Question 643879

The length of a rectangle is 6cm more than its width. The area is 216 sq cm. What are the dimensions? I have w= width, l= w+ 6cm = length A= 216cm^2 
I set it up like this: 216cm^2= w+ 6cm(w)= 216cm^2= w^2+ 6cmw= 216cm^2- w^2-6cmw=0
and that's when i just got stuck I'm not sure if i was doing it right in the first place. Can I please have a bit of guidance?


Let width be W
Then length = W + 6
A, or area = LW
Since area = 216, then: 216 = W(W + 6)


{{{216 = W^2 + 6W}}}


{{{W^2 + 6W - 216 = 0}}}


(W - 12) (W + 18) = 0 


W - 12 = 0, or W = - 18 (ignore as measurement CANNOT be negative)


W, or width = {{{highlight_green(12)}}}


Length = 12 + 6, or {{{highlight_green(18)}}} cm


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