AMC10 2013 B

AMC10 2013 B · Q16

AMC10 2013 B · Q16. It mainly tests Triangles (properties), Pythagorean theorem.

In $\triangle ABC$, medians $\overline{AD}$ and $\overline{CE}$ intersect at $P$, $PE = 1.5$, $PD = 2$, and $DE = 2.5$. What is the area of $\triangle AEDC$?

在 $\triangle ABC$ 中，中线 $\overline{AD}$ 和 $\overline{CE}$ 相交于 $P$，$PE = 1.5$，$PD = 2$，且 $DE = 2.5$。求 $\triangle AEDC$ 的面积。

(A) 13 13

(B) 13.5 13.5

(D) 14.质5 14.5

(E) 15 15

Answer

Correct choice: (B)

正确答案：（B）

Solution

The ratio of PE:PD:DE is 3:4:5. Hence by the converse of the Pythagorean Theorem, $\triangle DPE$ is a right triangle. Therefore CE is perpendicular to AD, and the area of $\triangle AEDC$ is one-half the product of its diagonals. Because P is the centroid of $\triangle ABC$, it follows that CE = 3(PE) = 4.5 and AD = 3(PD) = 6. Therefore the area of $\triangle AEDC$ is 0.5(4.5)(6) = 13.5.

$PE:PD:DE=3:4:5$。因此由勾股定理的逆命题，$\triangle DPE$ 是直角三角形。因此 $CE$ 垂直于 $AD$，$\triangle AEDC$ 的面积是其对角线乘积的一半。因为 $P$ 是 $\triangle ABC$ 的质心，故 $CE=3(PE)=4.5$，$AD=3(PD)=6$。因此 $\triangle AEDC$ 的面积为 $0.5(4.5)(6)=13.5$。

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