AMC10 2005 B

AMC10 2005 B · Q23

AMC10 2005 B · Q23. It mainly tests Area & perimeter, Polygons.

In trapezoid ABCD we have AB parallel to DC, E as the midpoint of BC, and F as the midpoint of DA. The area of ABEF is twice the area of FECD. What is AB/DC?

梯形ABCD中，AB平行的DC，E为BC中点，F为DA中点。区域ABEF的面积是FECD面积的两倍。AB/DC是多少？

(A) 2 2

(B) 3 3

(D) 6 6

(E) 8 8

Answer

Correct choice: (C)

正确答案：（C）

Solution

(C) First note that $FE=\dfrac{AB+DC}{2}$. Because trapezoids $ABEF$ and $FECD$ have the same height, the ratio of their areas is equal to the ratio of the averages of their parallel sides. Since \[ AB+\dfrac{AB+DC}{2}=\dfrac{3AB+DC}{2} \] and \[ \dfrac{AB+DC}{2}+DC=\dfrac{AB+3DC}{2}, \] we have \[ 3AB+DC=2(AB+3DC)=2AB+6DC,\quad \text{and}\quad \dfrac{AB}{DC}=5. \]

（C）首先注意到 $FE=\dfrac{AB+DC}{2}$。因为梯形 $ABEF$ 和 $FECD$ 的高相同，它们的面积之比等于它们平行边平均数之比。由于 \[ AB+\dfrac{AB+DC}{2}=\dfrac{3AB+DC}{2} \] 并且 \[ \dfrac{AB+DC}{2}+DC=\dfrac{AB+3DC}{2}, \] 因此有 \[ 3AB+DC=2(AB+3DC)=2AB+6DC,\quad \text{并且}\quad \dfrac{AB}{DC}=5. \]

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