Question 1010091
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The diagonal of a rectangle measures 10 inches. If the length is 2 inches more than the width, find the dimensions of the rectangle.
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Let x be the length of the rectangle, in inches.

Then the width is x-2 inches, and the diagonal is {{{sqrt(x^2 + (x-2)^2)}}}.

An equation is

{{{sqrt(x^2 + (x-2)^2)}}} = {{{10}}}.

The (3,4,5) triangle came into the mind with the legs of 6 inches and 8 inches, but I will complete the solution algebraically.

Square both sides of the equation (1), then simplify step by step:

{{{x^2 + (x-2)^2}}} = {{{100}}},

{{{2x^2 - 2x + 4 - 100}}} = {{{0}}},

{{{x^2 - x - 48}}} = {{{0}}}.


Factor the left size:

{{{x^2 - x - 48}}} = (x+6)*(x-8) = 0.

Now it is clear that x = 8 inches is the solution of the problem.

The rectangle has the dimensions 8 inches and 6 inches.
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