Question

What is the height reached by a 12-ft diagonal brace on a bridge, given the horizontal distance it spans is 3 ft longer than the height?

Answer 1

Let's denote the height reached by the diagonal brace as \( h \) (in feet). According to the problem, the horizontal distance that the diagonal brace spans is \( h + 3 \) feet.

Since the diagonal brace forms a right triangle with the height and the horizontal distance, we can use the Pythagorean theorem. The lengths of the sides of the triangle are as follows:

- The height \( h \)
- The horizontal distance \( h + 3 \)
- The diagonal length (the brace) is \( 12 \) ft

According to the Pythagorean theorem:

\[
(h + 3)^2 + h^2 = 12^2
\]

Now we can expand and simplify this equation:

1. Expand \( (h + 3)^2 \):

\[
(h + 3)^2 = h^2 + 6h + 9
\]

2. Substitute this back into the Pythagorean theorem:

\[
h^2 + 6h + 9 + h^2 = 144
\]

3. Combine like terms:

\[
2h^2 + 6h + 9 = 144
\]

4. Rearranging gives:

\[
2h^2 + 6h + 9 - 144 = 0
\]
\[
2h^2 + 6h - 135 = 0
\]

5. Simplify the equation by dividing by 2:

\[
h^2 + 3h - 67.5 = 0
\]

Now we can use the quadratic formula to find \( h \):

\[
h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = 3 \), and \( c = -67.5 \).

6. Calculate the discriminant:

\[
b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot (-67.5) = 9 + 270 = 279
\]

7. Now, substituting into the quadratic formula:

\[
h = \frac{-3 \pm \sqrt{279}}{2}
\]

8. We only consider the positive solution since height cannot be negative:

\[
h = \frac{-3 + \sqrt{279}}{2}
\]

To approximate \( \sqrt{279} \):

\[
\sqrt{279} \approx 16.73
\]

Thus,

\[
h \approx \frac{-3 + 16.73}{2} \approx \frac{13.73}{2} \approx 6.865
\]

Rounding to two decimal places, the height \( h \) is approximately:

\[
\boxed{6.87 \text{ ft}}
\]