document.write( "Question 329404: A rectangle and a square have the same area. The length of the rectangle is seventy feet more than two times its width. The length of a side of the square is thirty feet. Which equations would help you solve for the dimensions of the rectangle?
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document.write( "What are the dimensions of the rectangle?\r
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document.write( "I really need help on this any help is appreciated !!! \n" );
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Algebra.Com's Answer #236082 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! A rectangle and a square have the same area. \n" ); document.write( "The length of the rectangle is seventy feet more than two times its width. \n" ); document.write( "--- \n" ); document.write( "Let width be W. \n" ); document.write( "Then length = 2W+70 \n" ); document.write( "And Area = W(2W+70) = 2W^2+70W \n" ); document.write( "------ \n" ); document.write( "The length of a side of the square is thirty feet. \n" ); document.write( "Area of square = 30^2 = 900 sq ft. \n" ); document.write( "----------- \r \n" ); document.write( "\n" ); document.write( "Which equations would help you solve for the dimensions of the rectangle? \n" ); document.write( "What are the dimensions of the rectangle? \n" ); document.write( "--- \n" ); document.write( "Area of rectangle = Area of square \n" ); document.write( "2W^2+70W = 900 \n" ); document.write( "W^2 + 35W - 450 = 0 \n" ); document.write( "Factor: \n" ); document.write( "(W+45)(W-10) = 0 \n" ); document.write( "Positive solution: \n" ); document.write( "W = 10 ft (width of the rectangle) \n" ); document.write( "Since length = 2W+70, length = 20+70 = 90 ft \n" ); document.write( "================================================= \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "============ \n" ); document.write( " \n" ); document.write( " |