A 125-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 25 ft longer than the height that it reaches on the side of the bridge. Find the horizontal and vertical distances spanned by this brace.

Check 4-30-11,9:40pm post for solution

125^2=height^2 + horiz^2

125^2=height^2+ (height-25)^2
solve for height first.

To find the horizontal and vertical distances spanned by the diagonal brace, we can use the Pythagorean theorem since we have a right triangle formed by the diagonal brace, the horizontal distance, and the vertical distance.

Let's call the horizontal distance "x" and the vertical distance "y."

According to the problem, the diagonal brace has a length of 125 ft. This means that we have the equation:

x^2 + y^2 = 125^2

We are also given that the horizontal distance is 25 ft longer than the height, so we can write:

x = y + 25

Now we can substitute this value of x into the first equation:

(y + 25)^2 + y^2 = 125^2

Expanding and simplifying the equation, we get:

y^2 + 50y + 625 + y^2 = 15625

Combining like terms, we have:

2y^2 + 50y + 625 = 15625

Rearranging the equation, we get:

2y^2 + 50y - 15000 = 0

To solve this quadratic equation, we can use the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 2, b = 50, and c = -15000. Plugging in these values, we get:

y = (-50 ± √(50^2 - 4(2)(-15000))) / (2(2))

Simplifying further, we have:

y = (-50 ± √(2500 + 120000)) / 4

y = (-50 ± √122500) / 4

y = (-50 ± 350) / 4

Now we have two possible values for y:

y = (-50 + 350) / 4 = 75

or

y = (-50 - 350) / 4 = -100

However, since the height cannot be negative, we can discard the negative value.

Therefore, the vertical distance spanned by the brace is y = 75 ft.

Now we can substitute this value of y into the equation x = y + 25 to find the horizontal distance:

x = 75 + 25 = 100 ft

So, the horizontal distance spanned by the brace is x = 100 ft, and the vertical distance spanned is y = 75 ft.