.

In a right-angled triangle ABF, we determine the length of the leg AF. tg60 = BF / AF.

AF = BF / tg60 = 4 / √3 cm.

Determine the length of the base AD.

AD = AF + FE + DE = 4 / √3 + 3 + 4 = 7 + 4 / √3 cm.

Determine the area of the trapezoid.

Savsd = (BC + AD) * CE / 2 = (3 + 7 + 4 / √3) * 4/2 = (10 + 4 / √3) * 2 = 20 + 8 / √3 cm2.

Answer: The area of the trapezoid is 20 + 8 / √3 cm2.




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