12.2 – surface area of prisms and cylinders
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12.2 – Surface Area of Prisms And Cylinders. Prism:. Polyhedron with two parallel, congruent bases. Named after its base. Sum of the area of each face of the solid. Surface area:. Front. Back. Top. Left. Right. Bottom. Sum of the area of each face of the solid. Surface area:. - PowerPoint PPT PresentationTRANSCRIPT
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12.2 – Surface Area of Prisms And Cylinders
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Polyhedron with two parallel, congruent basesNamed after its base
Prism:
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Surface area: Sum of the area of each face of the solid
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Surface area: Sum of the area of each face of the solid
BackLeft
Top
Bottom
Front Right
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Lateral area: Area of each lateral face
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Right Prism: Each lateral edge is perpendicular to both bases
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Oblique Prism: Each lateral edge is NOT perpendicular to both bases
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Cylinder: Prism with circular bases
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Net: Two-dimensional representation of a solid
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Surface Area of a Right Prism:
SA = 2B + PH
B = area of one base
P = Perimeter of one base
H = Height of the prism
H
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Surface Area of a Right Cylinder:
22 2SA r rHπ π= +
H
SA = 2B + PH
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1. Name the solid that can be formed by the net.
Cylinder
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1. Name the solid that can be formed by the net.
Triangular prism
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1. Name the solid that can be formed by the net.
rectangular prism
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2. Find the surface area of the right solid.
SA = 2B + PH
SA = 2(30) + (22)(7)
B = bhB = (5)(6)
B = 30
P = 5 + 6 + 5 + 6P = 22
SA = 60 + 154
SA = 214 m2
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2. Find the surface area of the right solid.
SA = 2B + PH
SA = 2(30) + (30)(10)
P = 5 + 12 + 13P = 30
SA = 60 + 300
SA = 360 cm2
1
2B bh=
1(12)(5)
2B =
30B =
c2 = a2 + b2
c2 = (5)2 + (12)2
c2 = 25 + 144
c2 = 169
c = 13
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2. Find the surface area of the right solid.
22 2SA r rHπ π= +22 (2) 2 (2)(6)SA π π= +
2 (4) 2 (12)SA π π= +
8 24SA π π= +
32SA π= cm2
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2. Find the surface area of the right solid.
22 2SA r rHπ π= +22 (4) 2 (4)(12)SA π π= +
2 (16) 2 (48)SA π π= +
32 96SA π π= +
128SA π= ft2
12ft8ft
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6ft8ft
9ft
2. Find the surface area of the right solid.
SA = 2B + PH
SA = 2(24) + (24)(9)
P = 6 + 8 + 10P = 24
SA = 48 + 216
SA = 264 ft2
1
2B bh=
1(6)(8)
2B =
24B =
c2 = (6)2 + (8)2
c2 = 36 + 64
c2 = 100
c = 10
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A cylindrical bass drum has a radius of 5 inches and a depth of 12 inches. Find the surface area.
2. Find the surface area of the right solid.
5in
12in
22 2SA r rHπ π= +
SA =2π(5)2 + 2π(5)(12)
SA =2π(25) + 2π(60)
SA =50π +120π
SA =170π in2