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• Right Rectangular Prisms– Surface Area
– Volume
• Right Rectangular Prisms– Surface Area
– Volume
Construction GeometryConstruction Geometry
Rectangular PrismsRectangular 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 PrismsRectangular 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.
Surface AreaSurface Area• Surface area can be thought of as
the amount of wrapping paper, with no overlap, needed to cover a box.
Surface AreaSurface 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”8 “
6”
6”10”
10”
10 in
10 in
8 in8 in A = 80 sq in
10 in
6 inA= 60 sq in
6 in
8 inA = 48 sq in
Surface AreaSurface Area 2(80) = 160 sq. in. 2(60) = 120 sq. in 2(48) = 96 sq. in 160+ 120+ 96 = 376 in2
10”
8”8”8 “
6”
6”10”
10”
10 in
10 in
8 in8 in A = 80 sq in
10 in
6 inA= 60 sq in
6in
8 inA = 48 sq in
CubeCube• 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
Surface AreaSurface 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.
Surface Area Surface Area
• Formula for a rectangular prism• SA = 2(LW+ LH + WH)
LengthW
idth
He
igh
t
Practice #1Practice #1• Determine the surface area of
the right rectangular prism using the formula.
• SA = 2(LW+ LH + WH)
2 m
m
5 mm10 mm
Practice #1Practice #1• SA = 2(LW + LH + WH)
2(10x2 + 10x5 + 2x5)
2(20 + 50 + 10)
2(80)
SA = 160 mm2
2 m
m
5 mm10 mm
ApplicationApplication• Building wrap is commonly
used in construction on exterior walls.
ApplicationApplication• Exterior
wrapping protects the structure from exterior water and air penetration. In
teri
or s
pace
ApplicationApplication• But it also
allows moisture from inside the building to escape.
Ext
erio
r sp
ace
insi
de
mo
istu
re
Practice #2Practice #2• Determine how much moisture
wrap is needed for this structure.
12’
10’
22’
Practice #2Practice #2• 2(10x12) = 240
• 2(10x22) = 440
• 1(12x22) = 264
• SA = 944 ft2
12 ft
10 ft
22 ft
10 ft
22 ft
12 ft
VolumeVolume• 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.
VolumeVolume• Volume is the number of cubic
units that a solid can hold.• 1 cubic yard =
27 cubic feet
21 20 19
24 23 22
27 26 25
3 feet
3 feet
3 feet
VolumeVolume• The volume of a rectangular prism
has the formula:• V = L*W*H• This formula is
found on the Math Reference Sheet.
VolumeVolume• 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
leng
th
Practice #3Practice #3• Determine the amount of concrete
needed to replace this damaged slab. V = (L x W x H)
12’ 12’1’ thick
Practice #3Practice #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’
• The footing is the most vital part of a foundation.
ApplicationApplication
• The foundation wall transfers weight to the footing.
ApplicationApplication
• The footing transfers the weight of the structure to the ground.
ApplicationApplication
ApplicationApplication• The foundation wall thickness is
determined by the anticipated load of the structure.
wall thickness
ApplicationApplication• The heavier the load of the
structure, the thicker the wall should be.
wall thickness
ApplicationApplication• The thickness of the footing is
then determined by the wall thickness.
2X
X
Xfooting
Foundationwall
ApplicationApplication• Steel reinforces
the concrete.• A footing
should be poured in one piece for best results.
Practice #4Practice #4• Determine the number of cubic feet
of concrete needed for this footing.
52’
22’
2’ deep1’ thick
Practice #4Practice #4• Solve by adding the volumes of 4 separate
sections OR outer section volume - inner section volume.
52’
22’
2’ deep1’ thick
Practice #4Practice #4• Volume (outer) = 52(22)(2) = 2288 ft3
• Volume (inner) = 50(20)(2) = 2000 ft3
52’
22’
2’ deep1’ thick
50’
20’
Practice #4Practice #4• Volume (outer) - Volume (inner) =• 2288 ft3 - 2000 ft3 = 288 ft3
52’
22’
2’ deep1’ thick
50’
20’2000 ft3 2288 ft3
Practice #5Practice #5• Determine the volume and
surface area for each of the cubes.
5’
9’
Practice #5Practice #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
5’
9’
• You are now ready for the practice problems for this lesson.
• After completion and review, take the assessment for this lesson.
Practice & Assessment Materials
Practice & Assessment Materials
