146
Areas of Parallelograms and Triangles 7-1
Areas of Parallelograms and
Triangles
7-1
Parallelogram
A parallelogram is a quadrilateral where the opposite sides are congruent and parallel.
A rectangle is a type of parallelogram, but we often see parallelograms that are not rectangles (parallelograms without right angles).
Area of a Parallelogram
Any side of a parallelogram can be
considered a base. The height of a
parallelogram is the perpendicular distance
between opposite bases.
The area formula is A=bh
A=bh A=5(3) A=15m
2
Area of a Triangle
A triangle is a three sided polygon. Any
side can be the base of the triangle. The
height of the triangle is the perpendicular
length from a vertex to the opposite base.
A triangle (which can be formed by splitting
a parallelogram in half) has a similar area
formula: A = ½ bh.
Example
A= ½ bh A= ½ (30)(10) A= ½ (300) A= 150 km
2
Complex Figures
Use the appropriate formula to find the area
of each piece.
Add the areas together for the total area.
Example
| 27 cm |
10 cm
24 cm
Split the shape into a rectangle and triangle.
The rectangle is 24cm long and 10 cm wide.
The triangle has a base of 3 cm and a height of 10
cm.
Solution
Rectangle
A = lw A = 24(10) A = 240 cm
2
Triangle A = ½ bh A = ½ (3)(10) A = ½ (30) A = 15 cm
2
Total Figure A = A1 + A2
A = 240 + 15 = 255 cm2
Practice!
Pg. 353-355 # 1-14 all
# 29, 36-43 all
7-2 The Pythagorean Theorem
and Its Converse
Parts of a Right Triangle
In a right triangle, the side opposite the right
angle is called the hypotenuse.
It is the longest side.
The other two sides are called the legs.
The Pythagorean Theorem
Pythagorean Triples
A Pythagorean triple is a set of nonzero
whole numbers that satisfy the Pythagorean
Theorem.
Some common Pythagorean triples include:
3, 4, 5 5, 12, 13 8, 15, 17 7, 24, 25
If you multiply each number in the triple by the
same whole number, the result is another
Pythagorean triple!
Finding the Length of the
Hypotenuse
What is the length of the hypotenuse of
ABC? Do the sides form a Pythagorean
triple?
The legs of a right triangle have
lengths 10 and 24. What is the length
of the hypotenuse? Do the sides form
a Pythagorean triple?
Finding the Length of a Leg
What is the value of x? Express your
answer in simplest radical form.
The hypotenuse of a right triangle
has length 12. One leg has length 6.
What is the length of the other leg?
Express your answer in simplest
radical form.
Triangle Classifications
Converse of the Pythagorean Theorem: If the square of the
length of the longest side of a triangle is equal to the sum of
the squares of the lengths of the other two sides, then the
triangle is a right triangle.
If c2 = a2 + b2, than ABC is a right triangle.
Theorem 8-3: If the square of the length of the longest side of a
triangle is great than the sum of the squares of the lengths of
the other two sides, then the triangle is obtuse.
If c2 > a2 + b2, than ABC is obtuse.
Theorem 8-4: If the square of the length of the longest side of a
triangle is less than the sum of the squares of the lengths of the
other two sides, then the triangle is acute.
If c2 < a2 + b2, than ABC is acute.
Classifying a Triangle
Classify the following triangles as acute,
obtuse, or right.
85, 84, 13
6, 11, 14
7, 8, 9
Practice!!
Pg. 360-363 #1-35 odd
#36-38 all, 45
# 66 5 extra credit points!
Two Special Right Triangles
45°- 45°- 90°
30°- 60°- 90°
7-3
45°- 45°- 90°
The 45-45-90
triangle is based
on the square
with sides of 1
unit.
1
1
1
1
45°- 45°- 90°
If we draw the
diagonals we
form two
45-45-90
triangles.
1
1
1
1
45°
45°
45°
45°
45°- 45°- 90°
Using the
Pythagorean
Theorem we can
find the length of
the diagonal.
1
1
1
1
45°
45°
45°
45°
45°- 45°- 90°
12 + 12 = c2
1 + 1 = c2
2 = c2
2 = c
1
1
1
1
45°
45°
45°
45°
2
45°- 45°- 90°
Conclusion:
the ratio of the
sides in a 45-
45-90 triangle
is
1-1-2
1
1 2
45°
45°
45°- 45°- 90° Practice
4
4 2
SAME leg*2
4
45°
45°
45°- 45°- 90° Practice
9
9 2
SAME leg*2
9
45°
45°
45°- 45°- 90° Practice
2
2 2
SAME leg*2
2
45°
45°
45°- 45°- 90° Practice
14
SAME leg*2
7
7
45°
45°
45°- 45°- 90° Practice
45°- 45°- 90° Practice
3 2
hypotenuse
2
45°
45°
45°- 45°- 90° Practice
3 2
2 = 3
45°- 45°- 90° Practice
3 2
hypotenuse
2
45°
45°
3 SAME
3
45°- 45°- 90° Practice
6 2
hypotenuse
2
45°
45°
45°- 45°- 90° Practice
6 2
2 = 6
45°- 45°- 90° Practice
6 2
hypotenuse
2
45°
45°
6 SAME
6
45°- 45°- 90° Practice
11 2
hypotenuse
2
45°
45°
45°- 45°- 90° Practice
11 2
2 = 11
45°- 45°- 90° Practice
112
hypotenuse
2
45°
45°
11 SAME
11
45°- 45°- 90° Practice
8
hypotenuse
2
45°
45°
45°- 45°- 90° Practice
8
2
2
2 * =
82
2 = 42
45°- 45°- 90° Practice
8
hypotenuse
2
45°
45°
42 SAME
42
45°- 45°- 90° Practice
4
hypotenuse
2
45°
45°
45°- 45°- 90° Practice
4
2
2
2 * =
42
2 = 22
45°- 45°- 90° Practice
4
hypotenuse
2
45°
45°
22 SAME
22
45°- 45°- 90° Practice
6
Hypotenuse
2
45°
45°
45°- 45°- 90° Practice
6
2
2
2 * =
62
2 = 32
45°- 45°- 90° Practice
6
hypotenuse
2
45°
45°
32 SAME
32
30°- 60°- 90°
The 30-60-90
triangle is based
on an equilateral
triangle with sides
of 2 units.
2 2
2
60° 60°
2 2
2
60° 60°
30°- 60°- 90°
The altitude (also
the angle bisector
and median) cuts the
triangle into two
congruent triangles. 1 1
30° 30°
30°
60°
This creates the
30-60-90
triangle with a
hypotenuse a
short leg and a
long leg.
30°- 60°- 90°
Short Leg
Long L
eg
60°
30°
30°- 60°- 90° Practice
1
2
We saw that the
hypotenuse is
twice the short leg.
We can use the
Pythagorean
Theorem to find
the long leg.
60°
30°
30°- 60°- 90° Practice
1
2 3
A2 + B2 = C2
A2 + 12 = 22
A2 + 1 = 4
A2 = 3
A = 3
30°- 60°- 90°
Conclusion:
the ratio of the
sides in a 30-
60-90 triangle
is
1- 2 - 3
60°
30°
3
1
2
60°
30°
30°- 60°- 90° Practice
4
8
Hypotenuse =
short leg * 2
43
The key is to find
the length of the
short side.
Long Leg =
short leg * 3
60°
30°
30°- 60°- 90° Practice
5
10 Hypotenuse =
short leg * 2 53
Long Leg =
short leg * 3
60°
30°
30°- 60°- 90° Practice
7
14 Hypotenuse =
short leg * 2 73
Long Leg =
short leg * 3
60°
30°
30°- 60°- 90° Practice
3
23 Hypotenuse =
short leg * 2 3
Long Leg =
short leg * 3
60°
30°
30°- 60°- 90° Practice
10
210 Hypotenuse =
short leg * 2
30
Long Leg =
short leg * 3
30°- 60°- 90° Practice
60°
30°
30°- 60°- 90° Practice
11
22 Short Leg =
Hypotenuse 2 113
Long Leg =
short leg * 3
60°
30°
30°- 60°- 90° Practice
2
4 Short Leg =
Hypotenuse 2 23
Long Leg =
short leg * 3
60°
30°
30°- 60°- 90° Practice
9
18 Short Leg =
Hypotenuse 2 93
Long Leg =
short leg * 3
60°
30°
30°- 60°- 90° Practice
15
30 Short Leg =
Hypotenuse 2 153
Long Leg =
short leg * 3
60°
30°
30°- 60°- 90° Practice
23
46 Hypotenuse =
Short Leg * 2 233
Short Leg =
Long leg 3
60°
30°
30°- 60°- 90° Practice
14
28 Hypotenuse =
Short Leg * 2 143
Short Leg =
Long leg 3
60°
30°
30°- 60°- 90° Practice
16
32 Hypotenuse =
Short Leg * 2 163
Short Leg =
Long leg 3
60°
30°
30°- 60°- 90° Practice
3 3
63 Hypotenuse =
Short Leg * 2 9
Short Leg =
Long leg 3
60°
30°
30°- 60°- 90° Practice
4 3
83 Hypotenuse =
Short Leg * 2 12
Short Leg =
Long leg 3
60°
30°
30°- 60°- 90° Practice
9 3
183 Hypotenuse =
Short Leg * 2 27
Short Leg =
Long leg 3
60°
30°
30°- 60°- 90° Practice
7 3
143 Hypotenuse =
Short Leg * 2 21
Short Leg =
Long leg 3
60°
30°
30°- 60°- 90° Practice
113
223 Hypotenuse =
Short Leg * 2 33
Short Leg =
Long leg 3
Practice!!
Pg. 369-370
# 1-30 all
#32
Areas of Trapezoids, Rhombuses,
and Kites
7-4
Trapezoids:
leg leg
b1 = base 1
b2 = base 2
h = height
Height – distance
between the 2 bases.
* Must be
A = ½ h(b1 + b2)
Area of
trapezoid
Height
base base
Find the area of the
following trapezoid.
20in
18in
36in
30in
A = ½ h(b1 + b2)
= ½ (18in)(36in + 20in)
= ½ (18in)(56in)
= 504in2
This is the height!!
Find the area of following
trapezoid.
60
5cm
7cm
A = ½ h(b1 + b2)
= ½ (3.5cm)(5cm + 7cm)
= ½ (3.5cm)(12cm)
= 20.8cm2
Need to find h first!
Short side = 2cm
h = 2√3
h = 3.5cm
This is a 30-60-90 Δ
h
Area of a Rhombus or a Kite
Rhombus
4 equal sides.
Diagonals bisect each
other.
Diagonals are .
Kite
Adjacent sides are .
No sides //.
Diagonals are .
A = ½ d1d2
Area of
Kites or
Rhombi
Diagonal One Diagonal Two
Find the Area of the
following Kite.
3m 3m
4m
5m
A = ½ d1d2
= ½ (6m)(9m)
=27m2
Find the area of the following
Rhombus
12m
12m
b
A = ½ d1d2
= ½ (24m)(18m)
= 216m2
d1 = 24m
d2 = 18m
a2 + b2 = c2
122 + b2 = 152
144 + b2 = 225
b2 = 81
b = 9
15m
15m
What have I learned??
Area of Trapezoid
A = ½ h(b1 + b2)
Area of Rhombus or Kite
A = ½ d1d2
Practice!!!
Pg. 376-377 #1-35 all
7-5 Area of Regular
Polygon
Apothem
Find the Area of an Equilateral Triangle
Area is ½ ab a is Altitude
b is Base
34a
88
8b
316
344
8342
1
A
A
A
Find the Area of an Equilateral Triangle
(there is an easier way)
Theorem
s is Side of triangle
34a
88
8b
234
1sisArea
316
3644
1
834
1 2
A
A
A
Find “s” of an equilateral triangle
with area of
s is Side of triangle
234
1sisArea
325
Find “s” of an equilateral triangle
with area of
s is Side of triangle
234
1sisArea
325
10
100
100
4
125
34
1325
2
2
2
s
s
s
s
s
Finding the Area of a Regular
Hexagon inscribed in a circle.
Parts of the inscribed hexagon
Center
nAngleCentral
360
sidetocenterfrom
Apothem
Finding the Area of a Regular
Hexagon inscribed in a circle.
Parts of the inscribed hexagon
606
360
AngleCentral
sidetocenterfrom
Apothem
30
60 60
60
Finding the Area of a Regular
Hexagon inscribed in a circle.
Parts of the inscribed hexagon
606
360
AngleCentral
sidetocenterfrom
Apothem
30
60 60
60
Perimeterisp
apothemisa
apArea2
1
Finding the Area of a Regular
Hexagon inscribed in a circle.
Perimeterisp
apothemisa
apArea2
1
10
1010
35
Finding the Area of a Regular
Hexagon inscribed in a circle.
Perimeterisp
apothemisa
apArea2
1
10
1010
35
60106
35
Perimeter
apothem
3150
3530
60352
1
A
A
A
Finding the Area of a Regular Octagon
inscribed in a circle.
Sides of 4, what the
Central Angle 4
Perimeterisp
apothemisa
apArea2
1
4
4
4
45
8
360AngleCentral
4
4
4
4
How do you find the Apothem
Sides of 4 a
5.67 5.67
5.22
22
a
25.22
83.4
5.22
2
25.22
Tana
aTan
Finding the Area of a Regular Octagon
inscribed in a circle.
Sides of 4, what the
Central Angle 4
Perimeterisp
apothemisa
apArea2
1
4
4 4
3248
83.4
Perimeter
apothem
45
8
360AngleCentral
4
4 4
2 2
Finding the Area of a Regular Octagon
inscribed in a circle.
Sides of 4
4
Perimeterisp
apothemisa
apArea2
1
4
4
3248
83.4
Perimeter
apothem
45
8
360AngleCentral
4
4 4
2 2
28.77
3283.42
1
A
A
Find the Area of a 12-gon
Sides of 1.2; Radius of 2.3
Apothem 3.2
6.022.1
a
22.2
6.03.2
6.03.2
3.26.0
22
222
222
a
a
a
a
4.14122.1 Perimeter
Find the Area of a 12-gon
Sides of 1.2; Radius of 2.3
Apothem
4.14
22.2
p
a
776.344.1483.42
1Area
Practice!!
Pg. 382-383
# 1-32 all
and
Objective: Find the measures
of central angles and arcs.
A CIRCLE is the set of all points equidistant from a given point called
the center.
This is circle P for Pacman.
Circle P
P
A CENTRAL ANGLE of a circle is an angle with its vertex at the
center of the circle.
Central angle
An arc is a part of a circle. In this case it is the part Pacman would
eat.
Arc
One type of arc, a semicircle, is half of a circle.
P
A
C Semicircle ABC
m ABC = 180
B
A minor arc is smaller than a semicircle. A major arc is greater
than a semicircle.
R
S
P
L
N
M
RS is a minor arc.
mRS = m RPS.
O
LMN is a major arc. mLMN
=
360 – mLN
A C
D
Identify the following in circle O:
1) the minor arcs
E O
A C
D
Identify the following in circle O:
2) the semicircles
E O
A C
D
Identify the following in circle O:
3) the major arcs containing point A
E O
The measure of a central angle is equal to its intercepted
arc.
53o
53o
O P
R
Find the measure of each arc.
1. BC = 32
2. BD = 90
3. ABC = 180
4. AB = 148
Here is a circle graph that shows how people
really spend their time. Find the measure of
each central angle in degrees.
1. Sleep
2. Food
3. Work
4. Must Do
5. Entertainment
6. Other
Practice!!!
Pg. 389-392 #1-14 all
# 27-41 odd and #59
*59 may be turned in for 3 extra credit points!!
7-7 Areas of Circles and Sectors
Quick Review
What is the circumference of a circle?
• What is the area of a circle?
• The interior angle sum of a circle is ?
• What is the arc length formula?
2r
r2
360o
mA) B
3602r
mA) B
360C
Sector of a Circle
Formula is very similar to arc length
Notation is slightly different! - The center pt is used when describing a sector. - The is not used for sectors.
How can we find area, based on
what we already know?
mA) B
360r2Area of a sector =
1. Find the area of the shaded sector 2. Find the arc length of the shaded sector.
Segment
any ideas of how to find
the shaded area?
Finding the Area of a Segment of a Circle
– =
Area of
Sector
Area of
Triangle
Area of
Segment minus equals
Find the area of segment RST to the nearest hundredth.
Use formula for area of sector.
Substitute 4 for r and 90 for m.
= 4π m2
Step 1 Find the area of sector RST.
Simplify.
Step 2 Find the area of ∆RST.
Continued
Find the area of segment RST to the nearest hundredth.
ST = 4 m, and RS = 4m.
= 8 m2
Step 3
area of segment = area of sector RST – area of ∆RST
4.57 m2
Find the area of segment RST to the nearest hundredth.
Continued
= 4π – 8
A segment of a circle
Find the area of segment LNM to the nearest hundredth.
Finding the Area of a Segment
Use formula for area of sector.
Substitute 9 for r and 120 for m.
= 27π cm2
Step 1 Find the area of sector LNM.
Find the area of segment LNM to the nearest hundredth.
Continued
Step 2 Find the area of ∆LNM. Draw altitude NO.
Find the area of segment LNM to the nearest hundredth.
Continued
Step 3
area of segment = area of sector LNM – area of ∆LNM
= 49.75 cm2
Find the area of the shaded portion
Find the area of the red portion Of the Tube Sign.
Find the shaded area with r = 2 , 4 , & 10
1cm 3cm
5cm
What is the Area of the Black part?
If you have 2 circles A and B that intersect at 2 points and the distance between the centers is 10. What is the area of the intersecting region?
A B
10
Practice!!
Page 397- 398 7 - 27 odd 31-33 36-38
7-8 Geometric Probability
Finding a Geometric Probability
A probability is a number from 0 to 1 that
represents the chance an event will occur.
Assuming that all outcomes are equally
likely, an event with a probability of 0
CANNOT occur. An event with a
probability of 1 is just as likely to occur as
not.
Finding Geometric probability
continued . . .
In an earlier course, you may have evaluated
probabilities by counting the number of favorable
outcomes and dividing that number by the total
number of possible outcomes. In this lesson, you
will use a related process in which the division
involves geometric measures such as length or area.
This process is called GEOMETRIC
PROBABILITY.
Geometric Probability—
probability and length
Let AB be a segment that contains the
segment CD. If a point K on AB is chosen
at random, then the probability that that it is
on CD is as follows:
P(Point K is on CD) = Length of CD
Length of AB A
B
C
D
Geometric Probability—
probability and AREA
Let J be a region that contains region M. If
a point K in J is chosen at random, then the
probability that it is in region M is as
follows:
P(Point K is in region M) = Area of M
Area of J J
M
Ex. 1: Finding a Geometric
Probability
Find the geometric probability that a point
chosen at random on RS is on TU.
Practice!!!
Pg 404-405 #1-31 ODD
