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Volume and Surface Area of Cylinders (4)

Volume and Surface Area of Cylinders (4)

Assessment

Presentation

Mathematics

7th Grade

Medium

CCSS
8.G.C.9, HSG.GMD.A.3

Standards-aligned

Created by

Tami Wilson

Used 88+ times

FREE Resource

7 Slides • 4 Questions

1

Volume and Surface Area of Cylinders

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2

Volume =  π\pi  x  radius2radius^2  x height

V =  πr2h\pi r^2h  

3

Multiple Choice

 V=πr2hV=\pi r^2h  

What is V when r = 3 and h = 10?

Remember  π=3.14\pi=3.14  

1

282.6

2

94.2

3

90

4

How did I do that?

  •  V=πr2hV=\pi r^2h  

  •  V=(3.14)(32)(10)V=\left(3.14\right)\left(3^2\right)\left(10\right)   

  • V = 282.6

5

Multiple Choice

 V=πr2hV=\pi r^2h  

What is V when r = 5 and h = 8?

Remember  π=3.14\pi=3.14  

1

1004.8

2

628

3

125.6

6

How did I do that?

  •  V=πr2hV=\pi r^2h  

  •  V=(3.14)(52)(8)V=\left(3.14\right)\left(5^2\right)\left(8\right)   

  • V = 628

7

Surface Area = 2 x  π\pi  x  radius2 radius^2\  + 2 x  π\pi  x radius x height 

SA =  2πr2 + 2πrh2\pi r^2\ +\ 2\pi rh  

8

Multiple Choice

 SA =2πr2 + 2πrhSA\ =2\pi r^2\ +\ 2\pi rh 
What is SA when r = 3 and h = 10.
Remember  π = 3.14\pi\ =\ 3.14   


1

244.92

2

56.52

3

188.4

4

207.24

9

How did I do that?

  •  SA = 2πr2 + 2πrhSA\ =\ 2\pi r^2\ +\ 2\pi rh  

  • SA = (2)(3.14)( 323^2 ) + (2)(3.14)(3)(10)

  • SA = 56.52 + 188.4

  • SA = 244.92

10

Multiple Choice

 SA =2πr2 + 2πrhSA\ =2\pi r^2\ +\ 2\pi rh 
What is SA when r = 5 and h = 8.
Remember  π = 3.14\pi\ =\ 3.14   


1

157

2

25.12

3

408.2

4

282.6

11

How did I do that?

  •  SA = 2πr2 + 2πrhSA\ =\ 2\pi r^2\ +\ 2\pi rh  

  • SA = (2)(3.14)( 525^2 ) + (2)(3.14)(5)(18)

  • SA = 157 + 251.2

  • SA = 408.2

Volume and Surface Area of Cylinders

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