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Right Rectangular PrismsRight Rectangular Prisms –Surface Area –Volume Right Rectangular PrismsRight Rectangular Prisms –Surface Area –Volume Construction Geometry

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Rectangular Prisms Right rectangular prisms are 3 dimensional rectangles.Right rectangular prisms are 3 dimensional rectangles. We often think of them as closed boxes or, in construction, examples would be rectangular concrete slabs.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 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.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.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 10 8 8 8 6 6 1010 10 in 8 in A = 80 sq inA = 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 10 8 8 8 6 6 1010 10 in 8 in A = 80 sq inA = 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.A cube is a right rectangular prism. All its sides are congruent squares. All 6 faces have the same area.All 6 faces have the same area. So the surface area of a cube = 6 x (area of one face).So the surface area of a cube = 6 x (area of one face). Face = (4 x 4) = 16 ft 2 Surface area = 6(16) = 96 ft 2 4 ft

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

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

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

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Practice #1

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Application Building wrap is commonly used in construction on exterior walls.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.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.But it also allows moisture from inside the building to escape. Exterior space insidemoisture

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

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Practice #2 2(10x12) = 2402(10x12) = 240 2(10x22) = 4402(10x22) = 440 1(12x22) = 2641(12x22) = 264 SA = 944 ft 2SA = 944 ft 2 12 ft 10 ft 22 ft 10 ft 22 ft 12 ft

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Volume Volume is the measure of the amount of space occupied by an object.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.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.Volume is the number of cubic units that a solid can hold. 1 cubic yard = 27 cubic feet1 cubic yard = 27 cubic feet 21 20 19 24 23 22 27 26 25 3 feet

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Volume The volume of a rectangular prism has the formula:The volume of a rectangular prism has the formula: V = L*W*HV = L*W*H This formula is found on the Math Reference Sheet.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.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 un 3.Units for volume are cubic (cu) units or un 3. V = (L x W x H)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)Determine the amount of concrete needed to replace this damaged slab. V = (L x W x H)

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Practice #3 V = (L x W x H) = (12 x 12 x 1)V = (L x W x H) = (12 x 12 x 1) V = 144 ft 3 V = 144 ft 3 For cubic yards: 1 yd 3 = 27 ft 3For cubic yards: 1 yd 3 = 27 ft 3 144 = 5 yd 3 144 = 5 yd 3 27 27 12 1

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

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

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

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

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

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

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Application Steel reinforces the concrete.Steel reinforces the concrete. A footing should be poured in one piece for best results.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.Determine the number of cubic feet of concrete needed for this footing. 52 22 2 deep 1 thick

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

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Practice #4 52 22 2 deep 1 thick

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Practice #4 52 22 2 deep 1 thick

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

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Practice #5 Volume =Volume = 5 x 5 x 5 = 125 ft 35 x 5 x 5 = 125 ft 3 Surface area = (5x5) x 6 = 150 ft 2Surface area = (5x5) x 6 = 150 ft 2 Volume = 9 x 9 x 9 = 729 ft 3 Surface area = (9x9) x 6 = 486 ft 2 5 9

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

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