anonymous
  • anonymous
A 10-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 2 ft longer that the height that it reaches on the side of the bridge. Find the horizontal and vertical distances spanned by this brace.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
lol, is there an image?
anonymous
  • anonymous
no. which sucks.
anonymous
  • anonymous
damn lol I can't think right now, but if you had an image of it then it'll make my job easier :)

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anonymous
  • anonymous
The width of a rectangular gate is 2 meters (m) larger than its height. The diagonal brace measures √6m. Find the width and height. --------------------- Draw the picture Let the height be "x" meters Then the width is "x+2" meters ---------------- Draw the diagonal = sqrt(6) meters ----------------- EQUATION: Use Pythagoras to solve for "x": x^2 + (x+2)^2 = [sqrt(6)]^2 2x^2 + 4x + 4 = 6 2x^2 + 4x -2 = 0 x^2 + 2x -1 = 0 --------------------- Use the Quadratic Formula: x = [-2 +- sqrt(4 -4*1*-1)]/2 x = [-2 +- sqrt(8)]/2 Positive solution: x = [-2 + 2sqrt(2)]/2 x = [-1 + sqrt(2)] x = 0.414 meters (height of the rectangle) x+2 = 2.414 meters (width of the rectangle) ============================== Found it here - The width of a rectangular gate is 2 meters (m) larger than its height. The diagonal brace measures √6m. Find the width and height. --------------------- Draw the picture Let the height be "x" meters Then the width is "x+2" meters ---------------- Draw the diagonal = sqrt(6) meters ----------------- EQUATION: Use Pythagoras to solve for "x": x^2 + (x+2)^2 = [sqrt(6)]^2 2x^2 + 4x + 4 = 6 2x^2 + 4x -2 = 0 x^2 + 2x -1 = 0 --------------------- Use the Quadratic Formula: x = [-2 +- sqrt(4 -4*1*-1)]/2 x = [-2 +- sqrt(8)]/2 Positive solution: x = [-2 + 2sqrt(2)]/2 x = [-1 + sqrt(2)] x = 0.414 meters (height of the rectangle) x+2 = 2.414 meters (width of the rectangle) ============================== Found it here - The width of a rectangular gate is 2 meters (m) larger than its height. The diagonal brace measures √6m. Find the width and height. --------------------- Draw the picture Let the height be "x" meters Then the width is "x+2" meters ---------------- Draw the diagonal = sqrt(6) meters ----------------- EQUATION: Use Pythagoras to solve for "x": x^2 + (x+2)^2 = [sqrt(6)]^2 2x^2 + 4x + 4 = 6 2x^2 + 4x -2 = 0 x^2 + 2x -1 = 0 --------------------- Use the Quadratic Formula: x = [-2 +- sqrt(4 -4*1*-1)]/2 x = [-2 +- sqrt(8)]/2 Positive solution: x = [-2 + 2sqrt(2)]/2 x = [-1 + sqrt(2)] x = 0.414 meters (height of the rectangle) x+2 = 2.414 meters (width of the rectangle) ============================== I found it on a website. I have no rights to this solution. Hope it helps!
anonymous
  • anonymous
Alright, I'm not sure of my answer, but that's what I ended up with ^_^ : - bridge = rectangular shape. - diagonal is half way through the rectangle of length = 10ft - L = 2 + x - w = x. >_< LOL! I was abt to say this
anonymous
  • anonymous
thank you for your help LF ^_^
anonymous
  • anonymous
:)
anonymous
  • anonymous
(x+2)^2 + x^2=100. would that be right so far. buecause i am never good with these problems and never have been.
anonymous
  • anonymous
then i have 2x^2+2x+4=100. is that right or am i way off.
anonymous
  • anonymous
\[(x+2)^2 + x^2=100 \] = \[x^2 + 4 + 2(x)(2) + x^2 =100 \] using, \[(a + b) ^2 = a^2 + b^2 +2ab\]
anonymous
  • anonymous
then how would i go from there? never done a problem like this for awhile.
anonymous
  • anonymous
its a quadratic eqn in x..solve it for x using the quadratic formula
anonymous
  • anonymous
2x^2 + 4x -96=0 so x^2 +2x -48=0
anonymous
  • anonymous
its 8 feet and 6 feet
anonymous
  • anonymous
x is 6 ft
anonymous
  • anonymous
Use this formula, \[ax^2 + bx + c = 0\] \[=> x = -b \pm \sqrt{b^2 -4ac} / 2a\] the 2a term is dividing the entire term of \[-b \pm \sqrt{b^2 -4ac} \]

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