unit 3: geometry lesson #6: triangles & pythagorean theorem
TRANSCRIPT
Unit 3: GeometryLesson #6: Triangles & Pythagorean Theorem
LEARNING GOALS
• Find missing sides in right triangles• To determine if a triangle is right or not• To explain Pythagorean theorem• To explain how to and when to use the Pythagorean theorem
3
Classifying Triangles by Sides
Equilateral:
Scalene: A triangle in which all 3 sides are different lengths.
Isosceles: A triangle in which at least 2 sides are equal.
A triangle in which all 3 sides are equal.
AB
= 3
.02
cm
AC
= 3.15 cm
BC = 3.55 cm
A
B CAB =
3.47
cmAC = 3.47 cm
BC = 5.16 cmBC
A
HI = 3.70 cm
G
H I
GH = 3.70 cm
GI = 3.70 cm
4
Classifying Triangles by Angles
Acute:
Obtuse:
A triangle in which all 3 angles are less than 90˚.
A triangle in which one and only one angle is greater than 90˚& less than 180˚
108°
44°
28°B
C
A
57 ° 47°
76°
G
H I
5
Classifying Triangles by Angles
Right:
Equiangular:
A triangle in which one and only one angle is 90˚
A triangle in which all 3 angles are the same measure.
34
56
90B C
A
60
6060C
B
A
6
polygons
Classification by Sides with Flow Charts & Venn Diagrams
triangles
Scalene
Equilateral
Isosceles
Triangle
Polygon
scalene
isosceles
equilateral
7
polygons
Classification by Angles with Flow Charts & Venn Diagrams
triangles
Right
Equiangular
Acute
Triangle
Polygon
right
acute
equiangular
Obtuse
obtuse
Pythagoras
• Lived in southern Italy during the sixth century B.C.
• Considered the first true mathematician
• Used mathematics as a means to understand the natural world
• First to teach that the earth was a sphere that revolves around the sun
Right Triangles• Longest side is the
hypotenuse, side c (opposite the 90o angle)
• The other two sides are the legs, sides a and b
• Pythagoras developed a formula for finding the length of the sides of any right triangle
The Pythagorean Theorem
“For any right triangle, the sum of the areas of the two small squares is equal to the area of the larger.”
a2 + b2 = c2
It only works with right-angled triangles.
hypotenuse
The longest side, which is always opposite the right-angle, has a special name:
This is the name of Pythagoras’ most famous discovery.
Pythagoras’ Theorem
c
b
a
c²=a²+b²
Pythagoras’ Theorem
c
a
c
c
b
b
b a
a
c
ya
Pythagoras’ Theorem
c²=a²+b²
1m
8m
Using Pythagoras’ Theorem
What is the length of the slope?
1m
8m
c
b=
a=
c²=a²+ b²
c²=1²+ 8²
c²=1 + 64
c²=65
?
Using Pythagoras’ Theorem
How do we find c?
We need to use the
square root button on the calculator.It looks like this √
Press
c²=65
√ , Enter 65 =
So c= √65 = 8.1 m (1 d.p.)
Using Pythagoras’ Theorem
Example 1
c
12cm
9cm
a
bc²=a²+ b²
c²=12²+ 9²
c²=144 + 81
c²= 225
c = √225= 15cm
c
6m4m
s
ab
c²=a²+ b²
s²=4²+ 6²
s²=16 + 36
s²= 52
s = √52
=7.2m (1 d.p.)
Example 2
7m
5m
hc
a
b
c²=a²+ b²
7²=a²+ 5²
49=a² + 25?
Finding the shorter side
49 = a² + 25
We need to get a² on its own.Remember, change side, change sign!
Finding the shorter side
+ 25
49 - 25 = a²
a²= 24
a = √24 = 4.9 m (1 d.p.)
169 = w² + 36
c
w
6m
13m
a
b
c²= a²+ b²
13²= a²+ 6²
169 – 36 = a²
a = √133 = 11.5m (1 d.p.)
a²= 133
Example 1
169 = a² + 36
Change side, change sign!
Applications
• The Pythagorean theorem has far-reaching ramifications in other fields (such as the arts), as well as practical applications.
• The theorem is invaluable when computing distances between two points, such as in navigation and land surveying.
• Another important application is in the design of ramps. Ramp designs for handicap-accessible sites and for skateboard parks are very much in demand.
Baseball ProblemA baseball “diamond” is really a square.
You can use the Pythagorean theorem to find distances around a baseball diamond.
Baseball ProblemThe distance between
consecutive bases is 90
feet. How far does a
catcher have to throw
the ball from home
plate to second base?
Baseball Problem
To use the Pythagorean theorem to solve for x, find the right angle.
Which side is the hypotenuse?
Which sides are the legs?
Now use: a2 + b2 = c2
Baseball ProblemSolution• The hypotenuse is the
distance from home to second, or side x in the picture.
• The legs are from home to first and from first to second.
• Solution: x2 = 902 + 902 = 16,200 x = 127.28 ft
Ladder Problem A ladder leans against a
second-story window of a house. If the ladder is 25 meters long, and the base of the ladder is 7 meters from the house,
how high is the window?
Ladder ProblemSolution
• First draw a diagram that shows the sides of the right triangle.
• Label the sides: • Ladder is 25 m• Distance from house is 7
m
• Use a2 + b2 = c2 to solve for the missing side. Distance from house: 7 meters
Ladder ProblemSolution
72 + b2 = 252
49 + b2 = 625 b2 = 576 b = 24 m
How did you do?
A = 7 m
Success Criteria• I can identify a right-angle triangle• I can identify Pythagorean theorem• I can identify when to use Pythagorean theorem• I can use Pythagorean theorem to find the longest side• I can use Pythagorean theorem to find the shortest side• I can solve problems using Pythagorean theorem