# college algebra

posted by on .

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 - ,

Check this out:

http://en.wikipedia.org/wiki/Stop_sign

It looks like 2/9 of the original square gets removed.

• college algebra - ,

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

• correction: college algebra - ,

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 - ,

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.