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October 30, 2014

Posted by **algebrajunkie** on Tuesday, April 26, 2011 at 8:40pm.

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 9:10pmCheck 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 9:11pmEach 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

- correction: college algebra -
**Reiny**, Tuesday, April 26, 2011 at 9:19pmtypo!

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 9:23pmReiny 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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