document.write( "Question 690994: in triangle ABC, C is a right angle. Point M is the midpoint of AB, Point N is midpoint of AC and O is the midpoint of AM. Perimeter of traingle is 112 and ON is 12.5. What is the area of MNCB? \n" ); document.write( "

Algebra.Com's Answer #426590 by Edwin McCravy(20055) $\"\"$ $\"About$

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document.write( "Median ON of right triangle AMN\r\n" );
document.write( "is drawn to the hypotenuse AM,\r\n" );
document.write( "which divides right triangle AMN\r\n" );
document.write( "into two isosceles triangles.\r\n" );
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document.write( "Therefore ON=AO=OM=12.5.  (All\r\n" );
document.write( "three can be seen to be radii of\r\n" );
document.write( "the green semicircle which inscribes\r\n" );
document.write( "right angle ANM.)\r\n" );
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document.write( "Therefore AM = 2×12.5 = 25\r\n" );
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document.write( "By the Pythagorean theorem on \r\n" );
document.write( "right triangle AMN,\r\n" );
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document.write( "MN² + AN² = AM²\r\n" );
document.write( "  x² + y² = 25²\r\n" );
document.write( "  x² + y² = 625 \r\n" );
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document.write( "Each side of triangle AMN is one-\r\n" );
document.write( "half of the corresponding sides\r\n" );
document.write( "of triangle ABC, so its perimeter\r\n" );
document.write( "is one-half of the perimeter of\r\n" );
document.write( "triangle ABC, therefore\r\n" );
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document.write( "MN + AN + AM = ·112 = 56\r\n" );
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document.write( "  x + y + 25 = 56\r\n" );
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document.write( "       x + y = 31\r\n" );
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document.write( "So we solve the system of equations\r\n" );
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document.write( "by substitution and we get 2 \r\n" );
document.write( "solutions, \r\n" );
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document.write( "(x,y) = (24,7) and (x,y) = (7,24).\r\n" );
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document.write( "Using the first solution:\r\n" );
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document.write( "AC = 2·MN = 2x = 2(24) = 48.\r\n" );
document.write( "NC = y = 7.\r\n" );
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document.write( "So the two parallel side of trapezoid\r\n" );
document.write( "MNCB are AC=48 and MN=24, and the\r\n" );
document.write( "height is NC = 7, so the area, using\r\n" );
document.write( "the area formula for a trapezoid:\r\n" );
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document.write( "Area = \r\n" );
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document.write( "Area =  = 252.\r\n" );
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document.write( "(x,y) = (7,24).\r\n" );
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document.write( "Using the second solution:\r\n" );
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document.write( "AC = 2·MN = 2x = 2(7) = 14.\r\n" );
document.write( "NC = y = 24.\r\n" );
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document.write( "So the two parallel side of trapezoid\r\n" );
document.write( "MNCB are AC=14 and MN=24, and the\r\n" );
document.write( "height is NC = 24, so the area, using\r\n" );
document.write( "the area formula for a trapezoid:\r\n" );
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document.write( "Area = \r\n" );
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document.write( "Area =  = 252.\r\n" );
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document.write( "So both solutions gives the same area of 252.\r\n" );
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document.write( "The drawing above is drawn approximately to scale\r\n" );
document.write( "for the first solution,  Here is the way the\r\n" );
document.write( "drawing would look for the second solution.\r\n" );
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document.write( "So there are two drawings being talked about here,\r\n" );
document.write( "but both have the same area of MNCB, which is 252.\r\n" );
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document.write( "Edwin

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