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**Construction Geometry**

Right Rectangular Prisms Surface Area Volume

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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.

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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.

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

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**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

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**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

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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

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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.

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**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

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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

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**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

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

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

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

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

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**Practice #2 2(10x12) = 240 2(10x22) = 440 1(12x22) = 264 SA = 944 ft2**

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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.

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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

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**Volume The volume of a rectangular prism has the formula:**

V = L*W*H This formula is found on the Math Reference Sheet.

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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

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

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**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’

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

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Application The foundation wall transfers weight to the footing.

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

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

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

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

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**Application Steel reinforces the concrete.**

A footing should be poured in one piece for best results.

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

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**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’

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**Practice #4 Volume (outer) = 52(22)(2) = 2288 ft3**

Volume (inner) = 50(20)(2) = 2000 ft3 52’ 50’ 2’ deep 1’ thick 22’ 20’

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**Practice #4 Volume (outer) - Volume (inner) =**

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

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

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**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’

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**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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Volume Return to table of contents. Volume - The amount of space occupied by a 3-D Figure - The number of cubic units needed to FILL a 3-D Figure (layering)

Volume Return to table of contents. Volume - The amount of space occupied by a 3-D Figure - The number of cubic units needed to FILL a 3-D Figure (layering)

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