SOLUTION: I am having a nightmare with this one. Can someone please help me. A 75-ft diagonal brace on a bridge connects a support of the center of the bridge to a side support on the bri

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I am having a nightmare with this one. Can someone please help me. A 75-ft diagonal brace on a bridge connects a support of the center of the bridge to a side support on the bri      Log On


   

Question 384605: I am having a nightmare with this one. Can someone please help me.
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.
Found 2 solutions by stanbon, Earlsdon:
Answer by stanbon(75887) About Me  (Show Source):
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.
Draw this right triangle with hypotenuse = 75 ft
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The horizontal distance that it spans is 15 ft longer that the height that it reaches on the side of the bridge.
Let the height be of the triangle be x ft.
Then the horizontal distance is (x+15) ft
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Find the horizontal and vertical distances spanned by this brace.
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Use Pythgoras to find the missing sides of your triangle:
75^2 = x^2 + (x+15)^2
x^2 + x^2+30x+225 = 75^2
2x^2 + 30x - 5400 = 0
x^2 + 15x -2700 = 0
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(x-45)(x+60) = 0
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Positive solution:
x = 45 ft (vertical distance of the brace)
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x+15 = 60 ft (horizontal distance of the brace)
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Cheers,
Stan H.
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Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
You could use the Pythagorean theorem c%5E2+=+a%5E2%2Bb%5E2 to solve this problem.
You have a right triangle whose hypotenuse is the 75ft. diagonal and whose height is h. The base of this triangle is h+15ft.
You need to find h and h+15.
75%5E2+=+h%5E2%2B%28h%2B15%29%5E2
5625+=+h%5E2%2B%28h%5E2%2B30h%2B225%29 Rewrite as a quadratic equation in standard form.
2h%5E2%2B30h-5400+=+0 Divide by 2 to simplify a bit.
h%5E2%2B15h-2700+=+0 Solve by factoring.
%28h%2B60%29%28h-45%29+=+0 Apply the zero product rule.
h%2B60+=+0 or h-45+=+0
h+=+-60 or h+=+45 Discard the negative solution as the height, h, is a positive quantity.
h+=+45feet. This is the vertical distance.
h%2B15+=+60feet. This is the horizontal distance.
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I do hope this takes care of your nightmare!
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