algebra

A rectangle has a width of 8 meters. The length is twice as long as the width. What is the length of the diagonal? Show your work

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asked by Alexius
  1. W = 8
    L = 16

    x^2 = w^2 + L^2
    x^2 = 8^2 + 16^2 = (2^3)^2 + (2^4)^2
    x^2 = 2^6 + 2^8
    x^2 = 2^6 + 4*2^6
    x^2 = 5 * 2^6
    x^2 = sqrt 5 * 2^3
    x = 8 sqrt 5

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    posted by Damon
  2. The diagonal is c.

    a^2 + b^2 = c^2

    8^2 + 16^2 = c^2

    64 + 256 = c^2

    320 = c^2

    17.889 = c

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