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The length of a rectangle is two more than twice its width,
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14 Jun 2018, 14:31
14 Jun 2018, 14:31
Expert Reply
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Question Stats:
88% (01:22) correct
11% (01:28) wrong based on 34 sessions
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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?
Show: :: OA
28
Re: The length of a rectangle is two more than twice its width,
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05 Jul 2018, 01:59
05 Jul 2018, 01:59
Expert Reply
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!
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!
Re: The length of a rectangle is two more than twice its width,
[#permalink]
10 Jul 2018, 03:22
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.
