An arch is in the shape of a parabola whose axis is vertically downwords and measures 80 mts across its boltom on the ground. Its highest point is 24 mts. The measure of the horizontal beam across its cross section at a height or 18 mts is

Options

A. 50
B. 40
C. 45
D. 60

Answer

To solve the problem step by step, we will derive the equation of the parabola and then find the width of the horizontal beam at a height of 18 meters.

### Step 1: Understanding the Parabola
The parabola opens downwards, and we can represent it in the standard form as:
\[ x^2 = -4ay \]
where \( a \) is a constant that determines the distance from the vertex to the focus.

### Step 2: Identifying the Vertex and Points
The highest point (vertex) of the parabola is at the coordinates (0, -24) since it opens downwards and the height is given as 24 meters above the ground. The parabola is 80 meters wide at the ground level, meaning it spans from -40 to 40 on the x-axis.

### Step 3: Finding the Value of \( a \)
Since the parabola passes through the point (40, 0) (the ground level), we can substitute this point into the parabola's equation to find \( a \):
\[ 40^2 = -4a(0 + 24) \]
This simplifies to:
\[ 1600 = -4a(-24) \]
\[ 1600 = 96a \]
\[ a = \frac{1600}{96} = \frac{50}{3} \]

### Step 4: Writing the Equation of the Parabola
Now that we have \( a \), we can write the equation of the parabola:
\[ x^2 = -4 \left(\frac{50}{3}\right) y \]
This simplifies to:
\[ x^2 = -\frac{200}{3} y \]

### Step 5: Finding the Width at Height 18 Meters
To find the width of the beam at a height of 18 meters, we first need to find the corresponding y-coordinate:
\[ y = -24 + 18 = -6 \]
Now, substitute \( y = -6 \) into the parabola's equation:
\[ x^2 = -\frac{200}{3}(-6) \]
This simplifies to:
\[ x^2 = \frac{1200}{3} = 400 \]
Taking the square root gives:
\[ x = \pm 20 \]

### Step 6: Calculating the Width of the Beam
The total width of the beam at this height is:
\[ \text{Width} = 20 - (-20) = 20 + 20 = 40 \text{ meters} \]

### Final Answer
The measure of the horizontal beam across its cross-section at a height of 18 meters is **40 meters**.

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