geometry section 1-7 1112
TRANSCRIPT
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?
Friday, September 26, 14
Vocabulary1. Polyhedron:
2. Face:
3. Edge:
4. Vertex:
5. Prism:
6. Base:
Friday, September 26, 14
Vocabulary1. Polyhedron: A solid three-dimensional closed
figure with all flat surfaces
2. Face:
3. Edge:
4. Vertex:
5. Prism:
6. Base:
Friday, September 26, 14
Vocabulary1. Polyhedron: A solid three-dimensional closed
figure with all flat surfaces
2. Face: A flat surface of a polyhedron
3. Edge:
4. Vertex:
5. Prism:
6. Base:
Friday, September 26, 14
Vocabulary1. Polyhedron: A solid three-dimensional closed
figure with all flat surfaces
2. Face: A flat surface of a polyhedron
3. Edge: The line segment where two faces meet
4. Vertex:
5. Prism:
6. Base:
Friday, September 26, 14
Vocabulary1. Polyhedron: A solid three-dimensional closed
figure with all flat surfaces
2. Face: A flat surface of a polyhedron
3. Edge: The line segment where two faces meet
4. Vertex: The point where three or more edges meet
5. Prism:
6. Base:
Friday, September 26, 14
Vocabulary1. Polyhedron: A solid three-dimensional closed
figure with all flat surfaces
2. Face: A flat surface of a polyhedron
3. Edge: The line segment where two faces meet
4. Vertex: The point where three or more edges meet
5. Prism: Has two parallel, congruent polygonal bases with faces that are all parallelograms
6. Base:
Friday, September 26, 14
Vocabulary1. Polyhedron: A solid three-dimensional closed
figure with all flat surfaces
2. Face: A flat surface of a polyhedron
3. Edge: The line segment where two faces meet
4. Vertex: The point where three or more edges meet
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)
Friday, September 26, 14
Vocabulary7. Pyramid:
8. Cylinder:
9. Cone:
Friday, September 26, 14
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:
Friday, September 26, 14
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: A three-dimensional shape with congruent parallel circular bases which have one surface connecting them
9. Cone:
Friday, September 26, 14
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: 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
Friday, September 26, 14
Vocabulary10. Sphere:
11. Regular Polyhedron:
12. Platonic Solid:
Friday, September 26, 14
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:
Friday, September 26, 14
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: All faces are regular congruent polygons and all edges are congruent
12. Platonic Solid:
Friday, September 26, 14
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: 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
Friday, September 26, 14
Vocabulary13. Surface Area:
14. Volume:
Friday, September 26, 14
Vocabulary13. Surface Area: The total area of the surface of a
three-dimensional figure (square units)
14. Volume:
Friday, September 26, 14
Vocabulary13. Surface Area: The total area of the surface of a
three-dimensional figure (square units)
14. Volume: The amount of space that a three-dimensional object occupies (cubic units)
Friday, September 26, 14
Vocabulary13. Surface Area: The total area of the surface of a
three-dimensional figure (square 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?
Friday, September 26, 14
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.
Friday, September 26, 14
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
Friday, September 26, 14
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
r = radius, l = slant height
Friday, September 26, 14
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
r = radius, l = slant height
SA = π(5)(13)+ π(5)2
Friday, September 26, 14
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
r = radius, l = slant height
SA = π(5)(13)+ π(5)2
= 65π + 25π
Friday, September 26, 14
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
r = radius, l = slant height
SA = π(5)(13)+ π(5)2
= 65π + 25π = 90π
Friday, September 26, 14
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
r = radius, l = slant height
SA = π(5)(13)+ π(5)2
= 65π + 25π = 90π
≈ 282.74
Friday, September 26, 14
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
r = radius, l = slant height
SA = π(5)(13)+ π(5)2
= 65π + 25π = 90π
≈ 282.74 in2
Friday, September 26, 14
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
r = radius, l = slant height
SA = π(5)(13)+ π(5)2
= 65π + 25π = 90π
≈ 282.74 in2
V =
13π r2h
Friday, September 26, 14
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
r = radius, l = slant height
SA = π(5)(13)+ π(5)2
= 65π + 25π = 90π
≈ 282.74 in2
V =
13π r2h
r = radius, h = height
Friday, September 26, 14
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
r = radius, l = slant height
SA = π(5)(13)+ π(5)2
= 65π + 25π = 90π
≈ 282.74 in2
V =
13π r2h
r = radius, h = height
V =
13π(5)2(12)
Friday, September 26, 14
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
r = radius, l = slant height
SA = π(5)(13)+ π(5)2
= 65π + 25π = 90π
≈ 282.74 in2
V =
13π r2h
r = radius, h = height
= 100π V =
13π(5)2(12)
Friday, September 26, 14
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
r = radius, l = slant height
SA = π(5)(13)+ π(5)2
= 65π + 25π = 90π
≈ 282.74 in2
V =
13π r2h
r = radius, h = height
= 100π
≈ 314.16
V =
13π(5)2(12)
Friday, September 26, 14
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
r = radius, l = slant height
SA = π(5)(13)+ π(5)2
= 65π + 25π = 90π
≈ 282.74 in2
V =
13π r2h
r = radius, h = height
= 100π
≈ 314.16 in3
V =
13π(5)2(12)
Friday, September 26, 14
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?
Friday, September 26, 14
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
Friday, September 26, 14
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 or
Friday, September 26, 14
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 or SA = 2lw + 2wh + 2hl
Friday, September 26, 14
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 or SA = 2lw + 2wh + 2hlP = Perimeter of base, B = Area of the base
Friday, September 26, 14
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 or SA = 2lw + 2wh + 2hlP = Perimeter of base, B = Area of the base
SA = 2wl + 2lh + 2hw
Friday, September 26, 14
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 or SA = 2lw + 2wh + 2hlP = Perimeter of base, B = Area of the base
SA = 2wl + 2lh + 2hw = 2(20)(30)+ 2(30)(18)+ 2(18)(20)
Friday, September 26, 14
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 or SA = 2lw + 2wh + 2hlP = Perimeter of base, B = Area of the base
SA = 2wl + 2lh + 2hw
= 1200 +1080 + 720 = 2(20)(30)+ 2(30)(18)+ 2(18)(20)
Friday, September 26, 14
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 or SA = 2lw + 2wh + 2hlP = Perimeter of base, B = Area of the base
SA = 2wl + 2lh + 2hw
= 1200 +1080 + 720 = 2(20)(30)+ 2(30)(18)+ 2(18)(20)
= 3000 cm2
Friday, September 26, 14
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.
b. What is the volume of the box Matt would make?
Friday, September 26, 14
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.
b. What is the volume of the box Matt would make?
V = Bh
Friday, September 26, 14
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.
b. What is the volume of the box Matt would make?
V = Bh or
Friday, September 26, 14
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.
b. What is the volume of the box Matt would make?
V = Bh or V = lwh
Friday, September 26, 14
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.
b. What is the volume of the box Matt would make?
V = Bh or V = lwhB = Area of the base
Friday, September 26, 14
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.
b. What is the volume of the box Matt would make?
V = Bh or V = lwhB = Area of the base
V = lwh
Friday, September 26, 14
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.
b. What is the volume of the box Matt would make?
V = Bh or V = lwhB = Area of the base
V = lwh = (30)(20)(18)
Friday, September 26, 14
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.
b. What is the volume of the box Matt would make?
V = Bh or V = lwhB = Area of the base
V = lwh = (30)(20)(18) = 10800 cm3
Friday, September 26, 14
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.
Friday, September 26, 14
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
Friday, September 26, 14
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
100 = π r2h
Friday, September 26, 14
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
100 = π r2h
r2h =
100π
Friday, September 26, 14
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
100 = π r2h
r2h =
100π
r2h ≈ 31.38
Friday, September 26, 14
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
100 = π r2h
r2h =
100π
r2h ≈ 31.38
r h
Friday, September 26, 14
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
100 = π r2h
r2h =
100π
r2h ≈ 31.38
r h
3
Friday, September 26, 14
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
100 = π r2h
r2h =
100π
r2h ≈ 31.38
r h
3 r
2(3) ≈ 31.38
Friday, September 26, 14
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
100 = π r2h
r2h =
100π
r2h ≈ 31.38
r h
3 r
2(3) ≈ 31.383 3
Friday, September 26, 14
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
100 = π r2h
r2h =
100π
r2h ≈ 31.38
r h
3 r
2(3) ≈ 31.383 3 r
2 ≈ 10.46
Friday, September 26, 14
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
100 = π r2h
r2h =
100π
r2h ≈ 31.38
r h
3 r
2(3) ≈ 31.383 3 r
2 ≈ 10.46
r2 ≈ 10.46
Friday, September 26, 14
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
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
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
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
Friday, September 26, 14
Problem Set
Friday, September 26, 14
Problem Set
p. 71 #1-25 odd, 37, 38
“Nobody got anywhere in the world by simply being content.” - Louis L’Amour
Friday, September 26, 14