Question

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.

Answer 1

lol, is there an image?

Answer 2

no. which sucks.

Answer 3

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

Answer 4

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!

Answer 5

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

Answer 6

thank you for your help LF ^_^

Answer 7

:)

Answer 8

(x+2)^2 + x^2=100. would that be right so far. buecause i am never good with these problems and never have been.

Answer 9

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

Answer 10

\[(x+2)^2 + x^2=100 \] = \[x^2 + 4 + 2(x)(2) + x^2 =100 \]


using,
\[(a + b) ^2 = a^2 + b^2 +2ab\]

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

its 8 feet and 6 feet

Answer 15

x is 6 ft

Answer 16

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 \[-b \pm \sqrt{b^2 -4ac} \]