Given a retangle abcd, where ab=cd=A, and bc=da=B, the area of it is A*B. Cut it somehow and rearrange it to a square with an edge length sqroot(A*B) for the square should have the same area as that of the retangle. Here is the procedure to cut it.
Pass through the corner abc, draw a line with a length of sqroot(A*B), and join the edge cd at e. (1st cut) Then daw a line perpendicular to line be, and cross the corner dab (2nd cut) Then move them around to make a square with edge length of sqroot(A*B). www.ddhw.com
I did not use anything, no scissors, no compass, no cutter, even no paper and pencils. Use brain only!
www.ddhw.com
Imagine a retangle with edge lengthes A and B, and A > B (there was no specific numbers in my mind, except for the two "signs", A and B). So the retangle has an area A*B. Somehow to make it into a square with the equal area, the square's edge SHOULD be sqroot(A*B) (who cares what number it would be here).
www.ddhw.com
(For your convenience, I "print" the brain images here) Since parallel lines, to A and B, are out of the question, now IMAGINE a line passing (b) and joining the opposite long edge (A) at (e). Just LET length of this line, (be), be sqroot(A*B), for that such a line SHOULD exist, and MUST exist. This is beause
for B
Hence, Line (be) = sqroot(A*B) < length of line (bd) = sqroot(A^2 + B^2)
Also, from A>B, we have A*B > B^2, and hence
Line (be) = sqroot(A*B) > length of edge B
www.ddhw.com
If such a line is existed, I'll take it (So I have the line (be) with a length sqroot(A*B)). Still no scissors,......
Now imagine the second line which passes the point (a) and perpendicularly join line (be) at (g).
Since the triangle (bce) is similar to the triangle (agb) (angle bce= angle agb = 90 deg, and angle bec= angle gba), we have
line (ag)/A = B/sqroot(A*B)
that is, line (ag) =A*B/sqroot(A*B) = sqroot(A*B)
Ooooooooooops! www.ddhw.com
If line (be) = sqroot(A*B)> sqroot((A/2)^2 + B^2) (that is (ce) longer than A/2), the line (ag) will be short than sqroot(A*B). So a restrictive condition has to be posted here. That is, (A*B) has to be < or = (A/2)^2 + B^2, or A <=2B.
Now, imagine one can fold the retangle and tear it along line (be) first, and then along line (ag). Now you can shove the three pecies around until you gt the desired square.www.ddhw.com
Told you that I did not use scissors, compass, cutter, even paper or pencils
If you can draw a line of a given length (number) with a certain accuracy, then I can daw a square root of that number with the same accuracy. Reversely, if you can't daw a sqroot(2)= 1.41423... on a paper, you also can't draw a line of length 2.00000... with the same accuracy.
For 2nd question:
As long as there EXISTS such a perpendicular line with a length of sqroot(A*B), then the asked question CAN BE DONE. Once it has been proved can be done, my job is over.
哈哈哈哈 O(∩_∩)O哈哈~ 好久不来了，不小心看了一眼自己主页，发现自动关联了自己以往发过的帖，点进这个看了一眼，笑个半死。。。敢情我当初还发过这么搞笑的东西啊，我自己完全不记得了 
雷达卡
发表于 2008-9-19 07:40:41
收藏
分享
支持
反对
置顶卡
千斤顶
楼主