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Question
Please how to solve the problem
a) Trapezoid ABCD, AD is parallel to BC, angle A = angle D - 45degrees, while angle B = angle C = 135 degrees. If AB = 6 and the area of the ABCD is 30 find BC.
b) In trapezoid ABCD AD is parallel to BC, and AB=6, BC=7, CD=8, AD=17. If sides AB and CD are entended to meet at E find the angle E.
The trapezoid can be divided into one triangle at each end (for a total of two triangles) and a rectangle in the middle.
If the length of AB = 6, the areas of both triangles at either end can be found. Since they are both 45-45-90 triangles, the area of both of them together adds to a square with side 6/√2. The area of that square is 36/2 = 18.
That means the area of the remaining rectangle in the middle is
30 - 18 = 12. Since the height of this rectangle is the leg of a triangle, the height is 6/√2. We can say that height*width = area of that rectangle. The height is known to be 6/√2. The area is known to be 18. The width I will say is w.
We known that (6/√2)w = 12, so w = (12/6)√2 = 2√2.
Thus, the area of the middle rectangle is (6/√2)(2√2) = 12.
When this is added to the 18 (which is the area of the triangles),
the result is 30, which was given. Doing this, I can say the answer is checked, and therefore correct.
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