document.write( "Question 785524: express the hypotenuse h of a right triangle with area 25m^ as a function of its perimeter p (take base and height as a and b) \n" ); document.write( "
Algebra.Com's Answer #477776 by KMST(5328)\"\" \"About 
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\"a\"= length of one leg, in meters
\n" ); document.write( "\"b\"= length of the other leg, in meters
\n" ); document.write( "\"h\"= hypotenus of the triangle, in meters
\n" ); document.write( "\"ab%2F2=25\"= area of the triangle, in square meters
\n" ); document.write( "\"p=a%2Bb%2Bh\"= perimeter of the triangle, in square meters
\n" ); document.write( "
\n" ); document.write( "\"ab%2F2=25\"-->\"ab=25%2A2\"-->\"ab=50\"
\n" ); document.write( "\"p=a%2Bb%2Bh\"-->\"p-h=a%2Bb\"
\n" ); document.write( "According to the Pythagorean theorem:
\n" ); document.write( "\"h%5E2=a%5E2%2Bb%5E2\"
\n" ); document.write( "\"p%5E2=%28a%2Bb%2Bh%29%5E2=a%5E2%2Bb%5E2%2Bh%5E2%2B2ab%2B2ah%2B2bh\"
\n" ); document.write( "\"p%5E2-h%5E2=h%5E2%2B2ab%2B2ah%2B2bh\"
\n" ); document.write( "\"p%5E2-h%5E2=h%5E2%2B2%2A50%2B2ah%2B2bh\" (substituting \"ab=50\")
\n" ); document.write( "\"p%5E2=2h%5E2%2B100%2B2ah%2B2bh\"
\n" ); document.write( "\"p%5E2=2h%5E2%2B100%2B2h%28a%2Bb%29\"
\n" ); document.write( "\"p%5E2=2h%5E2%2B100%2B2h%28p-h%29\" (substituting \"p-h=a%2Bb\")
\n" ); document.write( "\"p%5E2=2h%5E2%2B100%2B2hp-2h%5E2\"
\n" ); document.write( "\"p%5E2=100%2B2hp\"
\n" ); document.write( "\"p%5E2-100=2hp\"
\n" ); document.write( "\"highlight%28h=%28p%5E2-100%29%2F2p%29\"
\n" ); document.write( "or something equivalent, like
\n" ); document.write( "\"highlight%28h=p%2F2-50%2Fp%29\" or \"highlight%28h=%28p%2B10%29%28p-10%29%2F2p%29\"
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\n" ); document.write( "EXTRA:
\n" ); document.write( "A more general expression, bases on perimeter and area \"A\" would be
\n" ); document.write( "\"highlight%28h=%28p%5E2-4A%29%2F2p%29\"
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