AMC10 2008 A

AMC10 2008 A · Q20

AMC10 2008 A · Q20. It mainly tests Similarity, Area & perimeter.

Trapezoid $ABCD$ has bases $\overline{AB}$ and $\overline{CD}$ and diagonals intersecting at $K$. Suppose that $AB = 9$, $DC = 12$, and the area of $\triangle AKD$ is 24. What is the area of trapezoid $ABCD$?

梯形 $ABCD$ 的底边为 $\overline{AB}$ 和 $\overline{CD}$，对角线交于点 $K$。已知 $AB = 9$，$DC = 12$，且 $\triangle AKD$ 的面积为 24。求梯形 $ABCD$ 的面积。

(A) 92 92

(B) 94 94

(D) 98 98

(E) 100 100

Answer

Correct choice: (D)

正确答案：（D）

Solution

Note that $\triangle ABK$ is similar to $\triangle CDK$. Because $\triangle AKD$ and $\triangle KCD$ have collinear bases and share a vertex $D$, $\frac{\text{Area}(\triangle KCD)}{\text{Area}(\triangle AKD)} = \frac{KC}{AK} = \frac{CD}{AB} = \frac{4}{3}$, so $\triangle KCD$ has area 32. By a similar argument, $\triangle KAB$ has area 18. Finally, $\triangle BKC$ has the same area as $\triangle AKD$ since they are in the same proportion to each of the other two triangles. The total area is $24 + 32 + 18 + 24 = 98$. OR Let $h$ denote the height of the trapezoid. Then $24 + \text{Area}(\triangle AKB) = \frac{9h}{2}$. Because $\triangle CKD$ is similar to $\triangle AKB$ with similarity ratio $\frac{12}{9} = \frac{4}{3}$, $\text{Area}(\triangle CKD) = \frac{16}{9} \text{Area}(\triangle AKB)$, so $24 + \frac{16}{9} \text{Area}(\triangle AKB) = \frac{12h}{2}$. Solving the two equations simultaneously yields $h = \frac{28}{3}$. This implies that the area of the trapezoid is $\frac{1}{2} \cdot \frac{28}{3} (9 + 12) = 98$.

注意 $\triangle ABK \sim \triangle CDK$。因为 $\triangle AKD$ 和 $\triangle KCD$ 有公共顶点 $D$ 且底边共线，故 $\frac{\text{Area}(\triangle KCD)}{\text{Area}(\triangle AKD)} = \frac{KC}{AK} = \frac{CD}{AB} = \frac{4}{3}$，所以 $\triangle KCD$ 面积为 32。类似地，$\triangle KAB$ 面积为 18。最后，$\triangle BKC$ 与 $\triangle AKD$ 面积相等，因为它们与其他两个三角形的比例相同。总面积为 $24 + 32 + 18 + 24 = 98$。或者，设梯形高度为 $h$。则 $24 + \text{Area}(\triangle AKB) = \frac{9h}{2}$。因为 $\triangle CKD \sim \triangle AKB$，相似比 $\frac{12}{9} = \frac{4}{3}$，故 $\text{Area}(\triangle CKD) = \frac{16}{9} \text{Area}(\triangle AKB)$，所以 $24 + \frac{16}{9} \text{Area}(\triangle AKB) = \frac{12h}{2}$。联立解得 $h = \frac{28}{3}$。梯形面积为 $\frac{1}{2} \cdot \frac{28}{3} (9 + 12) = 98$。

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