Question 108496
The Phytagorean Theorem can be applied!

Let a=x(the length)
____b=x-2(the width)
____c=10


Thus,

{{{x^2+(x-2)^2=10^2}}}
{{{x^2+(x(x-2)-2(x-2))=100}}}
{{{x^2+((x^2-2x)-(2x-4))=100}}}
{{{x^2+(x^2-2x-2x-4)=100}}}
{{{x^2+x^2-4x-4=100}}}
{{{2x^2-4x-4=100}}}
{{{2x^2-4x-104=0}}}

By the quadratic formula,

{{{x=(4+-sqrt(4^2-4*2*(-104)))/(2*2)}}}
{{{x=(4+-sqrt(16+832))/(4)}}}
{{{x=(4+-sqrt(848))/(4)}}}
{{{x=(4+- 29.120439557122)/(4)}}}

We will eliminate the {{{x=(4-29.120439557122)/(4)}}} part 'coz it will be only negative---an impossible result

Continuing,

{{{x=(4+29.120439557122)/(4)=33.120439557122/4=8.2801098892805}}}

Thus, the length is 8.2801098892805 units long

If x=8.2801098892805, then x+2=10.2801098892805

Thus, the width is 10.2801098892805 units long


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