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document.write( "If the sides of a square are decreased by 3cm, the area is decreased by 81cm^2.What were the dimensions of the original square?\r
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Algebra.Com's Answer #56823 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! If the sides of a square are decreased by 3cm, the area is decreased by 81cm^2.What were the dimensions of the original square? \n" ); document.write( ": \n" ); document.write( "Let 1 side of the original square = x \n" ); document.write( "Then the area of the original square = x^2 \n" ); document.write( ": \n" ); document.write( "The side of the smaller square would = (x-2); original side decreased by 2cm \n" ); document.write( "Area of the small square = (x-2)^2 \n" ); document.write( ": \n" ); document.write( "Problem states: \n" ); document.write( "Original square area - smaller square area = 81 sq/cm \n" ); document.write( " x^2 - (x-2)^2 = 81 \n" ); document.write( ": \n" ); document.write( "FOIL (x-2)(x-2) \n" ); document.write( "x^2 -(x^2 - 4x + 4) = 81 \n" ); document.write( ": \n" ); document.write( "Remove brackets, remember to change the signs inside the brackets: \n" ); document.write( "x^2 - x^2 + 4x - 4 = 81 \n" ); document.write( ": \n" ); document.write( "conveniently, the x^2's are eliminated: \n" ); document.write( "4x = 81 + 4 \n" ); document.write( "4x = 85 \n" ); document.write( "x = 85/4 \n" ); document.write( "x = 21.25 cm is the side of the original square: \n" ); document.write( ": \n" ); document.write( "The smaller square would be 21.25 - 2 = 19.25 \n" ); document.write( ": \n" ); document.write( "Check our solution by finding the area of each square: \n" ); document.write( "21.25^2 = 451.5625 \n" ); document.write( "19.25^2 = 370.5625 \n" ); document.write( "------------------- \n" ); document.write( "Differ by 81.00 sq/cm as given \n" ); document.write( ": \n" ); document.write( "Could you follow this OK? Any questions? \n" ); document.write( " \n" ); document.write( " |