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Chapter 10
Section 10.1 - Areas of
Parallelograms and Triangles
Objectives:
To find the area of a parallelogram
To find the area of a triangle
 Theorem 10.1 – Area of a Rectangle
 The area of a rectangle is the product of its base and height.
h
A = bh
b
 Base of a Parallelogram  any of its sides
 Altitude  a segment perpendicular to the line containing
that base, drawn from the side opposite the base.
 Height  the length of an altitude.
Altitude
Base
 Theorem 10.2 – Area of a Parallelogram
 The area of a parallelogram is the product of a base and the
corresponding height.
h
b
A = bh
 Base of a Triangle  any of its sides
 Height  the length of the altitude to the line containing
that base
 Theorem 10.3 – Area of a Triangle
 The area of a triangle is half the product of a base and the
corresponding height.
h
b
A=
1
bh
2
 Ex: Find the area of each Parallelogram
4.5 in
4.6 cm
4 in
5 in
2 cm
3.5 cm
 Ex: Find the area of each Triangle.
6.4 ft
10 ft
4 ft
5 cm
13 cm
12 cm
 When designing a building, you must be sure that the building
can withstand hurricane-force winds, which have a velocity of
73 mi/h or more. The formula F = 0.004A𝑣 2 gives the force F in
pounds exerted by a wind blowing against a flat surface. A is
the area of the surface in square feet, and v is the wind velocity
in miles per hour. How much force is exerted by a 73 mi/h wind
blowing against the side of the building shown below?
6 ft
12 ft
20 ft
Homework #18
Due Tuesday (March 12)
Page 536 – 538
# 1 – 27 odd
Section 10.2 – Areas of Trapezoids,
Rhombuses, and Kites
 Objectives:
To find the area of a trapezoid
To find the area of a rhombus or a kite
 Theorem 10.4 – Area of a Trapezoid
 The area of a trapezoid is half the product of the height and the
sum of the bases.
𝑏1
h
𝑏2
A=
1
h(𝑏1
2
+ 𝑏2 )
 Theorem 10.5 – Area of a Rhombus or a Kite
 The area of a rhombus or a kite is half the product of the
lengths of its diagonals.
𝑑1
𝑑2
A=
1
𝑑1 𝑑2
2
 Ex: Find the area of the trapezoid
12 cm
7 cm
15 cm
 What is the area of trapezoid PQRS? What would the area be
if <P was changed to 45°?
5m
S
R
60°
P
7m
Q
 Ex: Find the area of the kite.
3m
3m
2m
5m
 Find the area of the rhombus.
B
15 m
A
E
12 m
D
C
Homework #19
Due Wednesday (March 13)
Page 542 – 543
# 1 – 29 odd
Section 10.3 – Areas of Regular
Polygons
 Objectives:
To find the area of a regular polygon
 You can circumscribe a circle about any regular polygon.
 The center of a regular polygon is the center or the
circumscribed circle.
 The radius is the distance from the center to a vertex.
 The apothem is the perpendicular distance from the center to
a side.
Center
Radius
Apothem
 Ex: Finding Angle Measures
 The figure below is a regular pentagon with radii and apothem
drawn. Find the measure of each numbered angle.
360
= 72
5
1
= m<1 = 36
2
m<1 =
(Divide 360 by the number of sides)
m<2
(apothem bisects the vertex angle)
m<3 + 90 + 36 = 180 m<3 = 54
3
2
1
 Find the measure of each angle of the half of an octagon.
1
2
3
 Suppose you have a regular n-gon with side s. The radii divide
the figure into n congruent isosceles triangles. Each isosceles
1
triangle has area equal to as (a being apothem/s being side).
2
 Since there are n congruent triangles, the area of the n-gon is
1
A = n · as. The perimeter p of the n-gon is ns. Substituting p for
2
ns results in a formula for the area of the polygon.
 Theorem 10.6 – Area of a Regular Polygon
 The area of a regular polygon is half the product of the apothem
and the perimeter.
p
a
1
A = ap
2
 Ex: Find the area of each regular polygon.
A regular decagon with a 12.3 in. apothem and 8 in. sides
A regular pentagon with 11.6 cm sides and an 8 cm
apothem
 Ex: Find the area of the hexagon.
10 mm
5 mm
Homework #20
Due Thurs/Fri (March 14/15)
Page 548
# 1 – 23 all
Section 10.4 – Perimeters and Areas
of Similar Figures
 Objectives:
To find the perimeters and areas of similar figures
 Theorem 10.7 – Perimeters and Areas of Similar Figures
 If the similarity ratio of two similar figures is
𝑎
1. the ratio of their perimeters is
𝑏
𝑎2
2. the ratio of their areas is 2
𝑏
𝑎
, then
𝑏
 Ex: Finding Ratios in Similar Figures
 The trapezoids below are similar. Find the ratio of their perimeters
and ratio of their areas.
9m
6m
 Ex:
 Two similar polygons have corresponding sides in the ratio 5 : 7.
a. Find the ratio of their perimeters
b. Find the ratio of their areas
 Ex: Finding Areas Using Similar Figures
 The area of the smaller regular pentagon is 27.5 𝑐𝑚2 . What is the
area of the larger pentagon?
4 cm
10 cm
 Ex:
 The corresponding sides of two similar
parallelograms are in the ratio 3 : 4. The area of the
larger parallelogram is 96 𝒊𝒏𝟐 . Find the area of the
smaller parallelogram.
 Ex: Finding Similarity and Perimeter Ratios
 The areas of two similar triangles are 50 𝑐𝑚2 and 98 𝑐𝑚2 . What is
the similarity ratio? What is the ratio of their perimeters?
 The areas of two similar rectangles are 1875 𝑓𝑡 2 and 135 𝑓𝑡 2 . Find
the ratio of their perimeters.
Homework #21
Due Monday (March 18)
Page 555 – 556
# 1 – 23 all
Section 10.5 – Trigonometry and Area
 Objectives:
To find the area of a regular polygon using
trigonometry
To find the area of a triangle using trigonometry
 In the last lesson, we learned how to find the area of a regular
1
polygon by using the formula A = ap. By using this formula
2
and trigonometric ratios, you can solve other types of
problems.
 Trigonometry Review

𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
Sine =
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
 Cosine =
SOH
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
CAH
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
TOA
 Tangent =
 Ex: Find the area of the regular pentagon with 8cm sides.
8 cm
 Ex: Finding the Area and Perimeter
Find the area and perimeter of a regular octagon with radius 16m.
 Suppose we want to find the area of triangle ABC (below), but
you are only given m<A and lengths b and c. To use the formula
1
A = bh, you need to find the height. This can be found by using
2
the sine ratio:
sin A =
ℎ
𝑐
therefore
h = c(sin A)
B
c
h
A
Area =
a
C
b
1
bc(sin
2
A)
 Theorem 10.8 – Area of a Triangle Given SAS
 The area of a triangle is one half the product of the lengths of two
sides and the sine of the included angle.
B
1
Area ΔABC = bc(sin
2
a
c
A
b
C
A)
 Ex: Finding the area of a Triangle
 Two sides of a triangular building plot are 120 ft and 85 ft
long. They include an angle of 85°. Find the area of the
building plot to the nearest square foot.
Homework #22
Due Monday (March 18)
Page 561 – 562
# 1 – 27 odd
Quiz Tuesday (10.1 – 10.5)
Section 10.6 – Circles and Arcs
 Objectives:
To find the measures of central angles and arcs
To find circumference and arc length
 Circle  the set of all points equidistant from a given point called
the center.
 Radius  a segment that has one endpoint at the center and the
other endpoint on the circle.
 Congruent Circles  have congruent radii
 Diameter  a segment that contains the center of a circle and has
both endpoints on the circle
 Central Angle  an angle whose vertex is the center of the circle.
 Circle = ΘP
Central Angle = <CPA
 Radius = CP
Diameter = AB
C
A
B
P
 Semicircle  half of a circle (180°)
 Minor Arc  smaller than a semicircle (< 180°). Its measure is the
measure of its corresponding angle.
 Major Arc  greater than a semicircle (> 180°). Its measure is 360
minus the measure of its related minor arc.
 Adjacent Arcs  arcs of the same circle that have exactly one
point in common.
Ex: Identify the following in Θ O:
a. The minor arcs (4)
b. The semicircles (4)
c. The major arcs that contain point A (4)
A
C
O
D
E
 Postulate 10.1 – Arc Addition Postulate
 The measure of the arc formed by two adjacent arcs is the sum of
the measures of the two arcs.
B
C
A
mABC = mAB + mBC
 Circumference  the distance around a circle
 Pi (Π)  the ratio or the circumference of a circle to its diameter
 Theorem 10.9 – Circumference of a Circle
 The circumference of a circle is Π (pi) times the diameter
d
r
O
C = Πd
C
C = 2Πr
 Arc Length  a fraction of a circle’s circumference
 Theorem 10.10  Arc Length
 The length of an arc of a circle is the product of the ratio
𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑟𝑐
and the circumference of the circle.
360
A
r
Length of AB =
O
B
𝑚𝐴𝐵
360
· 2Πr
 Ex: Find the measure of each arc.
a. BC
b. BD
c. ABC
C
B
d. AB
58°
D
O
32°
A
 Ex:
 The diameter of a bicycle wheel is 22in. To the
nearest whole number, how many revolutions does
the wheel make when the bicycle travels 100 feet?
Homework #23
Due Thurs/Fri (March 21/22)
Page 569 – 570
# 1 – 39 odd
Section 10.7 – Areas of Circles and
Sectors
 Objectives:
To find the areas of circles, sectors, and
segments of circles.
 Theorem 10.11 – Area of a Circle
 The area of a circle is the product of Π (pi) and the square of the
radius.
r
O
A = Π𝑟 2
 Sector of a Circle  a region bounded by an arc of the circle
and the two radii to the arc’s endpoints. A sector is named
using one arc endpoint, the center of the circle, and the other
arc endpoint.
 Theorem 10-12 – Area of a Sector of a Circle
 The area of a sector of a circle is the product of the ratio
𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑟𝑐
and the area of the circle.
360
A
Area Sector AOB =
O
r
B
𝑚𝐴𝐵
360
· Π𝑟 2
 Segment of a circle  a part of a circle bounded by an arc and
the segment joining its endpoints. To find the area of a
segment, draw radii to form a sector. The area of the segment
equals the area of the sector minus the area of the triangle
formed.
T
 Ex:
You’re hungry one day and decide to go to Warehouse
Pizza to get some food. When you arrive, you check the menu
and notice they are having deals on 14-in and 12-in pizzas. If a 14in pizza costs $20.00 and a 12-in pizza costs $16.00, which price
gives you the most pizza for your dollar?
 Ex: Find the Area of a Segment of a Circle
A
T
Find the area of the circle
segment if the radius is 10-in and
central angle ATB forms a right
angle.
B
Section 10.8 – Geometric Probability
 Objectives:
To use segment and area models to find the
probabilities of events.
 You may remember that the probability of an event is the
ratio of the number of favorable outcomes to the number of
possible outcomes
P(event) =
𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
 Geometric Probability  a model in which points represent
outcomes. We find probabilities by comparing measurements
of sets of points.
𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑠𝑒𝑔𝑚𝑒𝑛𝑡
P(event) =
𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑒𝑛𝑡𝑖𝑟𝑒 𝑠𝑒𝑔𝑚𝑒𝑛𝑡
Homework #24
Due Monday (March 25)
Page 577 – 578
 # 1 – 27 odd
Chapter 10 Test Tuesday/Wednesday
(Notebooks also due)