SOLUTION: Hi, I really need help with this please? A 15-ft diagonal brace on a bridge connects a support of the center of the bridge to a side support on the bridge. The horizontal distan

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Question 492111: Hi, I really need help with this please?
A 15-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 3 ft longer than the height that it reaches on the side of the bridge. Find the horizontal and vertical distances spanned by this brace.
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
it forms a right triangle.

the hypotenuse is the diagonal.
one of the legs is the vertical.
one of the legs is the horizontal.

the length of the horizontal is x + 3
the length of the vertical is x.

from the pythagorean formula:

x^2 + (x+3)^2 = 15^2

this becomes:

x^2 + x^2 + 6x + 9 = 225 which becomes:

2x^2 + 6x + 9 = 225

subtract 225 from both sides of this equation to get:

2x^2 + 6x - 216 = 0

divide both sides of this equation by 2 to get:

x^2 + 3x - 108 = 0

factor this equation to get:

(x -9) * (x + 12) = 0

this makes x = 9 or x = -12.

since x can't be negative, then x has to be equal to 9.

the diagonal is 15 feet.
the horizontal is 9 + 3 = 12
the vertical is 9.

from the pythagorean formula, 9^2 + 12^2 = 81 + 144 = 225 = 15^2 which is correct.

that's your answer.
the horizontal distance is equal to 12 feet.
the vertical distance is equal to 9 feet.