What is the diagonal of a rectangle with: length = 3 + square root of 2 width = 3 - square root of 2
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For a rectangle with sides L and W the length of the diagonal is sqrt (L^2 + W^2)
In the question we have the following conditions:
L = 3 + sqrt 2
W = 3 - sqrt 2
Therefore teh length of the diagonal is
sqrt [ ( 3 + sqrt 2)^2 + ( 3 - sqrt 2)^2]
=> sqrt [ 9 + 2 + 6 sqrt 2 + 9 + 2 - 6 sqrt 2]
=> sqrt [ 22]
Therefore the diagonal is sqrt 22.
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To determine the diagonal of the rectangle, we'll apply the Pythagorean theorem in one of 2 right triangles formed when tracing a diagonal of the rectangle.
The length and the width are the legs of the right triangle and the diagonal is the hypotenuse.
According to Pythagorean theorem, we'll have:
hypotenuse^2 = leg^2 + leg^2
In our case:
diagonal^2 = length^2 + width^2
We'll substitute what we know:
diagonal^2 = (3+sqrt2)^2 + (3-sqrt2)^2
diagonal^2 = 9 + 6sqrt2 + 2 + 9 - 6sqrt2 + 2
We'll combine and eliminate like terms:
diagonal^2 = 18 + 4
diagonal^2 = 22
diagonal = sqrt 22
The diagonal of the rectangle is sqrt 22 units.
The diagonal d of the rectangle is give by: d = {l^2+w^2}^(1/2), where l = length and w = width.
Therefore d = {(3+2^(1/2))^2+(3-2^(1/2))^2}^(1/2)
d = {(3^2+2*3*2^(1/2)+2)+(3^2-2*3*2^(1/2)+2)}
d = {2*3^2+2*2}^(1/2).
d = {18+4}^(1/2).
d = 22^(1/2).
Therefore the diagonal of the rectangle is 22^(1/2).
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