Transcript Lesson 1 Contents

Lesson 12-1, 2, 7 & 13-2
3 D Figures
Nets
Spheres
Objectives
• Find the nets of 3-dimensional objects
• Find the Surface Area of Spheres
• Find the Volume of Spheres
Vocabulary
• Orthogonal Drawing – Two-dimensional view from
top, left, front and right sides
• Corner View – View of a figure from a corner
• Perspective View – same as a corner view
• Polyhedron – A solid with all flat surfaces that
enclose a single region of space
• Face – Flat surface of a polyhedron
• Edges – Line segments where two faces intersect
(edges intersect at a vertex)
• Bases – Two parallel congruent faces
• Prism – Polyhedron with two bases
• Regular Prism – Prism with bases that are regular
polygons
Vocabulary Cont
• Pyramid – Polyhedron with all faces (except for one
base) intersecting at one vertex
• Regular Polyhedron – All faces are regular
congruent polygons and all edges congruent
• Platonic Solids – The five types of regular polyhedra
(named after Plato)
• Cylinder – Solid with congruent circular bases in a
pair of parallel planes
• Cone – Solid with a circular base and a vertex (where
all “other sides” meet)
• Sphere – Set of points in space that are a given
distance from a given point (center)
• Cross Section – Intersection of a plane and a solid
• Reflection Symmetry – Symmetry with respect to
different planes (instead of lines)
Prisms & 3d-Terms
Triangular Prism
Pentagonal Prism
Prism – a polyhedron with
two parallel congruent faces
called bases. Other faces are
parallelograms.
Rectangular Prism
Faces (sides)
Vertexes
(corner pts)
Base (front and back)
Edges
(lines between
vertexes)
Other 3d Figures
Pyramid (Square)
h
Cylinder
l
h
Cone
l
h
Sphere
r
r
r
B
Pyramid – A
polyhedron with all
faces (except the
base) intersecting at
one vertex. Named
for their bases (which
can be any polygon).
l – slant height
Cylinder – A solid
with circular
congruent bases in
two parallel planes
(a can).
Cone – A solid
with circular
base and a
vertex.
l – slant height
Sphere – All points
equal distant from a
center point in 3space
Triangular Prism
Nets
Cylinder
Nets – cut a 3d figure on
its edges and lay it flat. It
can be folded into the
shape of the 3d figure with
no overlap
h
h
C
r
Square Prism (Cube)
Surface Area – Sum of each
area of the faces of the solid
Example 1
l
h
r
Which of the following represents the net of the cone above?
A.
B.
C.
D.
Example 2
h
r
Which of the following represents the net of the cylinder above?
A.
B.
C.
D.
Example 3
c h
c
l
b
Which of the following represents the net of the triangular prism above?
A.
B.
C.
D.
Spheres – Surface Area & Volume
Circles – Intersection between a plane
and a sphere
Great Circles – Intersections between a
plane passing through the center of the
sphere and the sphere. Great circles
have the same center as the sphere. The
shortest distance between two points on
a sphere lie on the great circle
containing those two points.
Hemisphere – a congruent half of a
sphere formed by a great circle. Surface
areas of hemispheres are half of the SA
of the sphere and the area of the great
circle. Volumes of hemispheres are half
of the volume of the sphere.
Sphere
r
V = 4/3 • π • r3
SA = 4π • r2
Sphere – All points
equal distant from a
center point in 3-space
Example 1:
Find the surface area and the
volume of the sphere to the right
10
SA = 4πr²
 need to find r
SA = 4π(10)² = 400π = 1256.64
V= 4/3πr³
 need to find r
V= 4/3π(10)³ = 4000π/3 = 4188.79
Example 2:
Find the surface area and the
volume of the sphere to the right
SA = 4πr²
 need to find r
r = ½ d = ½(18) = 9
SA = 4π(9)² = 324π = 1017.88
V= 4/3πr³
 need to find r
V= 4/3π(9)³ = 2916π/3 = 3053.63
18
Example 3:
16
Find the surface area and the volume
of the hemi-sphere to the right
½ of a sphere’s SA is just ½ SA = ½ 4πr²  NO!
We need to include the newly exposed “flat surface”
SA of a hemisphere = ½ 4π(r)² + π(r)² = 2πr² + πr² = 3πr²
SA = 3πr² = 3π(8)² = 192π = 603.19
Volume of ½ a sphere
½ V= ½ (4/3πr³) = 2/3πr³
V= 2/3π(8)³ = 1024π/3 = 1072.33
 need to find r
Find the surface area of a hemisphere with a radius
of 3.8 inches.
A hemisphere is half of a sphere. To find the surface area,
find half of the surface area of the sphere and add the
area of the great circle.
Surface area of a
hemisphere
Substitution
Use a calculator.
Answer: The surface area is approximately 136.1 sq inches.
Find the surface area of a ball with a circumference of
24 inches to determine how much leather is needed to
make the ball.
First, find the radius of the sphere.
Circumference of a circle
≈ 3.8
Next, find the surface area of the sphere.
Surface area of a sphere
Answer: The surface area is approximately 183.3 sq inches.
Find the volume of the sphere to the nearest tenth.
Volume of a sphere
r = 15
Use a calculator.
Answer: The volume of the sphere is approximately
14,137.2 cubic centimeters.
Find the volume of the sphere to the nearest tenth.
First find the radius of the sphere.
Circumference
of a circle
C 25
Solve for r.
Now find the volume.
Volume of a sphere
Use a calculator.
Answer: The volume of the sphere is approximately 263.9 cm³.
Summary & Homework
• Summary:
– Nets
• Every 3-dimensional solid can be represented by one or
more 2-dimensional nets (cardboard cutouts)
• Area of a net is the same as the surface area of the solid
– Sphere
• Volume: V = 4/3πr³
Surface Area: SA = 4πr²
– Hemi-sphere
• Volume: ½(sphere) = 2/3 πr³
Surface Area: SA = 3πr²
• A hemi-sphere’s surface consists of ½ SA of a sphere plus
area of exposed circle
• Homework:
– pg 704-705; 9-18