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Ch 12.2 A = (π)(4 2 ) = 8π 2 A = (π)(12 2 ) = 144π Find the area of the circle. Find the area of the sector. Find the area of the sector. A = (π)(22 2 )135 = 1089π 360

Ch 12.2 A = (π)(4 2 ) = 8π 2 A = (π)(12 2 ) = 144π Find the area of the circle. Find the area of the sector. A = (π)(22 2 )135 = 1089π 360

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Page 1: Ch 12.2 A = (π)(4 2 ) = 8π 2 A = (π)(12 2 ) = 144π Find the area of the circle. Find the area of the sector. A = (π)(22 2 )135 = 1089π 360

Ch 12.2

A = (π)(42) = 8π 2

A = (π)(122) = 144π

Find the area of the circle.

Find the area of the sector.

Find the area of the sector.A = (π)(222)135 = 1089π 360

Page 2: Ch 12.2 A = (π)(4 2 ) = 8π 2 A = (π)(12 2 ) = 144π Find the area of the circle. Find the area of the sector. A = (π)(22 2 )135 = 1089π 360

Ch 12.2Surface Areas of

Prisms & Cylinders

Standard 8.0Students know, derive, and solve problems

involving the area of common geometric figures.

Learning Target:I will be able to solve problems involving areas of prisms and cylinders.

Ch 10.5Ch 12.2

Page 3: Ch 12.2 A = (π)(4 2 ) = 8π 2 A = (π)(12 2 ) = 144π Find the area of the circle. Find the area of the sector. A = (π)(22 2 )135 = 1089π 360

Theorem 12-1

Ch 12.2

Lateral Area:The sum of the areas of the lateral faces of a solid.

Lateral Faces:The faces that are not bases.In a prism, the lateral faces are parallelograms.

Page 4: Ch 12.2 A = (π)(4 2 ) = 8π 2 A = (π)(12 2 ) = 144π Find the area of the circle. Find the area of the sector. A = (π)(22 2 )135 = 1089π 360

Lateral Area of a Prism

Find the lateral area of the regular hexagonal prism.

The bases are regular hexagons. So the perimeter of one base is 6(5) or 30 centimeters.

Answer: The lateral area is 360 square centimeters.

Lateral area of a prism

P = 30, h = 12

Multiply.

Ch 12.2

Page 5: Ch 12.2 A = (π)(4 2 ) = 8π 2 A = (π)(12 2 ) = 144π Find the area of the circle. Find the area of the sector. A = (π)(22 2 )135 = 1089π 360

A. 162 cm2

B. 216 cm2

C. 324 cm2

D. 432 cm2

Find the lateral area of the regular octagonal prism.

Ch 12.2

Lateral area of a prism

P = 24, h = 9

Multiply.

= (3 * 8) (9)

= 216

Page 6: Ch 12.2 A = (π)(4 2 ) = 8π 2 A = (π)(12 2 ) = 144π Find the area of the circle. Find the area of the sector. A = (π)(22 2 )135 = 1089π 360

Ch 12.2

Theorem 12-2

Surface Area:The sum of the areas of all surfaces of a solid figure.

Solid Figure:A figure that encloses a part of space.In a prism, the lateral faces are parallelograms.

Page 7: Ch 12.2 A = (π)(4 2 ) = 8π 2 A = (π)(12 2 ) = 144π Find the area of the circle. Find the area of the sector. A = (π)(22 2 )135 = 1089π 360

Surface Area of a Prism

Find the surface area of the rectangular prism.

Ch 12.2

Surface area of a prism

P = 24, h = 12, B = 36

Simplify.

= [(6 * 4) (12)] + 2 (6 * 6)

= 360

S = L + 2B

Page 8: Ch 12.2 A = (π)(4 2 ) = 8π 2 A = (π)(12 2 ) = 144π Find the area of the circle. Find the area of the sector. A = (π)(22 2 )135 = 1089π 360

A. 320 units2

B. 512 units2

C. 368 units2

D. 416 units2

Find the surface area of the triangular prism.

Ch 12.2

Surface area of a prism

P = 32, h=10, B = 48

Simplify.

= [(10+10+12)(10)] + 2 (½)(12*8)

= 416

S = L + 2B

Page 9: Ch 12.2 A = (π)(4 2 ) = 8π 2 A = (π)(12 2 ) = 144π Find the area of the circle. Find the area of the sector. A = (π)(22 2 )135 = 1089π 360

Ch 12.2

Theorem 12-3 & 12-4

Page 10: Ch 12.2 A = (π)(4 2 ) = 8π 2 A = (π)(12 2 ) = 144π Find the area of the circle. Find the area of the sector. A = (π)(22 2 )135 = 1089π 360

Lateral Area and Surface Area of a Cylinder

Find the lateral area and the surface area of the cylinder in terms of π

L = Ph Lateral area of a cylinder

= 2rh P = 2πr (circumference of a circle)

= 2(14)(18) r = 14 , h = 18.

≈ 504π Simplify.

Ch 12.2

S = L + 2B Surface area of a cylinder

= 504π + 2r2 L = 504π , B = π r2

≈ 504π + 2(14)2 r = 14

≈ 896π Simplify.

Page 11: Ch 12.2 A = (π)(4 2 ) = 8π 2 A = (π)(12 2 ) = 144π Find the area of the circle. Find the area of the sector. A = (π)(22 2 )135 = 1089π 360

A. lateral area ≈ 480π ft2 andsurface area ≈ 768π ft2

B. lateral area ≈ 480π ft2 andsurface area ≈ 384π ft2

C. lateral area ≈ 240π ft2 andsurface area ≈ 768π ft2

D. lateral area ≈ 240π ft2 andsurface area ≈ 384π ft2

Find the lateral area and the surface area of the cylinder in terms of π.

Ch 12.2

Page 12: Ch 12.2 A = (π)(4 2 ) = 8π 2 A = (π)(12 2 ) = 144π Find the area of the circle. Find the area of the sector. A = (π)(22 2 )135 = 1089π 360

Find Missing Dimensions

MANUFACTURING A soup can is covered with the label shown. What is the radius of the soup can?

L = Ph Lateral area of a cylinder

= 2rh P = 2π r (circumference of a circle)

125.6 = 2r(8) L = 15.7 × 8 , h = 8.

125.6 = 16r Simplify.

7.85/π = r Divide each side by 16.

Ch 12.2

Page 13: Ch 12.2 A = (π)(4 2 ) = 8π 2 A = (π)(12 2 ) = 144π Find the area of the circle. Find the area of the sector. A = (π)(22 2 )135 = 1089π 360

A. 12 inches

B. 16 inches

C. 18 inches

D. 24 inches

Find the diameter of a base of a cylinder if the surface area is 480 square inches and the height is 8 inches.

Ch 12.2