AMC10 2012 B

AMC10 2012 B · Q19

AMC10 2012 B · Q19. It mainly tests Similarity, Area & perimeter.

In rectangle ABCD, AB = 6, AD = 30, and G is the midpoint of AD. Segment AB is extended 2 units beyond B to point E, and F is the intersection of ED and BC. What is the area of BFDG ?

在矩形ABCD中，AB = 6，AD = 30，G为AD中点。将线段AB向B外延长2单位至点E，F为ED与BC交点。BFDG的面积是多少？

(A) 133/2 133/2

(B) 67 67

(D) 68 68

(E) 137/2 137/2

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Because $\triangle EBF$ is similar to $\triangle EAD$, it follows that $\dfrac{BF}{AD}=\dfrac{BE}{AE}$, or $\dfrac{BF}{30}=\dfrac{2}{8}$, giving $BF=\dfrac{15}{2}$. The area of trapezoid $BFDG$ is $\dfrac{1}{2}h(b_1+b_2)=\dfrac{1}{2}\cdot AB\cdot(BF+GD)=\dfrac{1}{2}\cdot 6\cdot\left(\dfrac{15}{2}+15\right)=\dfrac{135}{2}$.

答案（C）：因为$\triangle EBF$与$\triangle EAD$相似，所以有$\dfrac{BF}{AD}=\dfrac{BE}{AE}$，即$\dfrac{BF}{30}=\dfrac{2}{8}$，从而$BF=\dfrac{15}{2}$。梯形$BFDG$的面积为$\dfrac{1}{2}h(b_1+b_2)=\dfrac{1}{2}\cdot AB\cdot(BF+GD)=\dfrac{1}{2}\cdot 6\cdot\left(\dfrac{15}{2}+15\right)=\dfrac{135}{2}$。

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