Below is an interesting problem which I struggle to solve. Does anyone have any ideas?
if x is the side of the square with area 1, now if we increase it with y and area 2, we have
(x + y)(x + y) = 2,
x^2 + 2xy + y^2 = 2
but x^2 = 1 and x = 1, we have
1 + 2(1)(y) + y^2 = 2, re-arranging
1 - 2 + 2y + y^2 = 0
y^2 + 2y - 1 = 0,
Use quadratic formula:
y1 = sqrt 2 - 1
y2 = - sqrt 2 - 1
Notice that y1 is the only positive, so y = sqrt 2 - 1
length of the the new square = x + y = 1 + (sqrt 2 - 1) = sqrt 2.
If your square with side square root of 2, that is irrational . . . . hard to measure, right?
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NOTE: In general, doubling the area of ANY square with a side of length x, multiply it with sqrt 2, the SQUARE that,
New Area = [(x)(sqrt 2)]^2
To start, if all you have is a straight-edge,
how will you draw the 1st square itself?
Or do you start with a square already drawn ?
If so, do you have 2 choices: double it or get half of it?
Like, can a smaller square be drawn inside the original square?
If you haven't seen it done, HOW do you know?
Best I can do:
Draw the 2 corner to corner diagonals; they cross at square's center;
divides square in 4 identical triangles, each 1/4 of square's area.
Each "half-diagonal" is the required side length of a square equal to
half the original square.
Label square ABCD, A top left, going clockwise; diagonals cross at M.
Concentrate on right side of square, or side BC:
if you're able to draw a line from B perpendicular to DB,
and one from C perpendicular to AC, these 2 lines will cross, say at N.
Then you have the desired square: BMCN ; exactly half of ABCD.
But I know you'll say: "I have nothing I can use to make a 90degree angle!
Not even the end of my straight-edge..."
SO: go to corner A, tear off that corner! Since it's a 90degree corner,
use it at corner B and C to line up BN and CN...get it?
I know...I know: "I'm not allowed to use my fingers to tear off...."
Are you allowed a pencil with a square eraser?
this may help, make a square as much as you can, the hypotenuse will be the new side of the new square . . . . you may use a string to achieve it. the hypotenuse of the square if used as the side of the new square, doubles the AREA of the old square. i hope it helps you . . . .
or you may try this Methods of computing square roots - Wikipedia, the free encyclopedia
Thank you all so much for all the wonderful ideas! However, I have still found no way of doing this without measuring something (for example, with the option where you have to draw a perpendicular, you need a protractor to do that). All you are allowed is a straight edge to draw straight lines with - you are not allowed to measure anything!
Any more ideas?
Sure there is - check this out.
http://mei.org.uk/files/miotm_soluti...20apr%2009.pdf