Henry knows that the area of a rectangle is 30 square inches the perimeter is 12 inches if the length is 1 inch longer than the width what are the length and width Henry's rectangle explain how you know

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Henry knows that the area of a rectangle is 30 square inches the perimeter is 12 inches if the length is 1 inch longer than the width what are the length and width Henry's rectangle explain how you know

What you typed is impossible, the greatest possible area for a four sided polygon is a square. If the perimeter is 12, the sides of the square would be 3 inches. The area of that square would be 9 in^2. Work through this and see that it is impossible.P=2(W+L), we are told that P=12 so12=2(W+L)6=(W+L) and we are told that L=W+1 so6=2W+15=2WW=2.5, and then L=3.5 so the Area would be:A=2.5(3.5)= 8.75in^2 NOT 30 in^2Okay now that that crashed and burned, let's ignore the perimeter statement and just go with L=W+1Then A=LW and using L=W+1 in the area equation we get:A=W^2+W and since A=30 we have:30=W^2+WW^2+W-30=0(W+6)(W-5)=0 since W>0 to have any meaning...W=5 and L=6Okay that's fine, the area fits the length and width parameters...Now notice that P=2(5+6)=22in NOT 12 in:D So you mistyped???

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