Transcript 8.1 Prisms, Area and Volume • Prism – 2 congruent polygons lie in parallel planes – – – – corresponding sides are parallel. corresponding vertices are connected base edges.

8.1 Prisms, Area and Volume
• Prism – 2 congruent polygons lie in parallel
planes
–
–
–
–
corresponding sides are parallel.
corresponding vertices are connected
base edges are edges of the polygons
lateral edges are segments connecting
corresponding vertices
8.1 Prisms, Area and Volume
• Right Prism – prism in which the lateral
edges are  to the base edges at their points
of intersection.
• Oblique Prism – Lateral edges are not
perpendicular to the base edges.
• Lateral Area (L) – sum of areas of lateral
faces (sides).
8.1 Prisms, Area and Volume
• Lateral area of a right prism: L = hP
– h = height (altitude) of the prism
– P = perimeter of the base (use perimeter
formulas from chapter 7)
• Total area of a right prism: T = 2B + L
– B = base area of the prism (use area formulas
from chapter 7)
8.1 Prisms, Area and Volume
• Volume of a right rectangular prism (box) is
given by V = lwh where
– l = length
– w = width
– h = height
l
w
h
8.1 Prisms, Area and Volume
• Volume of a right prism is given by V = Bh
– B = area of the base (use area formulas from
chapter 7)
– h = height (altitude) of the prism
8.2 Pyramids Area, and Volume
• Regular Pyramid – pyramid whose base is a
regular polygon and whose lateral edges are
congruent.
– Triangular pyramid: base is a triangle
– Square pyramid: base is a square
8.2 Pyramids Area, and Volume
• Slant height (l) of a pyramid: The altitude of
the congruent lateral faces.
l a h
2
2
2
Slant height
height
apothem
8.2 Pyramids Area, and Volume
• Lateral area - regular pyramid with slant
height = l and perimeter P of the base is:
L = ½ lP
• Total area (T) of a pyramid with lateral area
L and base area B is: T = L + B
• Volume (V) of a pyramid having a base area
B and an altitude h is:
V  1 Bh
3
8.3 Cylinders and Cones
• Right circular cylinder: 2 circles in parallel planes
are connected at corresponding points. The
segment connecting the centers is  to both planes.
8.3 Cylinders and Cones
• Lateral area (L) of a right cylinder with
altitude of height h and circumference C
L  hC  2 rh
• Total area (T) - cylinder with base area B
T  L  2B  L  2( r )
2
• Volume (V) of a cylinder is V = B  h
V  Bh   r h
2
8.3 Cylinders and Cones
• Right circular cone – if the axis which
connects the vertex to the center of the base
circle is  to the plane of the circle.
8.3 Cylinders and Cones
• In a right circular cone with radius r,
altitude h, and slant height l (joins vertex to
point on the circle),
Slant height
l2 = r2 + h2
height
radius
8.3 Cylinders and Cones
• Lateral area (L) of a right circular cone is:
L = ½ lC = rl where l = slant height
• Total area (T) of a cone:
T = B + L (B = base circle area = r2)
• Volume (V) of a cone is:
V  Bh   r h
1
3
1
3
2
8.4 Polyhedrons and Spheres
• Polyhedron – is a solid bounded by plane regions.
A prism and a pyramid are examples of
polyhedrons
• Euler’s equation for any polyhedron: V+F = E+2
– V - number of vertices
– F - number of faces
– E - number of edges
8.4 Polyhedrons and Spheres
• Regular Polyhedron – is a convex polyhedron
whose faces are congruent polygons arranged in
such a way that adjacent faces form congruent
dihedral angles.
tetrahedron
8.4 Polyhedrons and Spheres
• Examples of polyhedrons (see book)
–
–
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Tetrahedron (4 triangles)
Hexahedron (cube – 6 squares)
Octahedron (8 triangles)
Dodecahedron (12 pentagons)
8.4 Polyhedrons and Spheres
• Sphere formulas:
– Total surface area (T) = 4r2
– Volume V  43  r 3
radius
9.1 The Rectangular Coordinate System
• Distance Formula: The distance between 2
points (x1, y1) and (x2,y2) is given by the
formula:
d
x2  x1    y 2  y1 
2
2
What theorem in geometry does this come
from?
9.1 The Rectangular Coordinate System
• Midpoint Formula: The midpoint M of the
line segment joining (x1, y1) and (x2,y2) is :
 x1  x2 y1  y2 
M 
,

2 
 2
• Linear Equation: Ax + By = C (standard
form)
9.2 Graphs of Linear Equations and Slopes
• Slope – The slope of a line that contains the
points (x1, y1) and (x2,y2) is given by:
y2  y1 rise
m

x2  x1
run
rise
run
9.2 Graphs of Linear Equations and Slopes
• If l1 is parallel to l2 then m1 = m2
• If l1 is perpendicular to l2 then: m1  m2  1
(m1 and m2 are negative reciprocals of each
other)
• Horizontal lines are perpendicular to
vertical lines
9.3 Preparing to do Analytic Proofs
To prove:
You need to show:
2 lines are parallel
m1 = m2, using
m
2 lines are perpendicular
y 2  y1
x 2  x1
m1 m2 = -1
2 line segments are congruent lengths are the same, using
d
A point is a midpoint
x2  x1 2   y 2  y1 2
 x  x2 y1  y 2 
M  1
,

2 
 2
9.3 Preparing to do Analytic Proofs
•
Drawing considerations:
1. Use variables as coordinates, not (2,3)
2. Drawing must satisfy conditions of the proof
3. Make it as simple as possible without losing
generality (use zero values, x/y-axis, etc.)
•
Using the conclusion:
1. Verify everything in the conclusion
2. Use the right formula for the proof
9.4 Analytic Proofs
• Analytic proof – A proof of a geometric
theorem using algebraic formulas such as
midpoint, slope, or distance
• Analytic proofs
– pick a diagram with coordinates that are
appropriate.
– decide on what formulas needed to reach
conclusion.
9.4 Analytic Proofs
• Triangles to be used for proofs are in:
table 9.1
• Quadrilaterals to be used for proofs are in:
table 9.2.
• The diagram for an analytic proof test
problem will be given on the test.
9.5 Equations of Lines
• General (standard) form: Ax + By = C
• Slope-intercept form: y = mx + b
(where m = slope and b = y-intercept)
• Point-slope form: The line with slope m
going through point (x1, y1) has the
equation: y – y1 = m(x – x1)
9.5 Equations of Lines
• Example: Find the equation in slope-intercept
form of a line passing through the point (-4,5) and
perpendicular to the line 2x+3y=6
(solve for y to get slope of line)
2 x  3 y  6  3 y  2 x  6
y
2
3
x2m
2
3
(take the negative reciprocal to get the  slope)
m 
3
2
9.5 Equations of Lines
3
m

• Example (continued):  2
Use the point-slope form with this slope and the
point (-4,5)
y  5  32 x  (4)
y  5  32 x  4  32 x  6
In slope intercept form:
y  32 x  11