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 so

12=2 (W+L)

6 = (W+L) and we are told that L=W+1 so

6=2W+1

5=2W

W=2.5, and then L=3.5 so the Area would be:

A=2.5 (3.5) = 8.75in^2 NOT 30 in^2

Okay now that that crashed and burned, let's ignore the perimeter statement and just go with L=W+1

Then 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+W

W^2+W-30=0

(W+6) (W-5) = 0 since W>0 to have any meaning ...

W=5 and L=6

Okay 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?