document.write( "Question 208383: Two frames are needed with the same perimeter. One frame is in the shape of a square and the other in the shape of a regular pentagon. Each side of the square is 7 inches longer than each side of the pentagon. Find the dimensions of each frame. \n" ); document.write( "
Algebra.Com's Answer #157673 by Theo(13342)![]() ![]() You can put this solution on YOUR website! side of the square is 35 inches long. \n" ); document.write( "side of the pentagon is 28 inches long. \n" ); document.write( "----- \n" ); document.write( "4 * 35 = 140 \n" ); document.write( "5 * 28 = 140 \n" ); document.write( "------- \n" ); document.write( "here's how it was done. \n" ); document.write( "----- \n" ); document.write( "a = side of the square \n" ); document.write( "b = side of the pentagon \n" ); document.write( "4a = p \n" ); document.write( "5b = p \n" ); document.write( "both equal to p (perimeter) so they are equal to each other. \n" ); document.write( "4a = 5b \n" ); document.write( "side of the square is 7 inches longer than side of the pentagon. \n" ); document.write( "a = b + 7 \n" ); document.write( "replace a with b + 7 in the equation to get: \n" ); document.write( "4 * (b+7) = 5b \n" ); document.write( "this becomes: \n" ); document.write( "4b + 28 = 5b \n" ); document.write( "this becomes: \n" ); document.write( "b = 28 \n" ); document.write( "----- \n" ); document.write( "if b = 28, then a = 28 + 7 = 35 \n" ); document.write( "a = 35 \n" ); document.write( "b = 28 \n" ); document.write( "----- \n" ); document.write( "side of square = 35 inches \n" ); document.write( "side of pentagon = 28 inches \n" ); document.write( "----- \n" ); document.write( " |