anonymous 5 years ago 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.

1. anonymous

lol, is there an image?

2. anonymous

no. which sucks.

3. anonymous

damn lol I can't think right now, but if you had an image of it then it'll make my job easier :)

4. 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!

5. 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

6. anonymous

thank you for your help LF ^_^

7. anonymous

:)

8. 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.

9. anonymous

then i have 2x^2+2x+4=100. is that right or am i way off.

10. anonymous

$(x+2)^2 + x^2=100$ = $x^2 + 4 + 2(x)(2) + x^2 =100$ using, $(a + b) ^2 = a^2 + b^2 +2ab$

11. anonymous

then how would i go from there? never done a problem like this for awhile.

12. anonymous

its a quadratic eqn in x..solve it for x using the quadratic formula

13. anonymous

2x^2 + 4x -96=0 so x^2 +2x -48=0

14. anonymous

its 8 feet and 6 feet

15. anonymous

x is 6 ft

16. 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 $-b \pm \sqrt{b^2 -4ac}$