Geometry Section 1-7 1112

Section 1-7Three-Dimensional Figures

Friday, September 26, 14

Essential Questions

✦ How do you identify and name three-dimensional figures?

✦ How do you find the surface area and volume?


Vocabulary1. Polyhedron:

2. Face:

3. Edge:

4. Vertex:

5. Prism:

6. Base:


Vocabulary1. Polyhedron: A solid three-dimensional closed

figure with all flat surfaces

2. Face:

3. Edge:

4. Vertex:

5. Prism:

6. Base:




2. Face: A flat surface of a polyhedron

3. Edge:

4. Vertex:

5. Prism:

6. Base:





3. Edge: The line segment where two faces meet

4. Vertex:

5. Prism:

6. Base:






4. Vertex: The point where three or more edges meet

5. Prism:

6. Base:







5. Prism: Has two parallel, congruent polygonal bases with faces that are all parallelograms

6. Base:







5. Prism: Has two parallel, congruent polygonal bases with faces that are all parallelograms

6. Base: The parallel sides of the surfaces of a prism; used to name the prism (triangular prism)


Vocabulary7. Pyramid:

8. Cylinder:

9. Cone:


Vocabulary7. Pyramid: Has one polygonal base with three or

more faces that meet at a common vertex; Also named after its base (square pyramid)

8. Cylinder:

9. Cone:




8. Cylinder: A three-dimensional shape with congruent parallel circular bases which have one surface connecting them

9. Cone:




8. Cylinder: A three-dimensional shape with congruent parallel circular bases which have one surface connecting them

9. Cone: Has one circular base which has a single solid curved surface that connects the base to a single vertex


Vocabulary10. Sphere:

11. Regular Polyhedron:

12. Platonic Solid:


Vocabulary10. Sphere: A solid three-dimensional figure that is

the set of all points in space that are all the same distance (radius) from its center; has no faces, bases, or vertices

11. Regular Polyhedron:

12. Platonic Solid:




11. Regular Polyhedron: All faces are regular congruent polygons and all edges are congruent

12. Platonic Solid:




11. Regular Polyhedron: All faces are regular congruent polygons and all edges are congruent

12. Platonic Solid: A special name for the five regular polyhedra, named after Plato


Vocabulary13. Surface Area:

14. Volume:


Vocabulary13. Surface Area: The total area of the surface of a

three-dimensional figure (square units)

14. Volume:




14. Volume: The amount of space that a three-dimensional object occupies (cubic units)




14. Volume: The amount of space that a three-dimensional object occupies (cubic units)

Where are some places you can find the formulas for surface area and volume?


Example 1Find the surface area and volume of a cone

whose base diameter is 10 in., height is 12 in., and slant height is 13 in.




SA = π rl + π r2




SA = π rl + π r2

r = radius, l = slant height




SA = π rl + π r2


SA = π(5)(13)+ π(5)2




SA = π rl + π r2


SA = π(5)(13)+ π(5)2

= 65π + 25π




SA = π rl + π r2


SA = π(5)(13)+ π(5)2

= 65π + 25π = 90π




SA = π rl + π r2


SA = π(5)(13)+ π(5)2

= 65π + 25π = 90π

≈ 282.74




SA = π rl + π r2


SA = π(5)(13)+ π(5)2

= 65π + 25π = 90π

≈ 282.74 in2




SA = π rl + π r2


SA = π(5)(13)+ π(5)2

= 65π + 25π = 90π

≈ 282.74 in2

V =

13π r2h




SA = π rl + π r2


SA = π(5)(13)+ π(5)2

= 65π + 25π = 90π

≈ 282.74 in2

V =

13π r2h

r = radius, h = height




SA = π rl + π r2


SA = π(5)(13)+ π(5)2

= 65π + 25π = 90π

≈ 282.74 in2

V =

13π r2h


V =

13π(5)2(12)




SA = π rl + π r2


SA = π(5)(13)+ π(5)2

= 65π + 25π = 90π

≈ 282.74 in2

V =

13π r2h


= 100π V =

13π(5)2(12)




SA = π rl + π r2


SA = π(5)(13)+ π(5)2

= 65π + 25π = 90π

≈ 282.74 in2

V =

13π r2h


= 100π

≈ 314.16

V =

13π(5)2(12)




SA = π rl + π r2


SA = π(5)(13)+ π(5)2

= 65π + 25π = 90π

≈ 282.74 in2

V =

13π r2h


= 100π

≈ 314.16 in3

V =

13π(5)2(12)


Example 2Matt Mitarnowski wants to make a box that fits into a space that is 20 cm wide, 30 cm long, and 18 cm tall.

a. What is the amount of cardboard he would need to make this box?




SA = Ph + 2B




SA = Ph + 2B or




SA = Ph + 2B or SA = 2lw + 2wh + 2hl




SA = Ph + 2B or SA = 2lw + 2wh + 2hlP = Perimeter of base, B = Area of the base





SA = 2wl + 2lh + 2hw





SA = 2wl + 2lh + 2hw = 2(20)(30)+ 2(30)(18)+ 2(18)(20)






= 1200 +1080 + 720 = 2(20)(30)+ 2(30)(18)+ 2(18)(20)






= 1200 +1080 + 720 = 2(20)(30)+ 2(30)(18)+ 2(18)(20)

= 3000 cm2



b. What is the volume of the box Matt would make?




V = Bh




V = Bh or




V = Bh or V = lwh




V = Bh or V = lwhB = Area of the base





V = lwh





V = lwh = (30)(20)(18)





V = lwh = (30)(20)(18) = 10800 cm3


Example 3Fuzzy Jeff is making a cylindrical can to hold his

water collection. He currently has 100 in3 of water that he needs to store. Explain how you could

determine three different possible dimensions for this can and provide your three sets of dimensions.





V = π r2h





V = π r2h

100 = π r2h





V = π r2h

100 = π r2h

r2h =

100π





V = π r2h

100 = π r2h

r2h =

100π

r2h ≈ 31.38





V = π r2h

100 = π r2h

r2h =

100π

r2h ≈ 31.38

r h





V = π r2h

100 = π r2h

r2h =

100π

r2h ≈ 31.38

r h

3





V = π r2h

100 = π r2h

r2h =

100π

r2h ≈ 31.38

r h

3 r

2(3) ≈ 31.38





V = π r2h

100 = π r2h

r2h =

100π

r2h ≈ 31.38

r h

3 r

2(3) ≈ 31.383 3





V = π r2h

100 = π r2h

r2h =

100π

r2h ≈ 31.38

r h

3 r

2(3) ≈ 31.383 3 r

2 ≈ 10.46





V = π r2h

100 = π r2h

r2h =

100π

r2h ≈ 31.38

r h

3 r

2(3) ≈ 31.383 3 r

2 ≈ 10.46

r2 ≈ 10.46





V = π r2h

100 = π r2h

r2h =

100π

r2h ≈ 31.38

r h

3 r

2(3) ≈ 31.383 3 r

2 ≈ 10.46

r2 ≈ 10.46

r ≈ 3.23 inFriday, September 26, 14




V = π r2h

100 = π r2h

r2h =

100π

r2h ≈ 31.38

r h

3 r

2(3) ≈ 31.383 3 r

2 ≈ 10.46

r2 ≈ 10.46

r ≈ 3.23 in

3.23


Problem Set


Problem Set

p. 71 #1-25 odd, 37, 38

“Nobody got anywhere in the world by simply being content.” - Louis L’Amour


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Geometry Section 1-7 1112

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