1. LOGO
Tutorial
A REA & VOLUME

2. MATHEMATICAL EQUATION
Triangular equation
GIVEN
calculation
a = 4.0 m
a
c
Area = √[S(S-a)(S-b)(S-c)]
where; S = ½ (a+b+c)
b = 3.5 m
c = 3.8 m
b
GIVEN
calculation
b = 3.5 m
h
Area = ½ (height x width)
= ½ (b x h)
h = 3.8 m
b
GIVEN
a
a = 4.5 m
calculation
Area = ½ a b sin c0
b = 5.0 m
c0
C = 300
b
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3. MATHEMATICAL EQUATION
Triangular equation
GIVEN
Answer
a = 4.0 m
a
c
S
= ½ (4.0+3.5+3.8) = 5.65
Area = √[5.65(5.65-4.0)(5.65-3.5)(5.65–3.8)]
= 6.089m2
b = 3.5 m
c = 3.8 m
b
GIVEN
Answer
Area = ½ (3.5 x 3.8)
= 6.65m2
b = 3.5 m
h
h = 3.8 m
b
GIVEN
a
a = 4.5 m
Answer
Area = ½ x 4.5 x 5.0 sin 300
= 5.625m2
b = 5.0 m
c0
C = 300
b
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4. MATHEMATICAL EQUATION
Rectangular equation
Trapezium equation
i)
Area = a x b
Area = ½ (a + b) x h
b
b
h
a
a
GIVEN
calculation
GIVEN
a = 4.0 m
a = 4.0 m
b = 3.5 m
calculation
b = 3.5 m
h = 2.5 m
Answer
Area = 4.0 x 3.5
= 14m2
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Answer
Area = ½ (4.0 + 3.5) x 2.5
= 9.375m2
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5. QUESTION 1
Determine volume of FIGURE 1 with trapezoid and Simpsons method.
AB = 10.5 m
CD = 16.0 m
PQ = 7.5 m
RS = 13.0 m
Area ABCD = ½ (10.5+16.0) x 5.2 = 68.9 m2
Area PQRS = ½ (7.5+13.0) x 4.0 = 41 m2
P
F
Length FG = ½ (AB + PQ) = ½ (10.5 + 7.5) = 9.0 m
Length HJ = ½ (CD + RS) = ½ (16.0 + 13.0) = 14.5 m
Height FGHJ = ½ (5.2 + 4.0) = 4.6 m
G
R
A
Q
h = 4.0m
S
B
Area FGHJ = ½ (9.0+14.5) x 4.6 = 54.05 m2
H
J
30 m
h = 5.2m
C
D
Trapezium equation
Area = ½ (a + b) x h
Trapezoid rule
Trapezoid rule
Volume = d/2 x [(AABCD + APQRS + 2 (AFGHJ)]
= 15/2 x [(68.9 + 41 + 2 (54.05)]
= 1634.85 m3
Simpson rule
Volume = d/3 x [(AABCD + APQRS + 4 (AFGHJ)]
= 15/3 x [(68.9 + 41 + 4 (54.05)]
The total area = d/2 x [(F + L) + 2 (other area)]
= 1630.3 m3
Simpson rule
The total area = 1 / 3 d [F + L + 4 (Es) + 2 (Os)]

6. QUESTION 2
Calculate the blank area at FIGURE 2 with trapezoid and simpson method.
Part A
Part B
Part D
RECTANGLE ABCD
20m
25m
Part C
Part E

7. O1
Part A
Part B
O2
O1
O2
O3
O4
O5
O6
O7
O3
O4
Trapezoid rule
Trapezoid rule
Simpson rule
AREA = d/3 x [(O1 + O7 + 4 (O2 + O4 + O6) + 2 (O3 + O5)]
= 5/3 x [(2.8 + 0 + 4 (5.7 + 5.0+ 3.5) + 2 (5.2 + 4.8)]
= 132.667 m2
Simpson rule
AREA = d/3 x [(O1 + O7 + 4 (O2 ) + 2 (O3)]
= 5/3 x [(0 + 0 + 4 (1.2) + 2 (1.7)]
= 13.667 m2

8. Part D
Part C
O1
O1
O2
Trapezoid rule
O2
O3
O4
O3
Trapezoid rule
Simpson rule
AREA = D/3 x [(O1 + O3 + 4 (O2)]
= 5/3 x [(0 + 5.2 + 4 (3.2)]
= 30 m2
Simpson rule
AREA = d/3 x [(O1 + O7 + 4 (O2 ) + 2 (O3)]
= 5/3 x [(0 + 0 + 4 (3.3) + 2 (2.5)]
= 30.333 m2

9. Part E
O1
RECTANGLE ABCD
O2
O3
O4
Trapezoid rule
20m
25m
O5
AREA = a x b
= 20 X 25
= 500 m2
Simpson rule
AREA = d/3 x [(O1 + O5 + 4 (O2 + O4 ) + 2 (O3)]
= 5/3 x [(6.5 + 0 + 4 (4.2 + 0.5) + 2 (2.8)]
= 51.5 m2

10. TOTAL OF BLANK AREA = Part A + Part B - Part C + Part D + Part E + Rectangle ABCD
Trapezoid rule
Part A
Part E
Rectangle
ABCD
TOTAL OF BLANK AREA = Part A
+ Part B - Part C + Part D +
Part E + Rectangle ABCD
= 696.25 m2
Part B
Simpson rule
TOTAL OF BLANK AREA = Part A
+ Part B - Part C + Part D +
Part E + Rectangle ABCD
= 698.167 m2
Part C
Part D
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11. QUESTION 3
Get the total area of the FIGURE 3 below.(Trapezoid method)
O5
O1
O2
0.5
0.4
O3
0.5
5
O4
0.8
5
O6
O7
1.0
0.9
350
0.7
6
c2 = a2 + b2 -2ab kos θ 0
c2 = 52 + 62 – {2 x 5 x 6 x kos 35 0 }
c = 3.443 m
6
Determine width of interval ????
s = 3.443 = 0.574 m
6
Trapezoid rule
5
6
Area = ½ a b sin c0
= ½ x 5 x 6 sin 350
= 8.604 m2
A
Determine length AB ????
Total area of the FIGURE
= rectilinear area + irregular area
= 8.604 + 2.411 =11.015 m2
B

12. QUESTION 4
Determine the volume of excavate to the uniform depth of 19.0 m above datum. Calculate the mean level of
the ground and the volume of earth to be excavated.
Stn.
S1
S2
S3
R.L (m)
26.5
25.7
22.9
Stn.
U1
U2
U3
U4
U5
R.L (m)
21.3
21.6
19.5
20.7
21.2
Stn.
T1
T2
T3
T4
R.L (m)
23.7
23.9
21.5
20.8
Stn.
V1
V2
V3
V4
V5
V6
R.L (m)
21.5
21.9
20.8
19.9
21.1
20.7
Note:
R.L = Stn reduced level (in unit meter)
Grid Size = 5m x 5m

13. Stn.
R.LG (m)
S1
S2
S3
T1
T2
T3
T4
U1
U2
U3
U4
U5
V1
V2
V3
V4
V5
V
26.5
25.7
22.9
23.7
23.9
21.5
20.8
21.3
21.6
19.5
20.7
21.2
21.5
21.9
20.8
19.9
21.1
20.7
R.LF (m)
19.0
Depth (m)
26.5 – 19.0 = 7.5
25.7 – 19.0 = 6.7
22.9 – 19.0 = 3.9
23.7 – 19.0 = 4.7
23.9 – 19.0 = 4.9
21.5 – 19.0 = 2.5
20.8 – 19.0 = 1.8
21.3 – 19.0 = 2.3
21.6 – 19.0 = 2.6
19.5 – 19.0 = 0.5
20.7 – 19.0 = 1.7
21.2 – 19.0 = 2.2
21.5 – 19.0 = 2.5
21.9 – 19.0 = 2.9
20.8 – 19.0 = 1.8
19.9 – 19.0 = 0.9
21.1 – 19.0 = 2.1
20.7 – 19.0 = 1.7
ƒN
2
3
2
3
6
6
3
3
6
6
6
3
1
3
3
3
3
1
Depth x N
15.0
20.1
7.8
14.1
29.4
15.0
5.4
6.9
15.6
3.0
10.2
6.6
2.5
8.7
5.4
2.7
6.3
1.7

14. Determine number of triangle corner ????
Stn.
S1
S2
S3
T1
T2
T3
T4
U1
U2
U3
U4
U5
V1
V2
V3
V4
V5
ƒN
2
3
2
3
6
6
3
3
6
6
6
3
1
3
3
3
3

15. 10 m
R.LG = Reduce Level (Ground level)
R.LF = Reduce Level (Floor level)
15 m
Area
= ½ (a + b) x h
= ½ (10 + 25) x 15
= 262.5 m2
Average of Depth
= ∑ DN / ∑ N
= 176.4 / 63
= 2.800 m
Calculated Area
= (5 x 5) x10.5
= 262.5 m2
Total Excavation Volume
= Calculated Area x Average of Depth
= 262.5 x 2.8
= 735.0 m3
25 m

16. EXAMPLE QUESTION 1
Figure below shown as a rectangular area, ABCD. Distance AB = DC = 48 m. Then
distance AD = BC is 12 m. Calculate area for the blank area using Simpson’s method.
48m
A
B
12m
2.25m
D
3.8m
4.2m
4.5m
4.0m
6m
3.95m
5.25m
4.85m
C

17. EXAMPLE QUESTION 2
Calculate the area of the plot shown in fig 2.1 if the offsets, scaled from the
plan at intervals of 10 m.
( 5 Marks )
O2
O1
Offset
Length (m) 16.76 19.81
O6
O7
Offset
Length (m) 17.68 17.68
O3
20.42
O8
17.37
O4
18.59
O9
16.76
O5
16.76
O10
17.68

18. EXAMPLE QUESTION 3
The figure below the proposed of the cross section area of embankments.
Calculate:
a)Width, W
b)Surface area
c)The volume of land which need to fill along the 75m route. Use the
Prismoidal rule.
9m
1:2.3
4.3 m
W
1:2.3