AMC8 1995

AMC8 1995 · Q24

AMC8 1995 · Q24. It mainly tests Triangles (properties), Area & perimeter.

In parallelogram ABCD, DE is the altitude to the base AB and DF is the altitude to the base BC. If DC = 12, EB = 4 and DE = 6, then DF =

在平行四边形 ABCD 中，DE 是底边 AB 的高，DF 是底边 BC 的高。若 DC = 12，EB = 4 且 DE = 6，则 DF =

(A) 6.4 6.4

(B) 7 7

(D) 8 8

(E) 10 10

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Since opposite sides of a parallelogram are equal, $AB = 12$. Then $AE = 12 - 4 = 8$. Using the Pythagorean Theorem gives $AD = \sqrt{8^2 + 6^2} = 10$, and then $BC = 10$ also. The area of a parallelogram is base $\times$ altitude. Using base $AB = 12$ and altitude $DE = 6$ gives an area of $12 \times 6 = 72$. Using base $BC = 10$ and altitude $DF$ must also give an area of $72$. Thus $DF = 72/10 = 7.2$.

答案（C）：由于平行四边形的对边相等，$AB = 12$。因此 $AE = 12 - 4 = 8$。利用勾股定理得 $AD = \sqrt{8^2 + 6^2} = 10$，于是 $BC = 10$。平行四边形的面积等于底 $\times$ 高。取底 $AB = 12$、高 $DE = 6$，面积为 $12 \times 6 = 72$。取底 $BC = 10$、高 $DF$，面积也应为 $72$。因此 $DF = 72/10 = 7.2$。

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