unit 9 solids. lesson 9.1 day 1: identifying solids
TRANSCRIPT
Unit 9Solids
Lesson 9.1Day 1:
Identifying Solids
Lesson 9.1 Objectives
Define a polyhedron Identify properties of polyhedra Utilize Euler’s Theorem Identify a prism Identify a cylinder Calculate the surface area of prisms and
cylinders (G1.8.1)
Polyhedron
A polyhedron is a solid made of polygons. Remember, polygons are 2-D shapes with line segments for
sides. The polygons form faces, or sides of the solid. An edge of a polyhedron is the line segment that is
formed by the intersection of 2 faces. Typically the sides of the polygon faces.
A vertex of a polyhedron is a point in which 3 or more edges meet Typically the corners of the polygon’s faces.
Example 9.1
Determine if the following figures are polyhedra.Explain your reasoning.1.
NoThere are no faces.There are
no polygons.
Yes All faces arepolygons.
NoOne of the facesis not a polygon.
2.
3.
4.
5.
6.
YesAll faces are
polygons.
NoThere are faces that
are not polygons.
YesAll faces are
polygons.
Convex v Concave Polyhedra
A polyhedron will be convex when any two points on its surface can be connected by a segment that lies completely in its interior.
Said another way, pick any two points on different edges and make sure that the segment connecting them stays inside the polyhedron.
All faces should be convex polygons.
A polyhedron will be concave when any two points on its surface can be connected by a segment that leaves the interior and returns.
Said another way, pick any two points on different edges and see that the segment goes outside the polyhedron and then back in.
If one face of the polyhedron is concave, the entire polyhedron is said to be concave.
Cross Sections
When you take a plane and cut through a solid, the resulting shape of the surface is called the cross section. When asked to identify a cross section,
you need to identify the polygon formed. The plane acts like a knife blade and cuts
through the solid.
Example 9.2
Identify the cross section.1.
2.
3.
4.
Pentagon
Circle(Oval)
Triangle
Rectangle
Theorem 12.1: Euler’s Theorem
The number of faces (F), the number of vertices (V), and the number of edges (E) in a polyhedron are related by
2F V E This formula can be used to solve for a missing
quantity For instance, it is sometimes hard to count the
edges in the picture because some are hidden in the back.
The BEST use of this formula is to check your work that you have counted correctly.
Example 9.3
Find the number of vertices, faces, and edges each polyhedron has.1.
FF + + VV = = EE + 2 + 2
FF + + 88 = = 1212 + 2 + 2
FF + + 88 = = 1414
FF = = 66
FF = 5 = 5
VV = 6 = 6
55 + + 66 = = EE + 2+ 2
11 = 11 = EE + 2+ 2
EE = 9= 9
FF = 6 = 6
VV = 6 = 6
66 + + 66 = = EE + 2+ 2
12 = 12 = E E + 2+ 2
EE = 10= 10
FF = 8 = 8
VV = 12 = 12
88 + + 1212 = = EE + 2+ 2
20 = 20 = EE + 2+ 2
EE = 18= 18
2.
3.
4.
You do not have to use the formula every time.However, the formula should be used to
check your work!
Regular Polyhedra
A regular polyhedron is a solid that uses a ALL regular polygons for faces. And that the same number of faces meet
at each vertex. There are only five regular polyhedra,
called Platonic solids. Named after Greek mathematician and
philosopher Plato
The only shapes used as faces are: Equilateral Triangles Squares Regular Pentagons
Platonic Solids
NameFace
ShapeFace
sVertice
sEdge
sExample
Tetrahedron Triangle 4 4 6
Cube Square 6 8 12
Octahedron Triangle 8 6 12
Dodecahedron
Pentagon
12 20 30
Icosahedron Triangle 20 12 30
Prisms
A prism is a polyhedron with two congruent faces that are parallel to each other.
The congruent faces are called bases. The bases must be parallel to each other.
The other faces are called lateral faces. These are always rectangles or parallelograms or
squares. When naming a prism, they are always named
by the shape of their bases. Hexagonal Prism
Right v Oblique
In a right prism, the length of the lateral edge is the height. A right prism is one
that stands up straight with the lateral edges perpendicular to the bases.
The height of a prism is the perpendicular distance between the bases.
An oblique prism is one that is slanted to one side or the other. The length of the
slanted lateral edge is called the slant height.
In an oblique prism, the height must be drawn in so that it is perpendicular to both bases.
Example 9.4
Name the solid1.
2.
3.
4.
RectangularPrism
TriangularPrism
HexagonalPrism
TriangularPrism
Cylinder
A cylinder is a solid with congruent and parallel circles for bases. The lateral surface is a rectangle that is wrapped
around the circles.
height
Pyramid
A pyramid is a polyhedron with one base and lateral faces that meet at one common vertex. The base must be a polygon.
Not necessarily a square! The lateral faces will always be triangles.
Name the pyramid by its base shape.
Cone
A cone has a circular base and a vertex that is not in the same plane as the base.
Example 9.5
Name the solid.1.
2.
3.
4.
PentagonalPyramid
Cone
HeptagonalPyramid
TriangularPrism
Sphere
A sphere is a set of points in space that are equidistant from one given point.A sphere is a shell of points that are
the same distance from the center.A sphere is a 3-dimensional circle.
Parts of a Sphere
The point inside the sphere where all points are equidistant to is called the center of the sphere.
A radius of the sphere is a segment drawn from the center to a point on the sphere.
A chord of a sphere is a segment that joins any two points on the sphere.
The diameter is also a chord. Any 2-dimensional circle that contains the
center of the sphere is called a great circle. The equator would be a great circle.
Every great circle of a sphere splits a sphere into two congruent halves called hemispheres.
Nets
A net is a two-dimensional drawing of a three-dimensional solid. If you were to unfold a solid, the net
would show what it looks like. Every solid has a net.
However, there are only certain ways to draw a net for each solid.
Example 9.6
Identify the solid formed by the given net.Remember: Solids have a full name (2 parts)
1.
TriangularPrism
Cube
Cylinder
2.
3.
4.
PentagonalPyramid
Homework 9.1a
Lesson 9.1 – Identifying Solids (Day 1)
Due Tomorrow
Lesson 9.1Day 2:
Surface Area of Prisms and Cylinders
Theorem 12.2: Surface Area of a Prism
The surface area (SA) of a right prism can be found using the following formula:
2SA B Ph
B = Area of the base
P = Perimeter of the base
h = Height of the prism
Remember: The heightis the distance betweenthe bases.
Area of the Base (B)
The area of the base will be calculated using the appropriate formula for the shape of the base.
2squareB s
rectangleB b h
triangle
1
2B b h
kite 1 2
1
2B d d
rhombus 1 2
1
2B d d
parallelogramB b h
trapezoid 1 2
1
2B b b h
2
equilateral triangle 34
sB
2
regular hexagon 3 64
sB
regular polygon
1
2B a n s
Example 9.7
Find the surface area of the following right prisms.1.
SASA = 2 = 2BB + P + PhhSASA = 2 = 2BB + P + Phh
2.
SASA = 2 = 2(b(b•h)•h) + P + Phh
SASA = 2 = 2(5(5•6)•6) + P + Phh
SASA = 2 = 2(5(5•6)•6) + (5+6+5+6) + (5+6+5+6)(7)(7)
SASA = 2 = 2(30(30)) + (22) + (22)(7)(7)
SASA = 60 + 154 = 60 + 154= 204 sq. meters= 204 sq. meters
Lateral Area
The surface area of a prism is the sum of the areas of all the faces and bases.
The lateral area of a prism is the sum of the areas of the lateral faces ONLY.
2SA B Ph
LA Ph
Example 9.8
Find the lateral area of the right prisms.1.
2.
3.
LALA = P = Phh LALA = P = Phh
LALA = P = Phh
Theorem 12.3: Surface Area of a Cylinder
The surface area (SA) of a right cylinder can be found using the following formula:
2SA B Ch
B = Area of the base C = Circumference of the base
h = Height of the cylinder
Remember: The heightis the distance betweenthe bases.
Area of a Circler2
Circumference of a Circle2rord
2SA B Ph
Example 9.9
Find the surface area of the following cylinder.Round your answer to the nearest tenth.
1.
SA = 2SA = 2BB + + CChh
SA = 2SA = 2rr22 + + CChh
SA = 2SA = 2rr22 + + 22rrhh
SA = 2SA = 2(9)(9)22 + + 22(9)(9)(7)(7)
SA= 2SA= 2(81)(81) + 2 + 2(63)(63)
SA = 162SA = 162 + 126 + 126SA = 288SA = 288
SA = 508.94 + 395.84SA = 508.94 + 395.84= 904.8= 904.8
ftft22
ftft22
2.
Homework 9.1b
Lesson 9.1 – Surface Area of Prism and Cylinders (Day 2)
Due Tomorrow
Lesson 9.2Surface Area of Pyramids and Cones
Lesson 9.2 Objectives
Identify a pyramid Calculate slant height Identify a cone Calculate the surface area of a pyramid
and cone. (G1.8.1)
Regular Pyramids
A regular pyramid has a regular polygon for a base, and the common vertex is directly above the center of the base.
The height of a pyramid is the perpendicular distance from the base to the common vertex. The height of a regular pyramid is the length of the
line drawn from the center of the base straight up to the common vertex.
The slant height only exists in regular pyramids and cones and it is the length of a line drawn from the base up the lateral face to the common vertex.
Finding Slant Height
To find the slant height, you must know or be able to calculate height of the pyramid apothem of the base of the pyramid
can be found knowing one side of the base.
The reason you need those quantities is because they form a hidden right triangle. Then Pythagorean Theorem can be used to find the
missing slant height.
Example 9.10
Find the slant height for the following pyramid.Remember, you must make a right triangle using height, slant height, and apothem.
1.
10
24x
c2 = a2 + b2
x2 = 102 + 242
x2 = 100 + 576
x2 = 676
x = √676= 26 cm
2.
Theorem 12.4: Surface Area of a Regular Pyramid
The surface area (SA) of a regular pyramid is found using the following formula:
SA B P
B = Area of the base
P = Perimeter of the base
= Slant height of the pyramid
Remember: The slant heightis drawn up the middle of a lateral face..
Area of the Base (B)
The area of the base will be calculated using the appropriate formula for the shape of the base.
2squareB s
rectangleB b h
triangle
1
2B b h
kite 1 2
1
2B d d
rhombus 1 2
1
2B d d
parallelogramB b h
trapezoid 1 2
1
2B b b h
2
equilateral triangle 34
sB
2
regular hexagon 3 64
sB
regular polygon
1
2B a n s
Example 9.11
Find the surface area of the following pyramid.1.
SASA = = BB + ½P + ½Pl
SASA = = (9(9•9)•9) + ½P + ½Pl
SASA = = (9(9•9)•9) + ½(9+9+9+9) + ½(9+9+9+9)l
SASA = = (9(9•9)•9) + ½(9+9+9+9) + ½(9+9+9+9)(10)(10)
SASA = = (81(81)) + ½(36) + ½(36)(10)(10)
SASA = = (81(81)) + ½(360) + ½(360)
SASA = = (81(81)) + 180 + 180
SASA = 261 m = 261 m22
2.
1
2SA B P
Theorem 12.5: Surface Area of a Cone
The surface area of a right cone is found using the following formula:
B = Area of the base C = Circumference of the base
= Slant height of the cone
Area of a Circler2
Circumference of a Circle2rord
2SA r r
Example 9.12
Find the surface area of the following pyramid.1.
SASA = = rr22 + + rrll
SASA = = (7)(7)22 + + rrll
SASA = = (7)(7)22 + + (7)(15)(7)(15)
SASA = = 4949+ 105+ 105
2.
SASA = 154 = 154= 403.8= 403.8 inin22
Homework 9.2
Lesson 9.2 – Surface Area of Pyramids and Cones
Due Tomorrow
Lesson 9.3Volume of Special Solids
Lesson 9.3 Objectives
Calculate the volume of a prism (G1.8.1)
Calculate the volume of a cylinder (G1.8.1)
Calculate the volume of a pyramid (G1.8.1)
Calculate the volume of a cone (G1.8.1)
Apply Cavalieri’s Principle Utilize the Volume Postulates
Volume
The volume of any solid is the amount of space contained in its interior.
The volume is measured in cubic units m3
cm3
ft3
in3
units3
Volume PostulatesPostulate 27: Volume of a Cube
The volume of a cube is the length of its side cubed.
3V s
Volume Theorems
Theorem 12.7:Volume of a Prism The volume (V) of a
prism is found using the following formula:
Theorem 12.8:Volume of a Cylinder The volume (V) of a cylinder
is found using the following formula:
V B h V B h
B = Area of the base
h = Height of the prism
Area of the Base (B)
The area of the base will be calculated using the appropriate formula for the shape of the base.
2squareB s
rectangleB b h
triangle
1
2B b h
kite 1 2
1
2B d d
rhombus 1 2
1
2B d d
parallelogramB b h
trapezoid 1 2
1
2B b b h
2
equilateral triangle 34
sB
2
regular hexagon 3 64
sB
regular polygon
1
2B a n s
Example 9.13
Find the volume of the following figures.1.
2.
3.
VV = = BBhh
VV = = (b(b•h)•h)hh
VV = = (7(7•3)•3)hh
VV = = (21(21))(5)(5)
VV = = 105105mm33
VV = = BBhh
VV = = BBhh
Volume Theorems
Theorem 12.9:Volume of a Pyramid The volume (V) of a pyramid
is found using the following formula:
Theorem 12.10:Volume of a Cone The volume (V) of a cone is
found using the following formula:
1
3V B h
1
3V B h
B = Area of the base
h = Height of the prismNotice that it
is height and notslant height
Example 9.14
Find the volume of the following solids.1.
VV = = 11//33BBhh
VV = = 11//33[[11//22bh]bh]hh
VV = = 11//33[[11//22(4)h](4)h]hh
VV = = 11//33[[11//22(4)(6)](4)(6)]hh
VV = = 11//33[[11//22(4)(6)](4)(6)](5)(5)
VV = = 11//33[12][12](5)(5)
VV = = 11//33(60)(60)= 20 m= 20 m33
VV = = 11//33BBhh
VV = = 11//33[[rr22]]hh
VV = = 11//33[[(3)(3)22]]hh
VV = = 11//33[[(3)(3)22]](8)(8)
VV = = 11//33[9[9]](8)(8)
VV = = 11//33(72(72))
VV = 24 = 24= 75.40 mm= 75.40 mm33
2.
3.
Volume Postulates
Postulate 28:Volume Congruence
If two polyhedra are congruent, then they have the same volume.
Postulate 29:Volume Addition
The volume of a solid is the sum of the volume of all its nonoverlapping parts.
Theorem 12.6: Cavalieri’s Principle
If two solids have the same height and the same cross-sectional area at every level, then they have the same volume. So whether the solid is tilted or straight up, the
volume is the same as long as the base area is the same size all the way up the solid.
Cavalieri’s Principle holds true for pyramids and cones as well.
Homework 9.3
Lesson 9.3 – Volume Special Solids Due Tomorrow
Lesson 9.4Spheres
Lesson 9.4 Objectives
Define a sphere Calculate the surface area of a
sphere (G1.8.1)
Calculate the volume of a sphere (G1.8.1)
Verify the effects of a dimensional change on surface area and volume. (G2.3.5)
Example 9.15
Identify the following characteristics:1. Name the center of the sphere.
1. T
2. Name a segment that is the radius of the sphere.
2. segment TSsegment TQsegment TP
3. Name a chord of the sphere.3. segment QR
segment PS
4. Find the circumference of the great circle.Write your final answers in terms of .
4. 14 m
Theorem 12.11: Surface Area of a Sphere
The surface area (SA) of a sphere is found using the following formula:
24SA r
r = Radius of the sphere
Example 9.16
Find the surface area of the following spheres.Round answer to the nearest tenth.
1.
SASA = 4 = 4rr22
SSA= 4A= 4(19)(19)22
SASA = 4 = 4(361)(361)
SASA = 1444 = 1444SASA = 4536.5 ft = 4536.5 ft22
2.
3.
Theorem 12.12: Volume of a Sphere
The volume (V) of a sphere is found using the following formula:
34
3V r
r = Radius of the sphere
Example 9.17
Find the volume of the following spheres.Round answer to the nearest tenth.
1.
VV = = 44//33rr33
VV = = 44//33(19)(19)33
VV = = 44//33(6859)(6859)
VV = ( = (27,43627,436//33))
VV = 28,730.9 ft = 28,730.9 ft33
2.
3.
Dimensional Changes AffectingVolume and Surface Area
What happens to a sphere when the radius changes?
What happens to the surface area when the radius doubles?
The surface area is 4 times larger. What happens to the volume when the
radius doubled? The volume got 8 times larger.
Why? When a measurement changes, you
must make that change in the formula as well.
Since the radius is squared in the formula for surface area, then any change to the radius will have a squared effect on the surface area.
Doubled2 = Quadrupled And since the radius is cubed in the
formula for volume, then any change to the radius will have a cubed effect on the volume.
Doubled3 = 8 times the effect
This applies toall formulas for
all solids!
Example 9.18
1. Find the surface area of the sphere below. What would the surface area be if the radius were multiplied by 3?
2. Find the volume of the sphere above. What would the volume be if the radius were multiplied by 3?
3 times larger
6 times larger
9 times larger
27 times larger
3 times larger
6 times larger
9 times larger
27 times larger
Homework 9.4
Lesson 9.4 – Spheres Due Tomorrow
Lesson 9.5Converting Units
Lesson 9.5 Objectives
Convert one-dimensional measurements (L3.1.1)
Convert various dimensional measurements
Basic Unit Equivalencies
The following are equivalent measurements from different systems of units:
100 cm 1 m 1000 m 1 km12 in 1 ft
3 ft 1 yd
5280 ft 1 mi
16 oz 1 lb
1000 mm 1 m
24 hrs 1 day
7 days 1 week
60 sec 1 min
60 min 1 hr
1000 mg 1 g 1000 g 1 kgThe prefix centi-
means 1/100.So a centimeter is
1/100 of a meter.The prefix milli-
means 1/1000.So a millimeter is
1/1000 of a meter.
The prefix kilo- means 1000.
So a kilometer is1000 meters.
2.54 cm 1 in
Unit Conversions
Does the actual width of the front desk change just because we use a different side of the measuring stick?
No But the numerical value changes, how does that make sense?
Because the scales marked differently. So, how do we mathematically change a number but keep the overall
value the same? What happens when you multiply any number by 1?
The number stays the same, does it not?
When converting units, we use our conversion factors to make a fraction that is equivalent to 1.
Remember, the numbers are different but the value they are measuring is the same. And when you divide a value by the same value, the fraction is equal to 1.
That way when we multiply during the conversion, we multiply by 1.
Dimensional (Unit) Analysis
Here are the steps to dimensional analysis.1. Write down the measurement given.2. Find a conversion factor with the same units as your given value.
2. Hopefully that conversion factor will also contain the units that you want to change into.
3. Write that conversion factor as a fraction with the denominator of the fraction containing the units you are attempting to change.
3. The fraction should contain the numerical values and their assigned units as well.
4. To verify the setup is correct, now is the time to analyze the units so that all units make their own fractions equivalent to 1.
4. We will cancel out all units that make fractions of 1, since multiplying by 1 does not change the value of the overall problem.
5. Now multiply (or divide) the numbers as you see them in the problem.
6. Finally, the units on your final answer are the only units that have not been cancelled out during step 4.
11,400 mm1 m
1000 mm
1 km
1000 m
.0114 km
Example 9.19
Convert the following 1. 432 cm m
2. 3.5 hrs sec
3. 2.1 km mm
4. 1 weeks min
Multi-Dimensional Unit Conversions
What are the factors of 7? 1 and 7
How about x2? xx
Using that logic, what are the factors of x3? xxx
Continuing with that logic, what are the factors of in3? ininin
So when we convert units that involve exponents, the exponent tells us how many times we have to convert from one unit of measurement to the other.
11,400 mm1 m
1000 mm
1 km
1000 m
Example 9.20
Convert the following 1. 914 in2 ft2
2. 121 cm3 m3
3. 3.8 mi2 ft2
4. 1 m3 mm3
Homework 9.5
Lesson 9.5 – Converting Units Due Tomorrow