Post on 30-Dec-2015
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12.2 – Surface Area of Prisms And Cylinders
Polyhedron with two parallel, congruent basesNamed after its base
Prism:
Surface area: Sum of the area of each face of the solid
Surface area: Sum of the area of each face of the solid
BackLeft
Top
Bottom
Front Right
Lateral area: Area of each lateral face
Right Prism: Each lateral edge is perpendicular to both bases
Oblique Prism: Each lateral edge is NOT perpendicular to both bases
Cylinder: Prism with circular bases
Net: Two-dimensional representation of a solid
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
Surface Area of a Right Cylinder:
22 2SA r rHπ π= +
H
SA = 2B + PH
1. Name the solid that can be formed by the net.
Cylinder
1. Name the solid that can be formed by the net.
Triangular prism
1. Name the solid that can be formed by the net.
rectangular prism
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
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
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
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
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
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