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the area of a rectangle is 45ft^2 and the length of the rectangle is 1 ft more than double the width. Find the dimensions of the rectangle.

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Answer:

The dimensions are 4.5 ft by 10 ft.

Explanation:

The formula for the area of a quadrilateral is A= lw (area = the length multiplied by the width). We are given the area of 45 sq. ft so A = 45.

Now we know that l is double (twice - two times!) the width plus another foot. That means that l = 2w+1. We can use the substitution rule to replace l with 2w+1. So we have 45=(2w+1)w. From here you can distribute the w and you should get 45=2w2+1w. From here you should be able to solve for w by factoring, and therefore also solve for l.

45 factors into 5 and 9, and we know that 2x5=10, and 10-9=1 so that looks like a good bet! The only other factors for 45 would be 1 and 45, and that isn't going to give us 1w, so it must be 5 and 9.

So now let's solve it:

2w2 + w = 45

2w2 + w - 45 = 0

(2w -9) (w+5) = 0

2w - 9 = 0

w + 5 = 0

So w = 9/2 or w = -5

We can't have a negative width, so the answer is 9/2! Which means our length of 2w+1 = 10.

So the dimensions are 4.5 ft by 10 ft. Hope this helps you.

User Daniel Nyamasyo
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