document.write( "Question 136320: The hypotenuse of a right angled triangle is 2 cm. more than its base and 1 cm. more than twice the perpendicular. Find the sides of the triangle? \n" ); document.write( "

Algebra.Com's Answer #99882 by vleith(2983) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let b = the length of the base. Let the other sides be h and p\r
\n" ); document.write( "\n" ); document.write( "h = b+2 ---> b = h-2
\n" ); document.write( "h = 2p+1 ---> p = (h-1)/2\r
\n" ); document.write( "\n" ); document.write( "we also know h^2 = b^2 + p^2\r
\n" ); document.write( "\n" ); document.write( " $\"h%5E2+=+%28h-2%29%5E2+%2B+%28%28h-1%29%2F2%29%5E2+\"$
\n" ); document.write( " $\"h%5E2+=+h%5E2+-4h+%2B+4+%2B+%28h%5E2+-2h+%2B+1%29%2F4+\"$
\n" ); document.write( " $\"4h%5E2+=+4h%5E2+-+16h+%2B+16+%2B+h%5E2+-2h+%2B1\"$
\n" ); document.write( " $\"0+=+h%5E2+-18h+%2B17\"$
\n" ); document.write( " $\"0+=+%28h-17%29%28h-1%29\"$
\n" ); document.write( "h = 17 or 1.
\n" ); document.write( "Since base is h-2, h cannot be one since then b would be negative\r
\n" ); document.write( "\n" ); document.write( "so h = 17
\n" ); document.write( "b = 15
\n" ); document.write( "p = 8\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

\n" );