SOLUTION: 9. A 150-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 30 ft longer than th

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: 9. A 150-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 30 ft longer than th      Log On


   

Question 507707: 9. A 150-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 30 ft longer than the height that it reaches on the side of the bridge. Find the horizontal and vertical distances spanned by this brace.


Calling all math genius please help me!
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = height, so x+30 is the horizontal distance


So use the pythagorean theorem to find the height

a^2 + b^2 = c^2

x^2 + (x+30)^2 = 150^2

x^2 + x^2+60x+900 = 22500

x^2 + x^2+60x+900 - 22500 = 0

2x^2+60x-21600 = 0

2(x^2+30x-10800) = 0

x^2+30x-10800 = 0

Now use the quadratic formula to solve for x


x = (-B +- sqrt( B^2-4AC ))/(2A)


x = (-(30) +- sqrt( (30)^2-4(1)(-10800) ))/(2(1))


x = (-30 +- sqrt( 900+43200 ))/(2(1))


x = (-30 +- sqrt( 44100 ))/(2(1))


x = (-30 +- 210)/(2)


x = (-30 + 210)/(2) or x = (-30 - 210)/(2)


x = (180)/(2) or x = (-240)/(2)


x = 90 or x = -120


Now throw out the negative solution to get the only solution of x = 90


So the height is 90 ft and the horizontal distance is 120 ft (since 90+30 = 120)

Let me know if you need more help or if you need me to explain a step in more detail.
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Thanks,

Jim