document.write( "Question 384605: I am having a nightmare with this one. Can someone please help me.\r
\n" ); document.write( "\n" ); document.write( "A 75-ft diagonal brace on a bridge connects a support of the center of the bridge to a side support on the bridge. The horizontal distance that it spans is 15 ft longer that the height that it reaches on the side of the bridge. Find the horizontal and vertical distances spanned by this brace.
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Algebra.Com's Answer #272256 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
A 75-ft diagonal brace on a bridge connects a support of the center of the bridge to a side support on the bridge.
\n" ); document.write( "Draw this right triangle with hypotenuse = 75 ft
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\n" ); document.write( "The horizontal distance that it spans is 15 ft longer that the height that it reaches on the side of the bridge.
\n" ); document.write( "Let the height be of the triangle be x ft.
\n" ); document.write( "Then the horizontal distance is (x+15) ft
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\n" ); document.write( "Find the horizontal and vertical distances spanned by this brace.
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\n" ); document.write( "Use Pythgoras to find the missing sides of your triangle:
\n" ); document.write( "75^2 = x^2 + (x+15)^2
\n" ); document.write( "x^2 + x^2+30x+225 = 75^2
\n" ); document.write( "2x^2 + 30x - 5400 = 0
\n" ); document.write( "x^2 + 15x -2700 = 0
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\n" ); document.write( "(x-45)(x+60) = 0
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\n" ); document.write( "Positive solution:
\n" ); document.write( "x = 45 ft (vertical distance of the brace)
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\n" ); document.write( "x+15 = 60 ft (horizontal distance of the brace)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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