Date post: | 17-Jan-2016 |
Category: |
Documents |
Upload: | myron-walker |
View: | 215 times |
Download: | 0 times |
Lesson 12-1, 2, 7 & 13-2
3 D FiguresNets
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
Faces (sides)
Vertexes(corner pts)
Edges (lines between vertexes)Base (front and back)
Prism – a polyhedron with two parallel congruent faces called bases. Other faces areparallelograms.
Triangular Prism Pentagonal Prism
Rectangular Prism
Other 3d Figures
Pyramid (Square) Cylinder
h
r
h
B
lh
r
l
Cone Sphere
r
Pyramid – A polyhedron with all faces (except the base) intersecting at one vertex. Named for their bases (which can be any polygon).
Cylinder – A solid with circular congruent bases in two parallel planes (a can).
Cone – A solid with circular base and a vertex.
Sphere – All points equal distant from a center point in 3-space
l – slant height
l – slant height
NetsTriangular Prism
Square Prism (Cube)
Cylinder
h
r
h
C
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
Surface Area – Sum of each area of the faces of the solid
Example 1
h
r
l
Which of the following represents the net of the cone above?
D.C.B.A. D.
Example 2
Which of the following represents the net of the cylinder above?
h
r
D.C.B.A. D.
Example 3
Which of the following represents the net of the triangular prism above?
lb
hc c
D.C.B.A.A.
Spheres – Surface Area & Volume
V = 4/3 • π • r3
Sphere
r
SA = 4π • r2
Sphere – All points equal distant from a center point in 3-space
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.
Example 1:
10
Find the surface area and the volume of the sphere to the right
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
18
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
Example 3:
Find the surface area and the volume of the hemi-sphere to the right
16
½ of a sphere’s SA is just ½ SA = ½ 4πr²
SA of a hemisphere = ½ 4π(r)² + π(r)² = 2πr² + πr² = 3πr²
Volume of ½ a sphere ½ V= ½ (4/3πr³) = 2/3πr³ need to find r
V= 2/3π(8)³ = 1024π/3 = 1072.33
NO!We need to include the newly exposed “flat surface”
SA = 3πr² = 3π(8)² = 192π = 603.19
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
Use a calculator.
Answer: The surface area is approximately 136.1 sq inches.
Substitution
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
Next, find the surface area of the sphere.
Surface area of a sphere
Answer: The surface area is approximately 183.3 sq inches.
≈ 3.8
Find the volume of the sphere to the nearest tenth.
Answer: The volume of the sphere is approximately 14,137.2 cubic centimeters.
Volume of a sphere
r = 15
Use a calculator.
Find the volume of the sphere to the nearest tenth.
First find the radius of the sphere.
Circumference of a circle
Solve for r.
C 25
Answer: The volume of the sphere is approximately 263.9 cm³.
Volume of a sphere
Use a calculator.
Now find the volume.
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