A water plane has a length of 60m. Its half-ordinates are given as: 0.75m, 2.8m, 4.0m, 4.0m, 3.8m, 3.2m, 0m at equal intervals. Calculate the area of the water plane using Simpson's First Rule.

Step 1: Identify the given data and calculate the interval length.
The length of the water plane is $L = 60 \, \text{m}$.
The half-ordinates ($y$) are given at equal intervals:
$y_0 = 0.75 \, \text{m}$ (AP)
$y_1 = 2.8 \, \text{m}$
$y_2 = 4.0 \, \text{m}$
$y_3 = 4.0 \, \text{m}$
$y_4 = 3.8 \, \text{m}$
$y_5 = 3.2 \, \text{m}$
$y_6 = 0 \, \text{m}$ (FP)
There are $n=7$ ordinates, which means there are $n-1=6$ intervals.
The interval length ($h$) is calculated as:

$$h = \frac{L}{n-1} = \frac{60 \, \text{m}}{6} = 10 \, \text{m}$$

Step 2: Apply Simpson's First Rule to find the sum of products.
Simpson's multipliers for 7 ordinates are 1, 4, 2, 4, 2, 4, 1.
We calculate the product of each half-ordinate and its corresponding Simpson's multiplier:

$$y_0 \times 1 = 0.75 \times 1 = 0.75$$ $$y_1 \times 4 = 2.8 \times 4 = 11.2$$ $$y_2 \times 2 = 4.0 \times 2 = 8.0$$ $$y_3 \times 4 = 4.0 \times 4 = 16.0$$ $$y_4 \times 2 = 3.8 \times 2 = 7.6$$ $$y_5 \times 4 = 3.2 \times 4 = 12.8$$ $$y_6 \times 1 = 0 \times 1 = 0$$

The sum of these products (SP) is:

$$SP = 0.75 + 11.2 + 8.0 + 16.0 + 7.6 + 12.8 + 0 = 56.35$$