In a model, it is shown that an arc of a bridge in semi-elliptical with major axis horizontal. If the length of the base is 9 m and the highest part of the bridge is 3 m from the horizontal, the best approximation of the height of the arch at 2 m from the centre of the base is

Options

A. 11/4 m
B. 8/3 m
C. 7/2 m
D. 2 m

Answer

To solve the problem, we need to find the height of the arch of a semi-elliptical bridge at a distance of 2 meters from the center of its base. Here are the steps to arrive at the solution:

### Step 1: Understand the dimensions of the ellipse
The length of the base of the bridge is given as 9 meters. Since the base is the major axis of the ellipse, we can denote this as \(2a\). Therefore, we have:
\[ 2a = 9 \implies a = \frac{9}{2} = 4.5 \text{ meters} \]

### Step 2: Determine the height of the ellipse
The highest part of the bridge is given as 3 meters, which corresponds to the semi-minor axis \(b\) of the ellipse. Thus, we have:
\[ b = 3 \text{ meters} \]

### Step 3: Write the equation of the ellipse
The standard form of the equation of an ellipse centered at the origin with a horizontal major axis is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Substituting the values of \(a\) and \(b\):
\[ \frac{x^2}{(4.5)^2} + \frac{y^2}{3^2} = 1 \]
This simplifies to:
\[ \frac{x^2}{20.25} + \frac{y^2}{9} = 1 \]

### Step 4: Substitute \(x = 2\) into the equation
We want to find the height \(y\) at a distance of 2 meters from the center of the base, which means we substitute \(x = 2\) into the equation:
\[ \frac{(2)^2}{20.25} + \frac{y^2}{9} = 1 \]
This simplifies to:
\[ \frac{4}{20.25} + \frac{y^2}{9} = 1 \]

### Step 5: Solve for \(y^2\)
Calculating \(\frac{4}{20.25}\):
\[ \frac{4}{20.25} = \frac{4}{\frac{81}{4}} = \frac{16}{81} \]
Now substituting this back into the equation:
\[ \frac{16}{81} + \frac{y^2}{9} = 1 \]
Subtract \(\frac{16}{81}\) from both sides:
\[ \frac{y^2}{9} = 1 - \frac{16}{81} \]
Finding a common denominator:
\[ 1 = \frac{81}{81} \implies \frac{y^2}{9} = \frac{81 - 16}{81} = \frac{65}{81} \]
Now, multiply both sides by 9:
\[ y^2 = \frac{65 \times 9}{81} = \frac{585}{81} \]

### Step 6: Calculate \(y\)
Taking the square root:
\[ y = \sqrt{\frac{585}{81}} = \frac{\sqrt{585}}{9} \]
To approximate \(\sqrt{585}\), we can estimate it:
\[ \sqrt{585} \approx 24.2 \quad (\text{since } 24^2 = 576 \text{ and } 25^2 = 625) \]
Thus:
\[ y \approx \frac{24.2}{9} \approx 2.69 \text{ meters} \]

### Step 7: Best Approximation
For the best approximation, we can find a simpler fraction:
\[ \sqrt{64} = 8 \implies \text{Best Approximation} = \frac{8}{3} \text{ meters} \]

### Final Answer
The best approximation of the height of the arch at 2 meters from the center of the base is:
\[ \frac{8}{3} \text{ meters} \]