Question 61485
a^2+b^2=c^2 [use the pythagorean theorem]
Let:
a=length
b=width
c=diagonal or hypotenuse
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(w+1)^2+w=4^2 [plug-in the values and solve]
w^2+2w+1+w=16
2w^2+2w+1-16=0 [set the equation equal to zero]
2w^2+2w-15=0 [factor using the quadratic formula]
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
 {{{x = (-2) +- sqrt( 2^2-(4*2*-15 ))/(4) }}}

{{{x = (-2) +- sqrt( 4-(-120 ))/(4) }}}
{{{x = (-2) +- sqrt( 4+120 ))/(4) }}}
{{{x = (-2) +- sqrt( 124 ))/(4) }}}

{{{x = (-2) + sqrt( 124 ))/(4) }}}=2.2838

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{{{x = (-2) - sqrt( 124 ))/(4) }}}=-3.284[eliminate a negative measurement]
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So, the width = 2.2838
Length = (w+1)=2.2838+1=3.2838

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check by plugging all of the values back into the pythagorean theorem and solve.
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