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. 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.
10” 8”
8 “ 6”
10 in
8 in
A = 80 sq in
10 in
6 in
A= 60 sq in
6 in
8 in
A = 48 sq in

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

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 = 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. Formula for a rectangular prism SA = 2(LW+ LH + WH)
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
moisture
inside

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

16. 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 = cubic feet 27 26 25 24 23 22
3 feet 21 20 19
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. 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)
height
width
length

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 52’ 2’ deep 1’ thick 22’
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 ft ft3 = 288 ft3
52’
50’
2’ deep
1’ thick
2288 ft3
2000 ft3
22’
20’

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

35. 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 9’ 5’

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.