AMC 8 2008 Problem 23

AMC8 2008 年第 23 题

四边形与多边形★★★☆☆

In square ABCE , AF=2FE and CD=2DE . What is the ratio of the area of

\triangle BFD

to the area of square ABCE ?

在正方形 ABCE 中，AF＝2FE，CD＝2DE，那么△BPD 的面积和正方形 ABCE 的面积的比值是多少？

A. $\displaystyle \frac{1}{6}$
B. $\displaystyle \frac{2}{9}$
C. $\displaystyle \frac{5}{18}$
D. $\displaystyle \frac{1}{3}$
E. $\displaystyle \frac{7}{20}$

正确答案 · Correct Answer

解析 · Solution

Solution 1

Because the answer is a ratio, it does not depend on the side length of the square. Let 𝐴 𝐹 = 2 AF=2 and 𝐹 𝐸 = 1 FE=1. That means square 𝐴 𝐵 𝐶 𝐸 ABCE has side length 3 3 and area 3 2 = 9 3 2 =9 square units. The area of △ 𝐵 𝐴 𝐹 △BAF is equal to the area of △ 𝐵 𝐶 𝐷 = 1 2 ⋅ 3 ⋅ 2 = 3 △BCD= 2 1 ⋅3⋅2=3 square units. Triangle 𝐷 𝐸 𝐹 DEF is an isosceles right triangle with leg lengths 𝐷 𝐸 = 𝐹 𝐸 = 1 DE=FE=1. The area of △ 𝐷 𝐸 𝐹 △DEF is 1 2 ⋅ 1 ⋅ 1 = 1 2 2 1 ⋅1⋅1= 2 1 square units. The area of △ 𝐵 𝐹 𝐷 △BFD is equal to the area of the square minus the areas of the three right triangles: 9 − ( 3 + 3 + 1 2 ) = 5 2 9−(3+3+ 2 1 )= 2 5 . So the ratio of the area of △ 𝐵 𝐹 𝐷 △BFD to the area of square 𝐴 𝐵 𝐶 𝐸 ABCE is 5 2 9 = 5 18 9 2 5 = 18 5 . Answer: 𝐶 C .

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