Question 102898
A rectangle has a diagonal that is 3.6 feet longer than the length and 7.1 feet longer than the width. What are the dimensions of the rectangle?
Let's write down what you know so far:
1. {{{D=3.6 + L}}}
2. {{{D=7.1 + W}}}
The diagonal of a rectangle also follows the Pythagorean theorem since the width and length form the sides of a right triangle with a hypotenuse that is the diagonal. Therefore,
3. {{{D^2 = L^2 + W^2}}}
You can use equations (1) and (2) to come up with a relationship between L and W and use that in equation (3). 
Since {{{3.6+L}}} and {{{7.1+L}}} are both equal to {{{D}}}, then they are equal to each other. 
{{{3.6+L=7.1+W}}}
{{{3.6-3.6+L=7.1-3.6+W}}} Additive inverse of 3.6
4.{{{L=3.5+W}}}
Now use equation (3) and substitute using this latest result (4) and equation (2) to get an equation that only involves the variable W. 
3. {{{D^2 = L^2 + W^2}}}
{{{(7.1+W)^2=(3.5+W)^2+W^2}}}Substitute.
{{{highlight(7.1^2+2*7.1*W+W^2)=highlight(3.5^2+2*3.5*W+W^2)+W^2}}}Expand.
{{{50.41+14.2*W+W^2= 12.25+7*W+2W^2}}} Simplify both sides. 
{{{(50.41-12.25)+(14.2-7)*W+(1-2)W^2=cross(12.25)+cross(7*W)+cross(2W^2)-cross(12.25)-cross(7*W)-cross(2W^2)}}}
{{{38.16+7.2*W-W^2=0}}}
Use the quadratic formula to find the roots of the equation:
{{{W = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
where {{{a = -1}}}, {{{b = 7.2}}}, and {{{c = 38.16}}}
{{{W = (-7.2 +- sqrt( 7.2^2-4*(-1)*38.16 ))/(2*(-1)) }}} 
{{{W = (-7.2 +- 14.3)/(-2) }}} 
{{{W=-3.55}}} and {{{W=10.75}}} solve the equation. 
Since a negative width does not make sense, we'll only use the positive result. 
{{{W=10.75}}}
Use equation (4) to get the length. 
4.{{{L=3.5+W}}}
{{{L=3.5+10.75=14.25}}}
and finally 
1. D=3.6+L
{{{D=3.6+14.25}}}
{{{D=17.85}}}
You can double-check your answer using the Pythagorean theorem and your length and width, 
{{{D^2 = L^2 + W^2}}}
{{{D^2 = 14.25^2 + 10.75^2}}}
{{{D^2 = 203.0625 + 115.5625=318.625}}}
{{{D = 17.85}}}
Good answer