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The problem involves a semi-ellipse that represents the arch of a bridge. The arch has a major axis of 12 feet across (horizontally) and a height (semi-minor axis) of 20 feet. We are asked to find how high above the water level the arch is at a point 20 feet horizontally from the center.

### Step-by-Step Solution:

1. **Equation of the Ellipse**:
   The standard form of an ellipse centered at the origin \((0, 0)\) is:
   \[
   \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
   \]
   where \(a\) is the semi-major axis and \(b\) is the semi-minor axis.

   Given:
   - Semi-major axis \(a = \frac{12}{2} = 6\) feet
   - Semi-minor axis \(b = 20\) feet

   Thus, the equation becomes:
   \[
   \frac{x^2}{6^2} + \frac{y^2}{20^2} = 1
   \]
   Simplifying further:
   \[
   \frac{x^2}{36} + \frac{y^2}{400} = 1
   \]

2. **Calculate the height at \(x = 20\) feet**:
   We substitute \(x = 20\) feet into the equation to solve for \(y\) (the height above the water level):
   \[
   \frac{20^2}{36} + \frac{y^2}{400} = 1
   \]
   \[
   \frac{400}{36} + \frac{y^2}{400} = 1
   \]
   \[
   \frac{100}{9} + \frac{y^2}{400} = 1
   \]

   Subtract \(\frac{100}{9}\) from both sides:
   \[
   \frac{y^2}{400} = 1 - \frac{100}{9}
   \]

   First, calculate \(1 - \frac{100}{9}\):
   \[
   1 - \frac{100}{9} = \frac{9}{9} - \frac{100}{9} = \frac{-91}{9}
   \]

   Now solve for \(y^2\):
   \[
   \frac{y^2}{400} = \frac{-91}{9}
   \]

   Finally, multiply both sides by 400 to solve for \(y^2\):
   \[
   y^2 = \frac{-91 \times 400}{9} = -4044.44 \, \text{(approx)}
   \]

However, the equation has provided a complex value, meaning the point \(x = 20\) feet is actually not within the arch of the ellipse. This suggests that this might be an error or misinterpretation in the problem statement. 

Please clarify the specific values, as it seems \(20\) feet from the center horizontally is beyond the range of this particular ellipse, meaning there would be no real number \(y\) (height) at that point.

### Would you like further clarification or a different approach?

Here are some related questions you might be interested in:

1. How do the dimensions of a semi-ellipse change when the semi-major or semi-minor axis changes?
2. How would the equation change if the ellipse were shifted horizontally or vertically?
3. How can we verify if a point lies on or within an ellipse?
4. What happens to the height if the semi-minor axis is shortened or lengthened?
5. How do the properties of an ellipse relate to real-world structures like bridges?

**Tip**: When dealing with ellipses, always check whether the values for \(x\) fall within the range of the ellipse's dimensions to ensure valid results.

The arch of the bridge is in the shape of a semi-ellipse, with its major axis at the water level. Suppose the arch is 20 ft high in the middle and 12 ft across the major axis. How high above the water level is the major arch at a point 20 ft (horizontally) from the center? Round off to 2 decimal places.

This problem involves determining the height of a semi-ellipse bridge arch above water level. Given that the bridge arch is 20 feet high in the middle and spans 12 feet across, the solution uses the ellipse formula to find the height at a horizontal distance of 20 feet from the center. Suitable for high school students (Grades 10-12).

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