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标题: gwd-25-31 [打印本页]

作者: phdmass 时间: 2009-8-30 21:57 标题: gwd-25-31

Q31. In the figure, each side of square A B C D has length 1, the length of line Segment CE is 1, and the length of line segment BE is equal to the length Of line segment DE. What is the area of the triangular region BCE?

图贴不上来，square ABCD，E在c的上方。

a. 1/3 b. (2^-2 )/4 c. 1/2
d. (2^-2)/2 e. 3/4

这个算面积的题目我不会，可能没有理解好题意，答案d,哪位还记得怎么算的？

作者: chouliulu 时间: 2009-8-31 06:44

Area of BCE
= (Area of BDE - Area of BCD) / 2
= [(BD * E to midpoint of BD / 2) - (1 * 1 / 2)] / 2
= {[2^0.5 * (1 + 0.5 * 2^0.5) / 2] - 0.5} / 2
= 2^0.5 / 4

B ?

作者: phdmass 时间: 2009-8-31 19:52

多谢，我老是把这个图看成立体的，其实应该按照平面图做。

作者: chouliulu 时间: 2009-9-1 06:25

I thought so too at the beginning. The original graph is very confusing. I always draw the graph out myself (if it's not too difficult) on the scratch paper to help myself go over all the conditions. You can try that.

作者: xiangdingdan 时间: 2009-9-3 06:54

哪位看看，这道题答案给的是d，我算的的b？

作者: dreamerp 时间: 2009-9-7 06:54

是不是答案写错了？

设BD中点为M

SBCE=SBEM-SBCM=(2^/2*(1+2^/2))/2-1/4=(2^/2+1/2)/2-1/4=2^/4=(2^/2)/2 d

为什么是减号？？

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