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 that the height that it reaches on the side of the bridge. Find the horizontal and vertical distances spanned by this brace.

We represent the prob. by a rt triangle:

h = height = ver side.
(h + 25) = hor. side.
hyp. = 125 ft.

(h + 25)^2 + h^2 = (125)^2,
h^2 + 50h + 625 + h^2 = 15625,
2h^2 + 50h - 15000 = 0
Divide both sides by 2:
h^2 + 25h - 7500 = 0,
(h - 75)(h + 100) = 0,

h - 75 = 0,
h = 75 Ft. = ver. dist.

h + 25 = 75 + 25 = 100 Ft. = hor. dist.

X + 100 = 0,
X = -100.

Use positive value of X:
X = 75 FT.

X + 25 = 75 + 25 = 100 Ft. =

horizontal distance =100 Ft

vertical distance=75 ft

To find the horizontal and vertical distances spanned by the brace, we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.

Let's denote the horizontal distance spanned by the brace as x and the vertical distance spanned by the brace as y. We know that the diagonal brace is 125 ft long, so we can write the following equation:

x^2 + y^2 = 125^2

We are also given that the horizontal distance is 25 ft longer than the vertical distance, 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:

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

Combining like terms:

2y^2 + 50y + 625 - 15625 = 0

Subtracting 15625 from both sides:

2y^2 + 50y - 15000 = 0

Dividing the equation by 2 to simplify:

y^2 + 25y - 7500 = 0

Now we have a quadratic equation in terms of y. We can solve this equation to find the possible values of y. Once we have the value of y, we can substitute it back into x = y + 25 to find the corresponding value of x.

Factoring or using the quadratic formula, we find two solutions for y: y = 50 and y = -150.

Since distance can't be negative, we discard the negative value, so y = 50 ft.

Now we can substitute this value of y back into x = y + 25:

x = 50 + 25 = 75 ft

Therefore, the horizontal distance spanned by the brace is 75 ft and the vertical distance spanned by the brace is 50 ft.