AMC10 2003 B

AMC10 2003 B · Q20

AMC10 2003 B · Q20. It mainly tests Coordinate geometry, Ratios in geometry.

In rectangle ABCD, AB = 5 and BC = 3. Points F and G are on CD so that DF = 1 and GC = 2. Lines AF and BG intersect at E. Find the area of △AEB. [Diagram shows rectangle ABCD with AB = 5 (top), BC = 3 (right side), points F and G on CD with DF = 1, GC = 2]

在矩形ABCD中，AB = 5，BC = 3。点F和G在CD上，使得DF = 1，GC = 2。直线AF和BG相交于E。求△AEB的面积。[图示：矩形ABCD，AB = 5（顶部），BC = 3（右侧），点F和G在CD上，DF = 1，GC = 2]

(A) 10 10

(B) 21/2 21/2

(D) 25/2 25/2

(E) 15 15

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) Let $H$ be the foot of the perpendicular from $E$ to $\overline{DC}$. Since $CD=AB=5$, $FG=2$, and $\triangle FEG$ is similar to $\triangle AEB$, we have \[ \frac{EH}{EH+3}=\frac{2}{5}, \] so $5EH=2EH+6$, and $EH=2$. Hence \[ \text{Area}(\triangle AEB)=\frac{1}{2}(2+3)\cdot 5=\frac{25}{2}. \]

（D）设 $H$ 为从 $E$ 到 $\overline{DC}$ 的垂足。由于 $CD=AB=5$，$FG=2$，且 $\triangle FEG$ 与 $\triangle AEB$ 相似，我们有 \[ \frac{EH}{EH+3}=\frac{2}{5}, \] 所以 $5EH=2EH+6$，并且 $EH=2$。因此 \[ \text{Area}(\triangle AEB)=\frac{1}{2}(2+3)\cdot 5=\frac{25}{2}. \]

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