Right Rectangular PrismsRight Rectangular Prisms –Surface Area –Volume Right Rectangular...

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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

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