12.2 – Surface Area of Prisms AndCylinders
Polyhedron with two parallel, congruent bases
Named after its base
Prism:
Surface area:
Sum of the area of each face ofsolid
Lateral area:
Area of each lateral face
Right Prism:
Each lateral edge isperpendicular to both bases
Oblique Prism:
Each lateral edge is NOTperpendicular 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:
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 = bh
B = (5)(6)
B = 30
P = 5 + 6 + 5 + 6
P = 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 + 13
P = 30
SA = 60 + 40
SA = 100
cm2
c2 = a2 + b2
c2 = (5)2 + (12)2
c2 = 25 + 144
c2 = 169
c = 13
2. Find the surface area of the right solid.
cm2
2. Find the surface area of the right solid.
in2
144in
3. Solve for x, given the surface area.
SA = 2B + PH
142 = 2(5x) + (2x + 10)(7)
B = bh
B = 5x
P = 5 + x + 5 + x
P = 2x + 10
142 = 10x + 14x + 70
142 = 24x + 70
72 = 24x
3ft = x
3. Solve for x, given the surface area.
12.2
806-808
3, 5, 6, 7-15 odd
HW Problems
#5
SA = 2B + PH
SA = 2(4156.8) + (240)(80)
P = 406
P = 240
SA = 8313.6 + 19200
SA = 27513.6 ft2
