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Construction Geometry

Construction Geometry. Right Rectangular Prisms Surface Area Volume. Rectangular Prisms. Right rectangular prisms are 3 dimensional rectangles. We often think of them as closed boxes or, in construction, examples would be rectangular concrete slabs. Rectangular Prisms.

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Construction Geometry

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  1. Construction Geometry Right Rectangular Prisms Surface Area Volume

  2. Rectangular Prisms • Right rectangular prisms are 3 dimensional rectangles. • We often think of them as closed boxes or, in construction, examples would be rectangular concrete slabs.

  3. Rectangular Prisms • A right prism has bases which meet the lateral faces at right angles. • A right rectangular prism has bases which are rectangles and form right angles with the other faces.

  4. Surface Area • Surface area can be thought of as the amount of wrapping paper, with no overlap, needed to cover a box.

  5. 10” 6” 10” 8” 8” 8 “ 6” 10” 10 in 8 in 8 in A = 80 sq in 10 in 10 in 6 in A= 60 sq in A = 48 sq in 8 in 6 in Surface Area • Split into 3 separate rectangles. • Front/back sides • Top/bottom sides • Right/left sides • Find the areas of each (LxW) and double. • Sum the areas.

  6. 10” 6” 10” 8” 8” 8 “ 6” 10” 10 in 8 in 8 in A = 80 sq in 10 in 10 in 6 in A= 60 sq in A = 48 sq in 8 in 6in Surface Area • 2(80) = 160 sq. in. • 2(60) = 120 sq. in • 2(48) = 96 sq. in • 160+ 120+ 96= 376 in2

  7. Cube • A cube is a right rectangular prism. All its sides are congruent squares. • All 6 faces have the same area. • So the surface area of a cube = 6 x (area of one face). Face = (4 x 4) = 16 ft2 Surface area = 6(16) = 96 ft2 4 ft

  8. Surface Area • The surface area of a rectangular prism can be found using a formula. • SA= 2(LW + LH + WH) • This formula is found on the Math Reference Sheet.

  9. Surface Area • Formula for a rectangular prism • SA = 2(LW+ LH + WH) Width Height Length

  10. Practice #1 • Determine the surface area of the right rectangular prism using the formula. • SA = 2(LW+ LH + WH) 2 mm 10 mm 5 mm

  11. Practice #1 • SA = 2(LW + LH + WH) 2(10x2 + 10x5 + 2x5) 2(20 + 50 + 10) 2(80) SA = 160 mm2 2 mm 10 mm 5 mm

  12. Application • Building wrap is commonly used in construction on exterior walls.

  13. Application • Exterior wrapping protects the structure from exterior water and air penetration. Interior space

  14. Application • But it also allows moisture from inside the building to escape. Exterior space inside moisture

  15. Practice #2 • Determine how much moisture wrap is needed for this structure. 10’ 12’ 22’

  16. 10 ft 12 ft 10 ft 22 ft 12 ft 22 ft Practice #2 • 2(10x12) = 240 • 2(10x22) = 440 • 1(12x22) = 264 • SA = 944 ft2

  17. Volume • Volume is the measure of the amount of space occupied by an object. • Volume can also be thought of as the amount that an object can hold.

  18. Volume • Volume is the number of cubic units that a solid can hold. • 1 cubic yard = 27 cubic feet 26 25 27 23 22 24 3 feet 20 19 21 3 feet 3 feet

  19. Volume • The volume of a rectangular prism has the formula: • V = L*W*H • This formula is found on the Math Reference Sheet.

  20. length height width Volume • Volume is determined by the product of the 3 dimensions of a rectangular prism: height, length, width. • Units for volume are “cubic” (cu) units or un3. • V = (L x W x H)

  21. Practice #3 • Determine the amount of concrete needed to replace this damaged slab. V = (L x W x H) 12’ 12’ 1’ thick

  22. Practice #3 • V = (L x W x H) = (12 x 12 x 1) V = 144 ft3 • For cubic yards: 1 yd3 = 27 ft3 144 = 5⅓ yd3 27 12’ 12’ 1’

  23. Application • The footing is the most vital part of a foundation.

  24. Application • The foundation wall transfers weight to the footing.

  25. Application • The footing transfers the weight of the structure to the ground.

  26. Application • The foundation wall thickness is determined by the anticipated load of the structure. wall thickness

  27. Application • The heavier the load of the structure, the thicker the wall should be. wall thickness

  28. Application • The thickness of the footing is then determined by the wall thickness. X Foundation wall X footing 2X

  29. Application • Steel reinforces the concrete. • A footing should be poured in one piece for best results.

  30. Practice #4 • Determine the number of cubic feet of concrete needed for this footing. 52’ 2’deep 1’ thick 22’

  31. Practice #4 • Solve by adding the volumes of 4 separate sections OR outer section volume - inner section volume. 52’ 2’deep 1’ thick 22’

  32. Practice #4 • Volume (outer) = 52(22)(2) = 2288 ft3 • Volume (inner) = 50(20)(2) = 2000 ft3 52’ 50’ 2’deep 1’ thick 22’ 20’

  33. Practice #4 • Volume (outer) - Volume (inner) = • 2288 ft3 - 2000 ft3 = 288 ft3 52’ 50’ 2’deep 1’ thick 2288 ft3 2000 ft3 22’ 20’

  34. 5’ 9’ Practice #5 • Determine the volume and surface area for each of the cubes.

  35. 5’ 9’ Practice #5 • Volume = • 5’ x 5’ x 5’ = 125 ft3 • Surface area = (5x5) x 6 = 150 ft2 • Volume = • 9’ x 9’ x 9’ = 729 ft3 • Surface area = (9x9) x 6 = 486 ft2

  36. Practice & Assessment Materials • You are now ready for the practice problems for this lesson. • After completion and review, take the assessment for this lesson.

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