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05 Jan 2021, 09:00
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A: \((\frac{P}{2}-W)W\)
L: (\(\frac{P}{2}-W\))
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
93% (02:02) correct
7%
(02:18)
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For all positive values of P, W, L, and A, consider the family of rectangles having perimeter P feet, width W feet, length L feet, and area A square feet.
In the first column of the table, select the expression in terms of P and W that is equivalent to A, and in the second column of the table select the expression in terms of P and W that is equivalent to L. Select only two expressions, one in each column.
In the first column of the table, select the expression in terms of P and W that is equivalent to A, and in the second column of the table select the expression in terms of P and W that is equivalent to L. Select only two expressions, one in each column.
| A | L | Expression |
| \(\frac{P^2}{W}\) | ||
| \(\frac{PW}{4}\) | ||
| (\(\frac{P}{2}-W\)) | ||
| \(\frac{(P-W)^2}{4}\) | ||
| \((\frac{P}{2}-W)W\) |
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Official Answer
A: \((\frac{P}{2}-W)W\)
L: (\(\frac{P}{2}-W\))
11 Jan 2021, 23:15
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Official Explanation
For all positive values of P, W, L, and A, consider the family of rectangles having perimeter P feet, width W feet, length L feet, and area A square feet.
In the first column of the table, select the expression in terms of P and W that is equivalent to A, and in the second column of the table select the expression in terms of P and W that is equivalent to L. Select only two expressions, one in each column.
Strategize
For any rectangle, A = LW, and P = 2L + 2W. Dividing both sides of the latter equation by 2 yields P/2 = L + W, and subtracting W from both sides yields \([\frac{P}{2} -W]\) = L. Therefore L = \([\frac{P}{2} -W]\) and
A = LW
= \((\frac{P}{2} -W)W\)
Alternatively, observe that, since A = LW, the expression for A is equivalent to the expression for L with an additional factor, W. And the only two expressions in the list that have this relationship are C, \((\frac{P}{2}-W)\), and E, \((\frac{P}{2} -W)W\)
RO1, A: Strategize
The correct answer is E, \((\frac{P}{2} -W)W\)
RO2, L: Strategize
The correct answer is C, \((\frac{P}{2} -W)\)
For all positive values of P, W, L, and A, consider the family of rectangles having perimeter P feet, width W feet, length L feet, and area A square feet.
In the first column of the table, select the expression in terms of P and W that is equivalent to A, and in the second column of the table select the expression in terms of P and W that is equivalent to L. Select only two expressions, one in each column.
Strategize
For any rectangle, A = LW, and P = 2L + 2W. Dividing both sides of the latter equation by 2 yields P/2 = L + W, and subtracting W from both sides yields \([\frac{P}{2} -W]\) = L. Therefore L = \([\frac{P}{2} -W]\) and
A = LW
= \((\frac{P}{2} -W)W\)
Alternatively, observe that, since A = LW, the expression for A is equivalent to the expression for L with an additional factor, W. And the only two expressions in the list that have this relationship are C, \((\frac{P}{2}-W)\), and E, \((\frac{P}{2} -W)W\)
RO1, A: Strategize
The correct answer is E, \((\frac{P}{2} -W)W\)
RO2, L: Strategize
The correct answer is C, \((\frac{P}{2} -W)\)
General Discussion
06 Jan 2021, 00:56
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For all positive values of P, W, L, and A, consider the family of rectangles having perimeter P feet, width W feet, length L feet, and area A square feet. In the first column of the table, select the expression in terms of P and W that is equivalent to A, and in the second column of the table select the expression in terms of P and W that is equivalent to L. Select only two expressions, one in each column.
For length L, and width W.
Area, A = LW
Perimeter, P = 2L + 2W
So, P/2 = L + W...(i)
Rearranging (i)
L = P/2 - W
Now A = LW = (P/2-W)W
For length L, and width W.
Area, A = LW
Perimeter, P = 2L + 2W
So, P/2 = L + W...(i)
Rearranging (i)
L = P/2 - W
Now A = LW = (P/2-W)W
