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10 May 2008, 03:02
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Please let me know what is wrong in my approach:
perimeter of the rectangle is=2(l+w)= 18 by squareroot2
since the diagonals of square are equal and the sum of both diagonals is =length of the rectangle and the vertical diagonal is equal to width of the rectangle, the width of the rectangle should be half of the length, correct ?
thus 2(l+l/2) =3l =18 by squareroot2 or l=6 by squareroot2 or width =3 by squareroot2
diagonal of the square is (a squareroot2 )= 3 by squareroot2
thus a=3/2, perimeter of the square=4a= 4.3/2 =6
perimeter of the rectangle is=2(l+w)= 18 by squareroot2
since the diagonals of square are equal and the sum of both diagonals is =length of the rectangle and the vertical diagonal is equal to width of the rectangle, the width of the rectangle should be half of the length, correct ?
thus 2(l+l/2) =3l =18 by squareroot2 or l=6 by squareroot2 or width =3 by squareroot2
diagonal of the square is (a squareroot2 )= 3 by squareroot2
thus a=3/2, perimeter of the square=4a= 4.3/2 =6
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10 May 2008, 06:48
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Since each angle in a square is 90 deg and all 4 sides (x) are equal, the diagonal line must be \(x\sqrt2\).
Perimeter of rectangle = 2 (L + W), where L = length and W = width.
L = \(2(x\sqrt2)\) and,
W = \(x\sqrt2\)
2 (\(2(x\sqrt2)\) + \(x\sqrt2\)) = \(18\sqrt2\)
\(4x\sqrt2\) + \(2x\sqrt2\) = \(18\sqrt2\)
\(6x\sqrt2\) = \(18\sqrt2\)
6x = 18
x = 3
Perimeter of a square = 4(x) = 4(3) = 12
Ans: B (12)
Correct?
Perimeter of rectangle = 2 (L + W), where L = length and W = width.
L = \(2(x\sqrt2)\) and,
W = \(x\sqrt2\)
2 (\(2(x\sqrt2)\) + \(x\sqrt2\)) = \(18\sqrt2\)
\(4x\sqrt2\) + \(2x\sqrt2\) = \(18\sqrt2\)
\(6x\sqrt2\) = \(18\sqrt2\)
6x = 18
x = 3
Perimeter of a square = 4(x) = 4(3) = 12
Ans: B (12)
Correct?
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