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

This is what i have so far

(x + 2)^2 + x^2 = 100
X^2 +4 + 2 (x) (2) +x^2 = 100
Using,
(a + b)^2 = a^2 + b^2 +2ab
2x^2 +4x -96 = 0
So x^2 + 2x -48 = 0
Its 8 feet and 6 feet.
X is 6 ft.
Using this formula,
Ax^2 +bx +c = 0
x = -b ± √(b^2- 4ac)/2a
8=-b±√(b^2-4ac)/2a

Answer 2

from here I have no clue.

Answer 3

or am i wrong.

Answer 4

Lol
Look at the pythagorean thereom Ill send a pic with it
Solve for y
[4y^2 + y^2] = (5^1/2)*y = 10
y = 10/(5)^1/2 = 4.5 ft (vertical)
and 9.0ft (horizontal)

Answer 5

I am sorry that is a mistake...I made the horizontal distance 2 times the vertical
new solution in a minute

Answer 6

ok

Answer 7

u there.