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- 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 www.themegallery.com Company Logo
- 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 www.themegallery.com Company Logo
- 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 www.themegallery.com Answer Area = ½ (4.0 + 3.5) x 2.5 = 9.375m2 Company Logo
- 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 www.themegallery.com Company Logo
- 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

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