A square card board of length 2+root of 8 has xcm cut off from each corner in order to make an?

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. o

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.

or: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.

or:Letsqrt(n) = square root of na = the length of the square card boardA 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) anda = b + 2xand you see immediately that b = 2, x = sqrt(2).But we can do this also with a bit of algebra:a = 2x + bb = sqrt (2x^2) = sqrt(2) * x ;; Pythagorassubstitute 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 xx = (2 + 2 sqrt(2)) / (2 + sqrt(2)) ;;; expand with denominator's conjugatex = (2 + 2 sqrt(2))(2 - sqrt(2)) / (2 + sqrt(2))(2 - sqrt(2))x = 2 sqrt(2) / 2x = sqrt(2) cm = 1.41... cm-------------------------------------b = sqrt(2) * x = 2 cm-------------------------------------

or: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 \u221a2/2:1:\u221a2/2 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\

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