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