Question

Consider an area having ordinates 1.5 m, 2.4 m, 3.3 m, 4.4 m. What would be the traverse area by using Simpson's rule given that d = 15 m?
1. 580.5 sq m
2. 129.75 sq m
3. 193.5 sq m
4. 200 sq m

Answer 1

Correct Answer - Option 2 : 129.75 sq m

Concept:

Simpson's rule:

• This rule is based on the assumption that the figures are trapezoids.
• In order to apply Simpson's rule, the area must be divided in an even number i.e., the number of ordinates must be odd i.e., n term in the last set 'O_n' should be odd. 

The area is given by Simpson's rule:

\(Area = \frac{d}{3}\left[ {({O_1} + {O_n}) + 4({O_2} + {O_4} + .. + {O_{n - 1}}) + 2({O_3} + {O_5} + ...{O_{n - 2}})} \right]\)

where O₁, O₂, O₃, .........O_n is the ordinates.

Note:- In the case of an even number of cross-sections, the end strip is treated separately and the area of the remaining strip is calculated by Simpson's rule. The area of the last strip can be calculated by either trapezoidal or Simpson's rule.

Calculation:

Given data

O₁ = 1.5 m, O₂ = 2.4 m,O₃ = 3.3 m,O₄ = 4.4 m, and d = 15 m

Now the given ordinates are even so we find the area in two-component

A₁ = by Simpson's rule form O₁ to O₃

A₂ = Area of last strip by trapezoidal rule

The traverse area by using Simpson's rule is given from O₁ to O₃

\(A_1 = \frac{d}{3}\left[ {({O_1} + {O_3}) + 4({O_2} ) } \right]\)

\(A_1 = \frac{15}{3}\left[ {({1.5} + {3.3}) + 4({2.4} )} \right]\) = 72 m²

The traverse area by using a trapezoidal rule for the last strip is given from O3 to O₄

\(A_2 = {d \over 2}(O_3+O_4)\)

\(A_2 = {15 \over 2}(3.3+4.4)\) = 57.75 m²

Hence, the traverse area by using Simpson's rule given is

A = A₁ + A₂ = 72 + 57.75 = 129.75 m²