Unit 11 Advanced Triangles

1
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)

2
Standards

• SPIs 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

3
Standards

• 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 Geometers
  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.

4
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

5
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
F
E
C
B
Triangle ABC triangle DEF
6
Example

• Are the triangles similar?

B
A
E
45
45
D
C
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 dont know the lengths of any sides to get
a ratio
7
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

8
Example
D
A
6
9
6
4
C
B
F
E
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
2/3
What is the similarity ratio?
If side EF is 8 long, how long is side BC?
It would be 2/3 of 8 or 5.33
9
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

10
Find the value of x
12
6
x
8

• There are a couple of ways to do this
• 6/8 x/12 therefore x 9
• 6/x 8/12 therefore x 9

11
Assignment

• Page 455 7-12
• Page 457 24-26

12
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
3
4
5
You would end up with two new right triangles.
The question now is Are they similar?
13
Creating Similar Triangles
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

• 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
• 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

14
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

15
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, the
  geometric mean of a and b is the positive number
  x such that a/x x/b
• Therefore x vab

16
An example of Geometric Mean

• Find the geometric mean of 4 and 18
• Write the proportion
• 4/x x/18
• Cross multiply
• XX 418 or X2 72
• X v72
• It should be noted, that this is in fact
  X vab or X v418, so if I can remember
  how this works, I can jump to this step
• So, X 6 v2
• Yep, more radicals.

17
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
v123 Therefore, x 6
18
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 ACD the ratio of the long side to the
  short side of triangle CBD

c
a
b
d
19
Example of Geometric Mean

• Remember a/x x/b
• Here well say a is the length of segment ab,
  bis 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 v20 or x 2 v5
• Therefore the altitude 2 v5

e
f
2 v5
X
4
a
d
b
c
5
20
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.
• AD/AC AC/AB
• DB/CB CB/AB

c
a
b
d
21
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
22
Geometric Means 1st Rule
Look at A/X X/B
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
23
Geometric Means 2nd Rule
Look at a/x x/(a b)
Now lets rotate the first triangle and compare
---
We see that we have a ratio of long is to
hypotenuse, as long is to hypotenuse or a/xx/(a
b)
24
Geometric Means 3rd Rule
Look at b/x x/(ab)
Lets rotate the first triangle
Here, short side to hypotenuse, is x to (ab),
or x/(ab)
Again, we set up the ratio b/x, which is short
sideto hypotenuse
x
So short side to hypotenuse is to short side is
to hypotenuse is the same as b/x x/(ab)!
b
25
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 a2b2c2

26
Example

• Solve for X and Y
• OK, we know that 12/Y Y/4, therefore Y v124
• Therefore Y v48, or v163 or 4 v3
• Now we add this 4/X X/(12 4)
• So, X2 4(124)
• X v64 or 8

e
f
X
Y
4
a
d
b
c
12
27
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
192
240
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 v(108
192)/108. Therefore Y 180
28
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, its a right triangle, so
  we
• 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 v180320 240
  meters

B
29
Assignment

• Page 468 17-22
• Geometric Means worksheet 1 and 2

30
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 Py-theorem to solve
  for x and y, you dont even need geometric means
• Problem 5 and 6 are ratios, set up the way we
  just discussed

31
Page 99 continued

• To solve problems 7 and 8 you need to draw a
  picture first
• You are asked to solve
• for x, but first you need
• to find a and b, so that
• you can get the ratio of
• 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

24
10
x
a b
26
32
Page 100

• The first 6 problems are straight geometric means
  where x vab
• Just solve as a decimal, it doesnt 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.

33
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

34
Proportions in Triangles
As a review, solve for x
84
30
X 15
X
7.5
84/x 45/15 So x 28
X/7.5 9/6 So x 11.25
3 6
35
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

36
Side Splitter Examples

• Solve for X

Since the lines are parallel, the sides are
proportional. Therefore x/16 5/10, or x 8
X 5
16 10
5/2.5 (x 1.5)/x Therefore x 1.5 Note The
difference 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
X 1.5 5
X 2.5
37
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
D
A
B
38
Example with a Boat )

• The panels in the sail are sewn in a parallel
  pattern
• Solve for X, and solve for Y

1.7
2
1.7
x
y

• 2/x 1.7/1/7 therefore x 2
• 3/2 y/1.7 therefore y 2.55

1.7
3
2
39
Solve for X and Y

• To solve for x
• X/30 15/26
• Therefore x 17.3

y
16.5

• To solve for y
• Y/16.5 26/15
• Therefore y 28.6

15 26
x
30
40
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

41
Example
Remember, this angle is bisected equally, but it
splits the opposite side proportionally-The
opposite side is not split exactly in half
6 x
X/6 8/5 X 9.6
5 8
Or X/8 6/5 X 9.6
42
Example

• Find the value of Y

Y/3.6 8/5 Therefore Y 5.76
3.6
Y
Or Y/8 3.6/5 Therefore Y 5.76
5
8
43
Assignment

• Page 475 9-22
• Page 476 25-35
• Worksheet 8-5