# 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

Algebra ->  Algebra  -> Test  -> Lessons -> 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      Log On

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 Click here to see ALL problems on test 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(3458)   (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.