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#2 (permalink) |
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Within my grasp!
![]() ![]() Join Date: Feb 2006
Posts: 102
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Let "x" be the side of square.
Area, A=x^2 Perimeter, P=4x Given, A=2P+9 substitute value of A and P x^2=2(4x)+9 x^2=8x+9 x^2-8x-9=0 Solving, we get (x-9) (x+1) = 0 Therefore, x=9 or x=-1 But side cannot be negative, therefore x=9 Perimeter= 4x=4*9=36 Answer B |
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#3 (permalink) | |
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Within my grasp!
![]() ![]() Join Date: Feb 2005
Posts: 118
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perimeter = p
side of the square = p/4 area, a = 2p +9 (p/4)(p/4) = 2p +9 solving the eq, p = -4, 36 p cannot be in -ve, so p = 36. Bow wow wow yippee yo yippee yay!!!!11 question Quote:
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