A square card board of length 2+root of 8 has xcm cut off from each corner in order to make an octagon. Find the value of x and length of each side.
Answers (2)
Let
sqrt(n) = square root of n
a = the length of the square card board
A right isosceles triangle with cathete length x cm has to be cut off from each corner, such that the hypotenuse b is a side of the resulting regular octagon:
a = 2 + sqrt(8) = 2 + 2 sqrt(2) and
a = b + 2x
and you see immediately that
b = 2, x = sqrt(2).
But we can do this also with a bit of algebra:
a = 2x + b
b = sqrt (2x^2) = sqrt(2) * x ;; Pythagoras
substitute
a = 2x + sqrt (2x^2)
a = x (2 + sqrt (2))
As the text says,
a = 2 + sqrt(8) = 2 + 2 sqrt(2)
As both equations equal a, we can build the equation:
x (2 + sqrt (2)) = 2 + 2 sqrt(2) ;; solve for x
x = (2 + 2 sqrt(2)) / (2 + sqrt(2)) ;;; expand with denominator's conjugate
x = (2 + 2 sqrt(2))(2 - sqrt(2)) / (2 + sqrt(2))(2 - sqrt(2))
x = 2 sqrt(2) / 2
x = sqrt(2) cm = 1.41... cm
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b = sqrt(2) * x = 2 cm
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If you take wood shop, a common project is to cut a square post into an octagon. When you draw a picture of that you see that the side of the octagon and the corners removed make the ratio √2/2:1:√2/2 which is very close to 7;10;7 and those numbers happen to add up to 24. So your lay your 24" ruler across the post at an angle and make marks at 7" and 17", slide the ruler along the post, and repeat the marking process. Draw lines between those marks and you have the exact places to make your cuts to make a perfect octagon.
The denominator in line 2 has to be in parentheses! CORRECTION:
x = (2 + 2 sqrt(2)) / (2 + sqrt(2)) ;;; expand with denominator's conjugate
x = (2 + 2 sqrt(2))(2 - sqrt(2)) / ((2 + sqrt(2))(2 - sqrt(2)))
x = 2 sqrt(2) / 2
x = sqrt(2) cm = 1.41... cm
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b = sqrt(2) * x = 2 cm
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