AMC8 2005

AMC8 2005 · Q19

AMC8 2005 · Q19. It mainly tests Pythagorean theorem, Area & perimeter.

What is the perimeter of trapezoid ABCD?

梯形ABCD的周长是多少？

(A) 180 180

(B) 188 188

(D) 200 200

(E) 204 204

Answer

Correct choice: (A)

正确答案：（A）

Solution

By the Pythagorean Theorem, $AE=\sqrt{30^2-24^2}=\sqrt{324}=18$. (Or note that triangle $AEB$ is similar to a 3-4-5 right triangle, so $AE=3\times 6=18$.) Also $CF=24$ and $FD=\sqrt{25^2-24^2}=\sqrt{49}=7$. The perimeter of the trapezoid is $50+30+18+50+7+25=180$.

由勾股定理，$AE=\sqrt{30^2-24^2}=\sqrt{324}=18$。（或者注意到三角形 $AEB$ 与 3-4-5 的直角三角形相似，所以 $AE=3\times 6=18$。）另外，$CF=24$，且 $FD=\sqrt{25^2-24^2}=\sqrt{49}=7$。该梯形的周长为 $50+30+18+50+7+25=180$。

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