Question

1. Using Simpson's one-third rule, find the area of a field having the following dimensions:
Ordinates: 2, 9, 18, 40,70 and common distance = 30 m.


2. Using Simpson's rule, find the area of a field having the following dimensions :
Ordinates: 23, 19, 14, 11, 12.5, 16, 19, 20, 20
Common distance = 1.5​

Answer 1

Answer:

7

Step-by-step explanation:

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Answer 2

Answer:

1. By using Simpson's one-third rule Area = 3040 $m^{2}$

2. By using Simpson's one-third rule Area = 199 $m^{2}$.

Concept:

It says that the first and last ordinates must be added together. Add four times the sum of the last remaining even ordinates and twice the sum of the last remaining odd ordinates. The required area is obtained by multiplying this amount by a factor of three-quarters of the ordinates' average separation.

Given:

1. Length of ordinates, O =  2, 9, 18, 40,70

Common distance, d = 30 m

2. Length of ordinates, O =  23, 19, 14, 11, 12.5, 16, 19, 20, 20

Common distance, d = 1.5 m

Find:

1. Find the area, A

2. Find the area, A

Solution:

1. They must still be included in Simpson's rule even if the first or last ordinate is zero.

The following offsets are measured along a chain line from the chain line's right side to an irregular border.

Area = $\frac{d}{3} [First term + last term + 4 (Sum of odd terms) + 2 (Sum of even terms)]$

Area = $\frac{30}{3}$ [ (2+70) + 4(9 + 40) + 2(18)]

Area = 3040 $m^{2}$.

2.  They must still be included in Simpson's rule even if the first or last ordinate is zero.

The following offsets are measured along a chain line from the chain line's right side to an irregular border.

Area = $\frac{d}{3} [First term + last term + 4 (Sum of odd terms) + 2 (Sum of even terms)]$

Area = $\frac{1.5}{3}$ [ (23+20) + 4(19 + 11 + 16 + 20) + 2(14 + 12.5 + 19)]

Area = 199 $m^{2}$.

So, 1. Area = 3040 $m^{2}$ and 2. Area = 199 $m^{2}$.

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