SOLUTION: I am looking for some help with this math problem. Thank you. A 5ft diagonal brace on a bridge connects a support of the center of the bridge to a side support on the bridge. Th

Question 407818: I am looking for some help with this math problem. Thank you.
A 5ft 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 1 ft longer than the height that it reaches on the side of the bridge. Find the horizontal and vertical distances spanned by this brace.
Found 2 solutions by ankor@dixie-net.com, lisapivaral:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A 5ft 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 1 ft longer than the height that it
reaches on the side of the bridge.
Find the horizontal and vertical distances spanned by this brace.
:
This just a pythag problem where
x = vert portion of the brace
(x+1) = the horizontal portion
5 ft is the hypotenuse
:
x^2 + (x+1)^2 = 5^2
x^2 + x^2 + 2x + 1 = 25
2x^2 + 2x + 1 - 25 = 0
2x^2 + 2x - 24 = 0
simplify, divide by 2
x^2 + x - 12 = 0
Factor this to
(x+4)(x-3) = 0
positive solution
x = 3 ft, the vert portion
and
3 + 1 = 4 ft, is the horizontal span of the brace
Answer by lisapivaral(1) (Show Source): You can put this solution on YOUR website!
This just a pythag problem where
x = vert portion of the brace
(x+1) = the horizontal portion
5 ft is the hypotenuse
:
x^2 + (x+1)^2 = 5^2
x^2 + x^2 + 2x + 1 = 25
2x^2 + 2x + 1 - 25 = 0
2x^2 + 2x - 24 = 0
simplify, divide by 2
x^2 + x - 12 = 0
Factor this to
(x+4)(x-3) = 0
positive solution
x = 3 ft, the vert portion
and
3 + 1 = 4 ft, is the horizontal span of the brace

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