document.write( "Question 500319: Problem:
\n" ); document.write( "ABC is a right triangle. The length of its perimeter is equal to 60 units and the size of its area is equal to 150 units squared. Find the two sides and hypotenuse.
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Algebra.Com's Answer #338363 by cleomenius(959)\"\" \"About 
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Area f triangle = 150 = 1/2ab
\n" ); document.write( "ab = 300
\n" ); document.write( "a^2 + b^2 = c^2.
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\n" ); document.write( "a + b + c = 60; the perimeter.
\n" ); document.write( "a+ b = 60 - c rearrangement.
\n" ); document.write( "square both sides.
\n" ); document.write( "(a + b)^2 = (60 - c)^2
\n" ); document.write( "a^2 + 2ab + b^2 = 3600 + c^2 - 120c
\n" ); document.write( "subtract the equation A^2 + b^2 = c^2 (simultaneous equations)
\n" ); document.write( "2ab = 3600 - 120c
\n" ); document.write( "ab is known to be 300, hence
\n" ); document.write( "600 = 3600 - 120c
\n" ); document.write( "-3000 = -120c
\n" ); document.write( "25 = c
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\n" ); document.write( "substitute c by 25 in the equation a + b + c = 60 to obtain
\n" ); document.write( "a + b + 25 = 60
\n" ); document.write( "a + b = 35
\n" ); document.write( "Since AB = 300 then b = 300/a which is substituted into the equation a + b = 35
\n" ); document.write( "a + 300/a - 35 = 0
\n" ); document.write( "multiply all terms by a.
\n" ); document.write( "a^2 -35a + 300 = 0
\n" ); document.write( "Solve our quadratic equation.
\n" ); document.write( "(x -15)(x - 20)
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\n" ); document.write( "15, 20, 25 are our dimensions.
\n" ); document.write( "15 + 20 + 25 = 60 units perimeter.
\n" ); document.write( "This does check.
\n" ); document.write( "15 * 20 = 300 units area.
\n" ); document.write( "This does check.
\n" ); document.write( "15^2 + 20^2 = 25^2
\n" ); document.write( "225 + 400 = 625 units.
\n" ); document.write( "This does check also.
\n" ); document.write( "Cleomenius
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