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The length of a rectangle is two more than twice its width,

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The length of a rectangle is two more than twice its width, [#permalink] New post 14 Jun 2018, 14:31
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Question Stats:

88% (00:52) correct 11% (00:50) wrong based on 9 sessions
The length of a rectangle is two more than twice its width, and the area of the rectangle is 40. What is the rectangle’s perimeter?

[Reveal] Spoiler: OA
28

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Sandy
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Re: The length of a rectangle is two more than twice its width, [#permalink] New post 05 Jul 2018, 01:59
Expert's post
Explanation

Convert this word problem into two equations with two variables. “The length is two more than twice the width” can be written as:

\(L = 2W + 2\)

Since the area is 40 and area is equal to length × width:

\(LW = 40\)

Since the first equation is already solved for L, plug (2W + 2) in for L into the second equation:

\((2W + 2)W = 40\)
\(2W^2 + 2W = 40\)

Since this is now a quadratic (there are both a W2 and a W term), get all terms on one side to set the expression equal to zero:

\(2W^2 + 2W - 40 = 0\)

Simplify as much as possible—in this case, divide the entire equation by 2—before trying to factor:

\(W^2 + W - 20 = 0\)

\((W - 4)(W + 5) = 0\)

W = 4 or –5

Since a width cannot be negative, the width is equal to 4. Since LW is equal to 40, the length must be 10. Now use the equation for perimeter to solve:

Perimeter = 2L + 2W

Perimeter = 2(10) + 2(4)

Perimeter = 28

Note that it might have been possible for you to puzzle out that the sides were 4 and 10 just by trying values. However, if you did this, you got lucky—no one said that the values even had to be integers!
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Re: The length of a rectangle is two more than twice its width, [#permalink] New post 10 Jul 2018, 03:22
We have system of equations: l=2w+2 and lw=40 -> w(2w+2)=40 -> w(w+1)=20 -> w=4 -> l=10 -> perimeter = 2(4+10) = 28.
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Re: The length of a rectangle is two more than twice its width, [#permalink] New post 05 Aug 2018, 17:22
sandy wrote:
Explanation

Convert this word problem into two equations with two variables. “The length is two more than twice the width” can be written as:

\(L = 2W + 2\)

Since the area is 40 and area is equal to length × width:

\(LW = 40\)

Since the first equation is already solved for L, plug (2W + 2) in for L into the second equation:

\((2W + 2)W = 40\)
\(2W^2 + 2W = 40\)

Since this is now a quadratic (there are both a W2 and a W term), get all terms on one side to set the expression equal to zero:

\(2W^2 + 2W - 40 = 0\)

Simplify as much as possible—in this case, divide the entire equation by 2—before trying to factor:

\(W^2 + W - 20 = 0\)

\((W - 4)(W + 5) = 0\)

W = 4 or –5

Since a width cannot be negative, the width is equal to 4. Since LW is equal to 40, the length must be 10. Now use the equation for perimeter to solve:

Perimeter = 2L + 2W

Perimeter = 2(10) + 2(4)

Perimeter = 28

Note that it might have been possible for you to puzzle out that the sides were 4 and 10 just by trying values. However, if you did this, you got lucky—no one said that the values even had to be integers!


I pretty much solved using this same method EXCEPT, I used the quadratic equation formula to solve for W. Gave me the same answer, took a little longer but thats what came to my mind immediately.
Re: The length of a rectangle is two more than twice its width,   [#permalink] 05 Aug 2018, 17:22
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