Suppose the right-angled trapezoid ABCD is folded three times, and the resulting figure is a right-angled trapezoid A'B'C'D', with an angle of 90 degrees and an angle of 60 degrees (you should be able to draw this picture).

From the question "If the area of the overlapping part between the right-angled trapezoid and the equilateral triangle obtained by the third folding is equal to half the area of the right-angled trapezoid", the area of the overlapping part is equal to half the area of the right-angled trapezoid ABCD, that is, the root number 3 of 2 1* is divided by 2 and then divided by 2 (as for how to calculate the area of this trapezoid, you should know).

Given that NC is equal to 8 and CC' is equal to 10, C'n is equal to 2.

Suppose that point M is on B'C' and the intersection of MP and C'D' is point Q.

It is known that the angle DCB is equal to 60 degrees and the angle D'C'B' is equal to 60 degrees. Because the triangle MPN is an equilateral triangle and the angle PMN is equal to 60 degrees, the triangle MC'Q is an equilateral triangle.

Let MN be a long, then the area of the triangle MQC' is (a-2) * the square of the root number 3 and then divided by 4 (you should calculate the area of this equilateral triangle).

According to the area of half of the right-angled trapezoid calculated earlier, we get:

The radical number 3 of (a-2) divided by 4=2 1* the radical number 3 divided by 2 divided by 2.

Solution a equals

The root sign is 2 1 and then +2 (by the way, your answer here is wrong, I calculated it myself, and the answer over there was right in our school, and now it's all my answer)