How to write 20 12 Wenzhou Mathematics 16? You have to understand for yourself. Don't analyze it, it's too complicated.

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That is to say, the point DE is vertical.

Triangle QFE, triangle EDA and triangle DGP are all similar (congruence angle formed by right angle and parallel line).

Because PD: QE = 9: 4

According to the similarity, GP: EF = 9: 4 (the similarity between triangle QFE and triangle DGP) can be obtained.

Let gp = 9k and ef = 4k.

The coordinates of ABCDE5 point are expressed by ef, and the shadow area is also expressed by K.

Then the equation is composed of EA:DG=DA:PG.

Available k 4 = 1/36

That is, k 2 = 1/6.

Then bring it into the calculation = 13/3.