Asked by Cathy
A square measures 18feet 6inches in length. If an octagon is inscribed in the square, then how long should each side of the octagon be?
Answers
Answered by
Reiny
Look at one of the corner triangles.
Let each side of the octogon be x
so the hypotenuse of that triangle is x
let the other two sides be y each
y^2 + y^2 = x^2
2y^2 = x^2
√2 y = x
y = x/√2
then x/√2 + x + x/√2 = 18.5
multiply each term by √2
x + √2 x + x = 18.5√2
x(1 + √2 + 1) = 18.5√2
x = 18.5√2/(2+√2) or appr. 7.66 ft or 7 feet 8 inches
Let each side of the octogon be x
so the hypotenuse of that triangle is x
let the other two sides be y each
y^2 + y^2 = x^2
2y^2 = x^2
√2 y = x
y = x/√2
then x/√2 + x + x/√2 = 18.5
multiply each term by √2
x + √2 x + x = 18.5√2
x(1 + √2 + 1) = 18.5√2
x = 18.5√2/(2+√2) or appr. 7.66 ft or 7 feet 8 inches
Answered by
Cathy
thank you very much
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