Posted by algebrajunkie on Tuesday, April 26, 2011 at 8:40pm.
Shellie pays $4.00 for a square piece of wood, which she makes into a stop sign by cutting the corners off. what is the cost of the wasted part?
I don't even know where to begin?
do I take $4.00*1/8....which equals half and subtract that from 4.00...getting $2.50

college algebra  drwls, Tuesday, April 26, 2011 at 9:10pm
Check this out:
http://en.wikipedia.org/wiki/Stop_sign
It looks like 2/9 of the original square gets removed.

college algebra  Reiny, Tuesday, April 26, 2011 at 9:11pm
Each of the cutoff corners would be an isosceles rightangled 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

correction: college algebra  Reiny, Tuesday, April 26, 2011 at 9:19pm
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

college algebra  drwls, Tuesday, April 26, 2011 at 9:23pm
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 cutoff corners amounted to 1/9 of the square. That would amoount to $0.89 wasted.
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