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AREA & AREA & VOLUMEVOLUME

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MATHEMATICAL EQUATIONMATHEMATICAL EQUATION

ac

b

GIVEN GIVEN

a = 4.0 ma = 4.0 m

b = 3.5 mb = 3.5 m

c = 3.8 mc = 3.8 m

calculationcalculation

a

c0

b

GIVEN GIVEN

b = 3.5 mb = 3.5 m

h = 3.8 mh = 3.8 m

b

h

GIVEN GIVEN

a = 4.5 ma = 4.5 m

b = 5.0 mb = 5.0 m

C = 30C = 3000

calculationcalculation

calculationcalculation

Area = [S(S-a)(S-b)(S-c)] where; S = ½ (a+b+c)

Area = ½ (height x width) = ½ (b x h)

Area = ½ a b sin c0

Triangular equationTriangular equation

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MATHEMATICAL EQUATIONMATHEMATICAL EQUATION

ac

b

GIVEN GIVEN

a = 4.0 ma = 4.0 m

b = 3.5 mb = 3.5 m

c = 3.8 mc = 3.8 m

a

c0

b

GIVEN GIVEN

b = 3.5 mb = 3.5 m

h = 3.8 mh = 3.8 m

b

h

GIVEN GIVEN

a = 4.5 ma = 4.5 m

b = 5.0 mb = 5.0 m

C = 30C = 3000

S = ½ (4.0+3.5+3.8) = 5.65Area = [5.65(5.65-4.0)(5.65-3.5)(5.65–3.8)] = 6.089m2

Area = ½ (3.5 x 3.8) = 6.65m2

Area = ½ x 4.5 x 5.0 sin 300

= 5.625m2

AnswerAnswer

AnswerAnswer

AnswerAnswer

Triangular equationTriangular equation

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Rectangular equation

b

a

Area = a x b

b

a

h

Trapezium equation

i) Area = ½ (a + b) x h

GIVEN GIVEN

a = 4.0 ma = 4.0 m

b = 3.5 mb = 3.5 m

GIVEN GIVEN

a = 4.0 ma = 4.0 m

b = 3.5 mb = 3.5 m

h = 2.5 mh = 2.5 m

calculationcalculationcalculationcalculation

AnswerAnswerArea = 4.0 x 3.5 = 14m2

AnswerAnswerArea = ½ (4.0 + 3.5) x 2.5 = 9.375m2

MATHEMATICAL EQUATIONMATHEMATICAL EQUATION

AB = 10.5 mCD = 16.0 mPQ = 7.5 mRS = 13.0 m

A

C

B

D

30 m

P

S

Q

R

h = 5.2m

h = 4.0m

QUESTION 1Determine volume of FIGURE 1 with trapezoid and Simpsons method.

F G

H J

Area ABCD = ½ (10.5+16.0) x 5.2 = 68.9 mArea ABCD = ½ (10.5+16.0) x 5.2 = 68.9 m22

Trapezium equationTrapezium equationArea = ½ (a + b) x h Area = ½ (a + b) x h

The total area = d/2 x [(F + L) + 2 (other area)]The total area = d/2 x [(F + L) + 2 (other area)]

Trapezoid ruleTrapezoid rule

Simpson ruleSimpson rule

The total area = 1 / 3 d [F + L + 4 (Es) + 2 (Os)]The total area = 1 / 3 d [F + L + 4 (Es) + 2 (Os)]

Area PQRS = ½ (7.5+13.0) x 4.0 = 41 mArea PQRS = ½ (7.5+13.0) x 4.0 = 41 m22

Length FG = ½ (AB + PQ) = ½ (10.5 + 7.5) = 9.0 mLength FG = ½ (AB + PQ) = ½ (10.5 + 7.5) = 9.0 mLength HJ = ½ (CD + RS) = ½ (16.0 + 13.0) = 14.5 mLength HJ = ½ (CD + RS) = ½ (16.0 + 13.0) = 14.5 mHeight FGHJ = ½ (5.2 + 4.0) = 4.6 m Height FGHJ = ½ (5.2 + 4.0) = 4.6 m

Area FGHJ = ½ (9.0+14.5) x 4.6 = 54.05 mArea FGHJ = ½ (9.0+14.5) x 4.6 = 54.05 m22

Volume = d/2 x [(AVolume = d/2 x [(AABCDABCD + A + APQRSPQRS + 2 (A + 2 (AFGHJFGHJ)])]

= 15/2 x [(68.9 + 41 + 2 (54.05)]= 15/2 x [(68.9 + 41 + 2 (54.05)] = 1634.85 m= 1634.85 m33

Trapezoid ruleTrapezoid rule

Simpson ruleSimpson rule

Volume = d/3 x [(AVolume = d/3 x [(AABCDABCD + A + APQRSPQRS + 4 (A + 4 (AFGHJFGHJ)])]

= 15/3 x [(68.9 + 41 + 4 (54.05)]= 15/3 x [(68.9 + 41 + 4 (54.05)] = 1630.3 m= 1630.3 m33

QUESTION 2

Calculate the blank area at FIGURE 2 with trapezoid and simpson method.

RECTANGLE ABCDRECTANGLE ABCD

Part BPart B Part CPart C

Part DPart D

Part APart A

20m20m

25m25m

Part EPart E

Part APart A Part BPart B

OO11 OO77OO22 OO44 OO66OO33 OO55

Trapezoid ruleTrapezoid rule

Simpson ruleSimpson rule

AREA = d/3 x [(AREA = d/3 x [(OO11 + + OO77 + 4 ( + 4 (OO22 + + OO44 ++ OO66) + 2 () + 2 (OO33 + + OO55)])]

= 5/3 x [(= 5/3 x [(2.82.8 + + 00 + 4 ( + 4 (5.75.7 + + 5.05.0++ 3.53.5) + 2 () + 2 (5.25.2 + + 4.84.8)])]

= 132.667 m= 132.667 m22

OO11

OO22

OO33

OO44

Trapezoid ruleTrapezoid rule

Simpson ruleSimpson rule

AREA = d/3 x [(AREA = d/3 x [(OO11 + + OO77 + 4 ( + 4 (OO22 ) + 2 ( ) + 2 (OO33)])]

= 5/3 x [(= 5/3 x [(00 + + 00 + 4 ( + 4 (1.21.2) + 2 () + 2 (1.71.7)])] = 13.667 m= 13.667 m22

Part CPart C Part DPart D

Trapezoid ruleTrapezoid rule

Simpson ruleSimpson rule

AREA = D/3 x [(AREA = D/3 x [(OO11 + + OO33 + 4 ( + 4 (OO22)])]

= 5/3 x [(0 + 5.2 + 4 (3.2)]= 5/3 x [(0 + 5.2 + 4 (3.2)] = 30 m= 30 m22

OO11 OO22

OO44

OO33

Trapezoid ruleTrapezoid rule

Simpson ruleSimpson rule

AREA = d/3 x [(AREA = d/3 x [(OO11 + + OO77 + 4 ( + 4 (OO22 ) + 2 ( ) + 2 (OO33)])]

= 5/3 x [(= 5/3 x [(00 + + 00 + 4 ( + 4 (3.33.3) + 2 () + 2 (2.52.5)])] = 30.333 m= 30.333 m22

OO33OO11 OO22

Part EPart E

RECTANGLE ABCDRECTANGLE ABCD

20m20m

25m25m

OO11

OO22

OO44

OO33

OO55Trapezoid ruleTrapezoid rule

Simpson ruleSimpson rule

AREA = d/3 x [(AREA = d/3 x [(OO11 + + OO55 + 4 ( + 4 (OO22 + + OO44 ) + 2 () + 2 (OO33)])]

= 5/3 x [(= 5/3 x [(6.56.5 + + 00 + 4 ( + 4 (4.24.2 + + 0.50.5) + 2 () + 2 (2.82.8)])] = 51.5 m= 51.5 m22

AREA = a x bAREA = a x b = 20 X 25= 20 X 25 = 500 m= 500 m22

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TOTAL OF BLANK AREA TOTAL OF BLANK AREA = Part A + Part B = Part A + Part B - Part C - Part C + Part D + Part E + Rectangle ABCD+ Part D + Part E + Rectangle ABCD

Part APart A

Part BPart B

Part CPart C

Part DPart D

Part EPart ERectangle Rectangle ABCDABCD

Trapezoid ruleTrapezoid rule

TOTAL OF BLANK AREA TOTAL OF BLANK AREA = Part A = Part A + Part B + Part B - Part C - Part C + Part D + + Part D + Part E + Rectangle ABCD Part E + Rectangle ABCD = 696.25 m= 696.25 m22

Simpson ruleSimpson rule

TOTAL OF BLANK AREA TOTAL OF BLANK AREA = Part A = Part A + Part B + Part B - Part C - Part C + Part D + + Part D + Part E + Rectangle ABCDPart E + Rectangle ABCD = 698.167 m= 698.167 m22

0.5 0.4 0.91.00.80.5 0.7

5 6350

Get the total area of the FIGURE 3 below.(Trapezoid method)

Area = ½ a b sin cArea = ½ a b sin c00

= ½ x 5 x 6 sin 35= ½ x 5 x 6 sin 3500

= 8.604 m= 8.604 m22

AA BB

Determine length AB ???? Determine length AB ????

cc22 = a = a22 + b + b22 -2ab kos -2ab kos 00

cc22 = 5 = 522 + 6 + 622 – {2 x 5 x 6 x kos – {2 x 5 x 6 x kos 00 } }cc = 3.443 m = 3.443 m

Determine Determine width of intervalwidth of interval ???? ???? s = s = 3.4433.443 = 0.574 m = 0.574 m 6 6

55 66

6655

Trapezoid rule

OO11 OO77OO22

OO44

OO66

OO33

OO55

Total area of the FIGURE = rectilinear area + irregular area= rectilinear area + irregular area= 8.604 + 2.411 =11.015 m= 8.604 + 2.411 =11.015 m22

QUESTION 3

QUESTION 4Determine 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. R.L (m) Stn. R.L (m) Stn. R.L (m)S1 26.5 T1 23.7 V1 21.5S2 25.7 T2 23.9 V2 21.9S3 22.9 T3 21.5 V3 20.8

T4 20.8 V4 19.9Stn. R.L (m) V5 21.1U1 21.3 V6 20.7U2 21.6 Note:

R.L = Stn reduced level (in unit meter)Grid Size = 5m x 5m

U3 19.5U4 20.7U5 21.2

Stn. R.LG (m) R.LF (m) Depth (m) N Depth x N

S1 26.5

19.0

26.5 – 19.0 = 7.5 2 15.0S2 25.7 25.7 – 19.0 = 6.7 3 20.1S3 22.9 22.9 – 19.0 = 3.9 2 7.8T1 23.7 23.7 – 19.0 = 4.7 3 14.1T2 23.9 23.9 – 19.0 = 4.9 6 29.4T3 21.5 21.5 – 19.0 = 2.5 6 15.0T4 20.8 20.8 – 19.0 = 1.8 3 5.4U1 21.3 21.3 – 19.0 = 2.3 3 6.9U2 21.6 21.6 – 19.0 = 2.6 6 15.6U3 19.5 19.5 – 19.0 = 0.5 6 3.0U4 20.7 20.7 – 19.0 = 1.7 6 10.2U5 21.2 21.2 – 19.0 = 2.2 3 6.6V1 21.5 21.5 – 19.0 = 2.5 1 2.5V2 21.9 21.9 – 19.0 = 2.9 3 8.7V3 20.8 20.8 – 19.0 = 1.8 3 5.4V4 19.9 19.9 – 19.0 = 0.9 3 2.7V5 21.1 21.1 – 19.0 = 2.1 3 6.3V6 20.7 20.7 – 19.0 = 1.7 1 1.7

N = 63 DN = 176.4

Stn. N

S1 2S2 3S3 2T1 3T2 6T3 6T4 3U1 3U2 6U3 6U4 6U5 3V1 1V2 3V3 3V4 3V5 3V6 1

N = 63

Determine Determine number of triangle corner number of triangle corner ????????

R.LG = Reduce Level (Ground level)R.LF = Reduce Level (Floor level)

Average of DepthAverage of Depth = = DNDN / / NN

= 176.4 / 63= 176.4 / 63= = 2.800 m2.800 m

Calculated AreaCalculated Area = (5 x 5) x10.5= (5 x 5) x10.5= = 262.5 m262.5 m22

Total Excavation VolumeTotal Excavation Volume = Calculated Area x Average of Depth= Calculated Area x Average of Depth= 262.5 x 2.8= 262.5 x 2.8= = 735.0 m735.0 m33

10 m10 m

15 m15 m

25 m25 m

Area Area = ½ (a + b) x h= ½ (a + b) x h= ½ (10 + 25) x 15= ½ (10 + 25) x 15= 262.5 m= 262.5 m22

Figure below shown as a rectangular area, ABCD. Distance AB = DC = 48 m.Figure below shown as a rectangular area, ABCD. Distance AB = DC = 48 m.  Then Then distance AD = BC is 12 m. Calculate area for the blank area using Simpsondistance AD = BC is 12 m. Calculate area for the blank area using Simpson’’s method.s method.

12m12m

BB

DD CC

2.25m2.25m 3.8m3.8m 4.2m4.2m 4.5m4.5m 4.0m4.0m 3.95m3.95m 5.25m5.25m 4.85m4.85m

6 m6 m

48m48m

AA

EXAMPLE QUESTION 1

Offset O1 O2 O3 O4 O5Length (m) 16.7616.76 19.8119.81 20.4220.42 18.5918.59 16.7616.76

Offset O6 O7 O8 O9 O10Length (m) 17.6817.68 17.6817.68 17.3717.37 16.7616.76 17.6817.68

Calculate the area of the plot shown in fig 2.1 if the offsets, scaled from Calculate the area of the plot shown in fig 2.1 if the offsets, scaled from the plan at intervals of 10 m.the plan at intervals of 10 m.

( 5 Marks )( 5 Marks )

EXAMPLE QUESTION 2

The figure below the proposed of the cross section area of embankments.The figure below the proposed of the cross section area of embankments.Calculate: Calculate:

a)a)Width, WWidth, Wb)b)Surface areaSurface areac)c)The volume of land which need to fill along the 75m route. Use the The volume of land which need to fill along the 75m route. Use the

Prismoidal rule.Prismoidal rule.

9m

4.3 m4.3 m 1:2.3

W

1:2.3

EXAMPLE QUESTION 3