Unit 11 Advanced Triangles
•This unit finishes the analysis of
triangles with Triangle Similarity (AA,
SAS, SSS).
•This unit also addressed Geometric
Means, and triangle angle bisectors,
and the side-splitter theorem.
•This unit also contains the complete
set of instructions addressing Right
Triangle Trigonometry (SOHCAHTOA).
(Different set of slides)
Standards
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SPI’s taught in Unit 9:
SPI 3108.1.1 Give precise mathematical descriptions or definitions of geometric shapes in the plane and space.
SPI 3108.4.7 Compute the area and/or perimeter of triangles, quadrilaterals and other polygons when one or more
additional steps are required (e.g. find missing dimensions given area or perimeter of the figure, using trigonometry).
SPI 3108.4.9 Use right triangle trigonometry and cross-sections to solve problems involving surface areas and/or
volumes of solids.
SPI 3108.4.15 Determine and use the appropriate trigonometric ratio for a right triangle to solve a contextual problem.
CLE (Course Level Expectations) found in Unit 9:
CLE 3108.1.4 Move flexibly between multiple representations (contextual, physical written, verbal, iconic/pictorial,
graphical, tabular, and symbolic), to solve problems, to model mathematical ideas, and to communicate solution
strategies.
CLE 3108.1.5 Recognize and use mathematical ideas and processes that arise in different settings, with an emphasis
on formulating a problem in mathematical terms, interpreting the solutions, mathematical ideas, and communication of
solution strategies.
CLE 3108.1.7 Use technologies appropriately to develop understanding of abstract mathematical ideas, to facilitate
problem solving, and to produce accurate and reliable models.
CLE3108.2.3 Establish an ability to estimate, select appropriate units, evaluate accuracy of calculations and
approximate error in measurement in geometric settings.
CLE 3108.4.8 Establish processes for determining congruence and similarity of figures, especially as related to scale
factor, contextual applications, and transformations.
CLE 3108.4.10 Develop the tools of right triangle trigonometry in the contextual applications, including the Pythagorean
Theorem, Law of Sines and Law of Cosines
Standards
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CFU (Checks for Understanding) applied to Unit 9:
3108.1.5 Use technology, hands-on activities, and manipulatives to develop the language and the concepts of
geometry, including specialized vocabulary (e.g. graphing calculators, interactive geometry software such as
Geometer’s Sketchpad and Cabri, algebra tiles, pattern blocks, tessellation tiles, MIRAs, mirrors, spinners,
geoboards, conic section models, volume demonstration kits, Polyhedrons, measurement tools, compasses,
PentaBlocks, pentominoes, cubes, tangrams).
3108.1.7 Recognize the capabilities and the limitations of calculators and computers in solving problems.
.. 3108.1.8 Understand how the similarity of right triangles allows the trigonometric functions sine, cosine, and tangent
to be defined as ratio of sides.
3108.4.11 Use the triangle inequality theorems (e.g., Exterior Angle Inequality Theorem, Hinge Theorem, SSS
Inequality Theorem, Triangle Inequality Theorem) to solve problems.
3108.4.27 Use right triangle trigonometry to find the area and perimeter of quadrilaterals (e.g. square, rectangle,
rhombus, parallelogram, trapezoid, and kite).
3108.4.36 Use several methods, including AA, SSS, and SAS, to prove that two triangles are similar.
3108.4.37 Identify similar figures and use ratios and proportions to solve mathematical and real-world problems (e.g.,
Golden Ratio).
3108.4.42 Use geometric mean to solve problems involving relationships that exist when the altitude is drawn to the
hypotenuse of a right triangle.
3108.4.47 Find the sine, cosine and tangent ratios of an acute angle of a right triangle given the side lengths.
3108.4.48 Define, illustrate, and apply angles of elevation and angles of depression in real-world situations.
3108.4.49 Use the Law of Sines (excluding the ambiguous case) and the Law of Cosines to find missing side lengths
and/or angle measures in non-right triangles.
Introduction
• Previously, we learned how to prove
triangles congruent
• Now we will look at how to prove
triangles are similar
• As a reminder, by definition triangles
are similar if they have congruent
angles, and sides which have a uniform
similarity ratio
Angle -Angle Similarity (AA~)
• Postulate: If two angles of one triangle
are congruent to two angles of another
triangle, then the triangles are similar
A
D
B
C
E
Triangle ABC ~ triangle DEF
F
Example
• Are the triangles similar?
B
A
E
45
45
C
D
Yes. Vertical angles are congruent, and angle c and angle
b are congruent (both are 45 degrees). Therefore triangle
AEC is congruent to triangle DEB
Can we write a similarity statement?
No. we don’t know the lengths of any sides to get a ratio
Side Angle Side Similarity
(SAS ~)
• If an angle of one triangle is congruent
to an angle of a second triangle, and the
sides which are connected to each
angle are proportional, then the
triangles are similar
• This makes sense if you look at it. The
“pivot” angle dictates the opposite side,
and if the two sides are in ratio, then the
opposite side would be as well
Example
D
A
6
B
9
6
4
C
E
F
Triangle abc is similar to triangle def because angle a is congruent
to angle d, and side ab is proportional to side de, and side ac is
proportional to side df (the same proportion)
We can also conclude that side bc is proportional to side ef
What is the similarity ratio? 2/3
If side EF is 8 long, how long is side BC?
It would be 2/3 of 8 or 5.33
Side-Side-Side Similarity
(SSS ~)
• If the corresponding sides of two triangles are
proportional, then the triangles are similar
• Again, this makes sense. If all sides are in
proportion, then the angles will necessarily be
equal, which is the definition of similar
• No matter how you try to put the second
triangle together, it will only fit one way, and
that way will produce a similar triangle
Find the value of x
12
6
x
8
There are a couple of ways to do this
1) 6/8 = x/12; therefore x = 9
2) 6/x = 8/12; therefore x = 9
Assignment
• Page 455 7-12
• Page 457 24-26
Similarity in Right Triangles
What if you took a right triangle, and drew an altitude
from the right angle to the opposite side?
2
7 6
1
4 5
3
You would end up with two new right triangles. The question now is: Are they similar?
Creating Similar Triangles
72
Since the triangles are similar,
we can also conclude that the
sides all have similar ratios –
which we can predict if we are
given enough information
1
6
•
If you look, you can see
that triangle 635 is the
same as triangle 132, and
triangle 174 is the same as
triangle 132 and triangle
635
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When we say “the same”
we mean similar, because
they have 2 matching
angles –a 900 angle, and
they share an angle
•
Therefore 1 altitude
created three similar
triangles
4 5
3
Right Triangles and Altitudes
• The altitude to the hypotenuse of a right
triangle divides the triangle into two
triangles that are similar to the original
triangle, and to each other
• A point to keep in mind: if you know the
measure of the original angles of the
first triangle, by default you will know
the measure of all the angles of all 3
triangles
Finding the Geometric Mean
• KNOW THIS
• Proportions in which means (averages) are
equal occur frequently in geometry
• For any 2 positive numbers -such as a and b,
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the geometric mean of a and b is
the positive number x such that
a/x = x/b
Therefore: x = √ab
An example of Geometric
Mean
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Find the geometric mean of 4 and 18
Write the proportion:
4/x = x/18
Cross multiply
X*X = 4*18 or X2 = 72
X = √72
It should be noted, that this is in fact
X = √a*b or X = √4*18, so if I can remember
how this works, I can jump to this step
• So, X = 6 √2
• Yep, more radicals….
Another Example of
Geometric Mean
• Find the geometric mean of 3 and 12
We can do it the long way 3/X = X/12
Or cut to the final step where X= √12*3
Therefore, x = 6
Geometric Means and
Altitudes: 1st Rule
• The length of the altitude to the hypotenuse
(same altitude as before) of a right triangle is
the geometric mean of the lengths of the
segments of the hypotenuse.
• AD/CD = CD/DB
• If you look at the two triangles, this is the ratio
of the long side to the short side of triangle
c
ACD = the ratio of the long side to the short
side of triangle CBD
a
d
b
Example of Geometric Mean
• Remember a/x = x/b
• Here we’ll say “a” is the length of segment ab, “b”is the length of
segment cd and remember the altitude is “x” in this ratio
• So ab/eb = eb/cd
• So if segment ab = 5 and segment cd = 4, and the altitude = x
we could state 5/x = x/4 or x2 = 20, or x = √20 or x = 2 √5
• Therefore the altitude = 2 √5
e
f
2 √5 X
a
5
bc
4 d
Hypotenuse and Geometric
Means: 2nd and 3rd Rule
• The altitude to the hypotenuse of a right
triangle separates the hypotenuse so that the
length of each leg of the triangle is the
geometric mean of the length of the adjacent
hypotenuse segment and the length of the
hypotenuse.
c
• AD/AC = AC/AB
• DB/CB = CB/AB
a
d
b
Geometric Means: 1st Rule
Original Right Triangle
Altitude to hypotenuse
creates 2 new triangles
x
a
b
3rd right triangle
2nd right triangle
The original geometric means that we learned was a/x = x/b
Geometric Means: 1st Rule
Look at: A/X = X/B
x
a
x
a
x
b
What if you rotated the first
triangle to look like the
second one?
Now we can see the two ratios
clearly: A is to X, and X is to B!
Or: Long is to Short as Long
is to Short
Geometric Means: 2nd Rule
Look at: a/x = x/(a + b)
x
Now lets rotate the first
triangle and compare ---
a
x
a
b
We see that we have a
ratio of long is to
hypotenuse, as long is to
hypotenuse
or a/x=x/(a + b)
Geometric Means: 3rd Rule
Look at: b/x
x
a
x
b
b
Again, we set up
the ratio b/x,
which is short
side”to
hypotenuse
= x/(a+b)
Let’s rotate the first triangle
Here, “short side to
hypotenuse”, is x to (a+b),
or x/(a+b)
So: short side to
hypotenuse is to short
side is to hypotenuse is
the same as:
b/x = x/(a+b)!
Conclusions
• 1st, remember we are comparing similar
triangles
• Previously, we had 3 out of 4 numbers, we
would create a ratio, and solve for x
• Now we have 2 out of four numbers, but we
can use geometric means, and still solve for
x, as long as we can find the proper
relationship
• Also, it seems clear that we often need the
length of the hypotenuse. We can find this by
using the Pythagorean theorem, or a2+b2=c2
Example
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Solve for X and Y
OK, we know that 12/Y = Y/4, therefore Y = √12*4
Therefore Y = √48, or √16*3 or 4 √3
Now we add this: 4/X = X/(12 + 4)
So, X2= 4*(12+4)
e
X = √64 or 8
f
X
Y
a
12
b c
4 d
Another Example
• At a golf course, Maria
drove her ball 192 yards
straight towards the cup
• Gabriel drove his ball 240
yards, but off to the left
• Find X and Y to determine
their remaining distance to
the cup
Y
X
240
192
So, 192/240 = 240/(192 + X)
So 2402= 192(192 + X) or 1922 + 192X. Therefore
192X = 2402-1922 or X = (2402-1922)/192. So X = 108
Now, X/Y = Y/(X + 192) or 108/Y = Y/(108 + 192)
So Y = √(108 + 192)/108. Therefore Y = 180
Another Example
• Marla walks from point C -the
parking lot- to the lake at point D.
From the parking lot (C ) to the
Information Center is 300 meters.
From The parking lot (C ) to the
canoe rental is 400 meters. How far
is Marla from the Information
Center?
A
300
400
•We know that AD/AC = AC/AB
How long is AB? Well, it’s a right triangle, so we
B
can use a Pythagorean triple. Therefore AB is 500m
•Now we have AD/300 = 300/500
•Therefore AD = 180 meters
•How far did Marla walk from the Parking Lot at C?
•180/CD = CD/(500-180) or CD = √180*320 = 240 meters
Assignment
• Page 468 17-22
• Geometric Means worksheet 1 and 2
Page 99 in workbook
• On problems 1, 2, and 3 you can use
the Py-theorem to solve for the
hypotenuse, or the missing side
• On problem 4 you can use the Pytheorem to solve for x and y, you don’t
even need geometric means
• Problem 5 and 6 are ratios, set up the
way we just discussed
Page 99 continued
•
To solve problems 7 and 8 you need to
draw a picture first:
• You are asked to solve
24
10
x
for x, but first you need
to find a and b, so that
a
b
you can get the ratio of
26
a/x = x/b
• Problem 8 is the same with different
numbers
• Problem 9 wants to see this set up as well,
just with letters instead of numbers
Page 100
• The first 6 problems are straight
geometric means where x = √a·b
• Just solve as a decimal, it doesn’t say to
solve in reduced radical form
• Problems 7-12 are expecting you to
look for those relationships between
long side, short side, and hypotenuse
between the 3 right triangles.
Page 100 continued
• Problems 13-22 expect you to find the
relationships between the right
triangles, and set up a solution
• Some of them, such as #18, expect you
to use the Py-theorem to find the length
of a missing side before you solve for
the variable
Proportions in Triangles
As a review, solve for x
30
X
84/x = 45/15
So x = 28
15
7.5
X/7.5 = 9/6
So x = 11.25
3
6
Side Splitter Theorem
• If a line is parallel to one side of a
triangle and intersects the other two
sides, then it divides the sides
proportionally
Side Splitter Examples
• Solve for X
X
16
Since the lines are parallel, the sides
are proportional. Therefore x/16 =
5/10, or x = 8
5
10
5/2.5 = (x+ 1.5)/x
Therefore x = 1.5
Note: The difference
X+
in this as compared to
what we did before is that we
used to find the total length of the
side. Now we just use a straight
ratio
1.5
X
5
2.5
Corollary to the Side Splitter
Theorem
• If 3 parallel lines intersect 2
transversals, then the segments
intercepted in the transversals are
proportional
• A/B = C/D
C
A
B
D
Example with a Boat :)
•The panels in the sail are
sewn in a parallel pattern
•Solve for X, and solve for Y
•2/x = 1.7/1/7 therefore x = 2
•3/2 = y/1.7 therefore y = 2.55
2
x
3
2
1.7
1.7
y
1.7
Solve for X and Y
•To solve for x:
•X/30 = 15/26
•Therefore x = 17.3
•To solve for y:
•Y/16.5 = 26/15
•Therefore y = 28.6
16.5
15
x
y
26
30
Triangle-Angle-Bisector
Theorem
• If a ray bisects an angle of a triangle,
then it divides the opposite side into two
segments that are proportional to the
other two sides of the triangle
Example
6
5
x
Remember, this angle
is bisected equally, but
it splits the opposite
side proportionallyThe opposite side is
not split exactly in
half
X/6 = 8/5
X = 9.6
8
Or:
X/8 = 6/5
X = 9.6
Example
• Find the value of Y
Y/3.6 = 8/5
Therefore Y = 5.76
3.6
Y
5
8
Or:
Y/8 = 3.6/5
Therefore Y = 5.76
Assignment
• Page 475 9-22
• Page 476 25-35
• Worksheet 8-5