Question 218423
The width of a rectangle is 6 less than twice its length. If the area of the rectangle is 113 cm^2, what is the length of the diagonal? 


Step 1.  Let L be the length.


Step 2.  Let 2L-6  be the width since the width of a rectangle is 6 less than twice its length.


Step 3.  Area A=113=L*(2L-6)


Step 4.  Solving for L yields the following steps


{{{113=2L^2-6L}}}


Subtract 113 from both sides of the equation to get a quadratic equation


{{{113-113=2L^2-6L-113}}}


{{{2L^2-6L-113=0}}}


To solve for L, we can use the quadratic formula given as


{{{L = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 


where a=2, b=-6 and c=-113


*[invoke quadratic "L", 2, -6, -113 ]


{{{L=9.165}}} selecting the positive length

and 

{{{2L-6=12.33}}}


Step 5. Let c be the diagonal.


Step 6. Use the Pythagorean Theorem to find the diagonal which says that the sum of the squares of the legs (length and width in our example) is equal to the sum of the squares of the hypotenuse (diagonal in this case).  That is,


{{{c^2=9.165^2+12.33^2=236.03}}}


Take the square root to both sides of the equation.


{{{sqrt(c^2)=sqrt(236.03)}}}


{{{c=15.36}}}


Step 7.  ANSWER:  The diagonal is 15.36 cm.


I hope the above steps were helpful.


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And good luck in your studies!


Respectfully,
Dr J