Posted by algebrajunkie on Tuesday, April 26, 2011 at 8:40pm.
Check this out:
http://en.wikipedia.org/wiki/Stop_sign
It looks like 2/9 of the original square gets removed.
Each of the cut-off corners would be an isosceles right-angled triangle.
The stop sign will be an octagon with each side equal to x.
Consider one of the triangles, the hypotenuse will be x.
by Pythagoras you can show that each of the equal sides must be x/√2
each side of the original square is then equal to
x/√2 + x + x/√2 = x(2+√2)/2
and the original area = x^2(2+√2)^2/4 = x^2(3+2√2)/2
area of one triangle = (1/2)(x/√2)(x/√2) = x^2/4
or the total wasted part is x^2
so the part wasted is x^2/(x^2(3+2√2)/2))($4)
= 8/(3+2√2) dollars or appr. $1.37
typo!
last part should be .....
each side of the original square is then equal to
x/√2 + x + x/√2 = x(2+√2)/√2
and the original area = x^2(2+√2)^2/2 = x^2(3+2√2)
area of one triangle = (1/2)(x/√2)(x/√2) = x^2/4
or the total wasted part is x^2
so the part wasted is x^2/(x^2(3+2√2))($4)
= 4/(3+2√2) dollars or appr. $0.69
Reiny is correct for a regular octagon. In looking at the figure of
http://en.wikipedia.org/wiki/Stop_sign
I erroneously assumed that each pair of cut-off corners amounted to 1/9 of the square. That would amoount to $0.89 wasted.
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