document.write( "Question 585308: A 60cm piece of wire is bent to form a right triangle. If the hypotenuse is 26cm long, what are legs of the right triangle? \n" ); document.write( "

Algebra.Com's Answer #373266 by Schaman_Dempster(26) $\"\"$ $\"About$

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Let the other sides of the right triangle be x and y.\r
\n" ); document.write( "\n" ); document.write( "Since the length of the wire was 60 cm, the whole 60 cm has been bent to form a right triangle and that means 60 cm is as well the perimeter of the triangle.\r
\n" ); document.write( "\n" ); document.write( "Therefore, x + y + 26 = 60
\n" ); document.write( "or x + y = 34
\n" ); document.write( "or y = 34 -x ---------------- (1)\r
\n" ); document.write( "\n" ); document.write( "Using the Pythagoras theorem,\r
\n" ); document.write( "\n" ); document.write( "x^2 + y^2 = 26^2
\n" ); document.write( "x^2 + (34 - x)^2 = 26^2 {Using equation (1)}
\n" ); document.write( "x^2 + 34^2 + x^2 - 68x = 26^2
\n" ); document.write( "2x^2 - 68x + 34^2 - 26^2 = 0
\n" ); document.write( "2x^2 - 68x + (34 -26)(34+26) = 0 { a^2 - b^2 = (a+b)(a-b)}
\n" ); document.write( "2x^2 - 68x + 8*60 = 0
\n" ); document.write( "2x^2 - 68x + 480 = 0
\n" ); document.write( "x^2 - 34x + 240 = 0 { Dividing the equation throughout by 2}\r
\n" ); document.write( "\n" ); document.write( "x^2 - 24x - 10x + 240 = 0
\n" ); document.write( "x (x -24) - 10(x-24) =0
\n" ); document.write( "(x -10) (x-24) = 0
\n" ); document.write( "x =10, x =24\r
\n" ); document.write( "\n" ); document.write( "Plugging back the value of x in equation (1), we get:\r
\n" ); document.write( "\n" ); document.write( "When x =10, y = 34 -10 = 24
\n" ); document.write( "When x= 24, y = 34-24 =10\r
\n" ); document.write( "\n" ); document.write( "Hence, the two legs of the right triangle are 10 cm and 24 cm.\r
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