2005 AMC 10A 第 8 题

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8.

图中,正方形 ABCDABCD 的边 ABAB 长为 50\sqrt{50},点 EEBBHH 之间,且 BE=1BE = 1。内正方形 EFGHEFGH 的面积是多少?

In the figure, the length of side ABAB of square ABCDABCD is 50,\sqrt{50}, EE is between BB and H,H, and BE=1.BE = 1. What is the area of the inner square EFGH?EFGH?

2525

3232

3636

4040

4242

答案:C
知识点:正方形(几何)勾股定理全等(几何)
难度评级:1280
小提示:

四个角上的三角形,如 ABH\triangle ABH,是全等直角三角形。

The four corner triangles, such as ABH,\triangle ABH, are congruent right triangles

大提示:

每个角上的三角形斜边为 50\sqrt{50},一条直角边为 BE=1BE = 1;内正方形边长是另一条直角边减去 11

Each corner triangle is right with hypotenuse 50\sqrt{50} and one leg BE=1;BE = 1; the inner square’s side is the other leg minus 11

解答:

三角形 ABHABHBCEBCECDFCDFDAGDAG 是全等直角三角形。在 BCE\triangle BCE 中,斜边 BC=50BC = \sqrt{50},且 BE=1BE = 1,所以 CE=501=7CE = \sqrt{50 - 1} = 7。因为 BH=CE=7BH = CE = 7,且 EEBHBH 上、BE=1BE = 1,所以内正方形边长 EH=71=6EH = 7 - 1 = 6,面积为 62=366^2 = 36

所以正确答案是 C

The triangles ABH,ABH, BCE,BCE, CDF,CDF, and DAGDAG are congruent right triangles. In BCE\triangle BCE the hypotenuse is BC=50BC = \sqrt{50} and BE=1,BE = 1, so CE=501=7.CE = \sqrt{50 - 1} = 7. Since BH=CE=7BH = CE = 7 and EE lies on BHBH with BE=1,BE = 1, the inner square’s side is EH=71=6,EH = 7 - 1 = 6, giving area 62=36.6^2 = 36.

Thus, the correct answer is C.

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