AMC10 2017 B

AMC10 2017 B · Q15

AMC10 2017 B · Q15. It mainly tests Similarity, Pythagorean theorem.

Rectangle ABCD has $AB = 3$ and $BC = 4$. Point E is the foot of the perpendicular from B to diagonal AC. What is the area of $\triangle ADE$?

矩形ABCD有$AB = 3$，$BC = 4$。点E是从B到对角线AC的垂足。$ riangle ADE$的面积是多少？

(A) 1 1

(B) \frac{42}{25} \frac{42}{25}

(D) 2 2

(E) \frac{54}{25} \frac{54}{25}

Answer

Correct choice: (E)

正确答案：（E）

Solution

Triangles ADE and ABE have the same area because they share the base AE and, by symmetry, they have the same height. By the Pythagorean Theorem, AC = 5. Because $\triangle ABE \sim \triangle ACB$, the ratio of their areas is the square of the ratio of their corresponding sides. Their hypotenuses have lengths 3 and 5, respectively, so their areas are in the ratio 9 to 25. The area of $\triangle ACB$ is half that of the rectangle, so the area of $\triangle ABE$ is (9/25) · 6 = 54/25. Thus the area of $\triangle ADE$ is also 54/25.

$ riangle ADE$和$ riangle ABE$有相同面积，因为它们共享底AE，且对称有相同高。由勾股定理，AC = 5。因为$ riangle ABE \sim \triangle ACB$，面积比为对应边比的平方。斜边分别为3和5，面积比9比25。$ riangle ACB$面积为矩形一半，故$ riangle ABE$面积为(9/25) · 6 = 54/25。因此$ riangle ADE$面积也为54/25。

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