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The offset (in m) taken to a chain line to curve boundary are given below
Ch. 0 5 10 15 20 25 35 45 55 65
Offsets(m) 2.5 3.8 8.4 7.5 10.5 9.3 5.8 7.8 6.9 8.4

Find the area between the chain lines the 1st and last ordinate and boundary by

1. Trapezoidal rule

2. Simpson’s rule

3. Co-ordinate rule (average ordinate rule)

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 Area by the Trapezoidal rule:

Ch.in m Interval (d) ordinate Offset in m
0 Y1 2.5
5 Y2 3.8
10 d1 = 5 Y3 8.4
15 Y4 7.5
20 Y5 10.5
25 Y6 9.3
25 Y7 9.3
35 Y8 5.8
45 d2 = 10 Y9 7.8
55 Y10 6.9
65 Y11 8.4

Area of the interval ${d_1}$ of 5m ${A_1} = {d_1}[{\frac{y_1 + y_2}{2} + y_2 + y_3 + y_4 + y_5}]$

$= 5[{\frac{2.5 + 3.5}{2} + 3.8 + 8.4 + 7.5 + 10.5}]$

= 180.5 $m^2$

Area of the interval ${d_2}$ of 10m

${A_2} = {d_2}[{\frac{y_7 + y_11}{2} + y_8 + y_9 + y_10}]$

$= 10[{\frac{9.3 + 8.4}{2} + 5.8 + 7.8 + 6.9}]$

= 293.5 $m^2$

Total area A = $A_1 + A_2 = 180.5+293.5 = 475 m^2$

Area by Simpson's rule:

Ch.in m Interval (d) ordinate Offset in m
0 Y1 2.5
5 Y2 3.8
10 d1 = 5 Y3 8.4
15 Y4 7.5
20 Y5 10.5
25 Y6 9.3
25 Y1 9.3
35 Y2 5.8
45 d2 = 10 Y3 7.8
55 Y4 6.9
65 Y5 8.4

$A_1 = {\frac{d_1}{3}}[{(y_1 + y_5) + 4(y_2 + y_4) + 2(y_3)}] + {\frac{d_1}{2}(y_5 + y_6)}$

$= {\frac{5}{3}}[{(2.5 + 10.5) + 4(3.8 + 7.5) + 2(8.4)}] + {\frac{5}{2}(10.5 + 9.3)}$

= 174.5 $m^2$

$A_2 = {\frac{d_2}{3}}[{(y_1 + y_5) + 4(y_2 + y_4) + 2(y_3)}]$

$= {\frac{10}{3}}[{(9.3 + 8.4) + 4(5.8 + 6.9) + 2(7.8)}]$

= 280.33 $m^2$

Total area A = $A_1 + A_2 = 174.5 + 280.33 = 454.83 m^2$

Area by coordinate rule

For $1^{st}$ section

1. Number of interval $(N_1 )=5$

2. Length of interval $(d_1 )=5m$

3. length = $N_1*d_1=5*5=25m$

$A_1 = [{\frac{2.5+3.8+8.4+7.5+10.5+9.3}{N_1 + 1}}] = 25[{\frac{42}{5+1}}] = 175 {m^2}$

For $2^{nd}$ section

1. Number of interval $(N_2 )=4$

2. Length of interval $(d_2 )=10m$

3. length = $N_2*d_2=4*10=40m$

$A_2 = [{\frac{9.3+5.8+7.8+6.9+8.4}{N_2 + 1}}] = 40[{\frac{38.2}{4+1}}] = 305.6 {m^2}$

Total area A = $A_1 + A_2 = 175 + 305.6 = 480.6 m^2$