document.write( "Question 1190163: How do I find the perimeter of a rectangle that's in a circle when all I know is the diameter (6 inches) and the area of the square (15). I've been at this for a week please help \n" ); document.write( "
Algebra.Com's Answer #821723 by ikleyn(52788)\"\" \"About 
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\n" ); document.write( "How do I find the perimeter of a rectangle that's in a circle when all I know is the diameter (6 inches) and the area of the square (15).
\n" ); document.write( "I've been at this for a week please help
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\n" ); document.write( "\n" ); document.write( "After reading your post,  I have a question: \r
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\n" ); document.write( "\n" ); document.write( "        does the problem talks about a  RECTANGLE  or about a  SQUARE ?\r
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\n" ); document.write( "\n" ); document.write( "When formulated in a correct way,  this problem has a nice, short, simple and elegant  \" exact \"  solution.\r
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document.write( "    |    Find the perimeter of a rectangle inscribed in a circle of the     |\r\n" );
document.write( "    |    diameter 6 inches, if the area of the rectangle is 15 sq. inches.  | \r\n" );
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document.write( "Let x and y be the dimensions of the rectangle.\r\n" );
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document.write( "Then you have these two equations\r\n" );
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document.write( "    x^2 + y^2 = 36      (1)    (The Pythagoras, applied to the legs and the hypotenuse.\r\n" );
document.write( "                                which is the diameter of the circle)\r\n" );
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document.write( "    xy        = 15      (2)    (area equation).\r\n" );
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document.write( "Multiply equation (2) by 2 (both sides) and add it to equation (1).  You will get\r\n" );
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document.write( "    x^2 + 2xy + y^2 = 36 + 2*15,\r\n" );
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document.write( "or\r\n" );
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document.write( "    (x + y)^2 = 66,   which implies  x + y = \"sqrt%2866%29\".      (3)\r\n" );
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document.write( "The perimeter of the rectangle is  2x + 2y = ( from equation (3) ) = \"2%2Asqrt%2866%29\" = 16.248  inches (approximately).\r\n" );
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document.write( "Thus you have BOTH \"exact\" solution and approximate (rounded) numerical value.\r\n" );
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