OVERVIEW
Using Similarity & Congruence to Solve & Prove
G.SRT.5
G.SRT.5
Use congruence and similarity criteria
for triangles to solve problems and to
prove relationships in geometric
figures.
This is quite vague... This is referring
to the geometric mean relationship
and to special right triangles. These
two relationships are founded in
similarity of right triangles and lead
nicely into trigonometry. There may
be other items that fit in this category
but focus on these two for now.
(1) The student will be able to derive
the three geometric mean
relationships.
(2) The student will be able to use the
geometric mean to solve for sides of
triangles.
(3) The student will use exact values in
solve for sides and angles in the
special right triangles.
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Of all the objectives in this unit, this
one is the most open ended. This
could represent a number of different
concepts but it seems to be referring
to geometric mean and special right
triangles. These are the normal bridge
concepts from similarity to
trigonometry and they use
congruence and similarity. Neither
one is specifically mentioned by name
anywhere in the geometry objectives
so this would seem to be the natural
place for them.
The geometric mean is a natural
extension of similar triangles and
creates a relationship that solves for
sides and can be used to prove the
Pythagorean Theorem.
Special right triangles link fixed ratios
to triangle sides and those ratios
bridge to the trigonometric ratios.
The concept of exact answers will be
established.
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Geometric mean and special right
triangles are concepts that lead nicely
into trigonometry. The geometric
mean relationships helps to prove the
Pythagorean Theorem and special
right triangles help establish
important values for trigonometry.
1 – The geometric mean has always
been a difficult one to demonstrate to
students. Approach it from similar
triangles by having student separate
the triangles and match up
corresponding sides.
2 – Difficulty with special right
triangles comes from multiplying and
dividing radical values, review this
process before you begin and things
will go much better.
2/8/2016
ASSESSMENT
Using Similarity & Congruence to Solve & Prove
G.SRT.5
CONCEPT 1 – Use congruence and similarity criteria for triangles to solve problems and to
prove relationships in geometric figures – Geometric Mean.
Given any right triangle, ABC.
B
Drop the altitude (the height) from the right angle (B) to the opposite side
AC . Let the altitude be called BD .
o
x
C
A
This forms three triangles.
Inner Left
ADB
Inner Right
BDC
B
o
x
A
Outer Whole
D
C
ABC
B
Look carefully at ABC notice that mB = 90, mA = o and mC = x, in
other words o + x + 90 = 180.
x
Knowing that the 3rd ’s of a must be  some of the missing angle values
in the diagram can be determined, mABD = x(180 - 90 - o) and that
mCBD = o(180 - 90 - x).
o
x
A
D
o
x
A
GEOMETRIC MEAN #1 -- Using the fact that ADB  ABC, set up
the proportion:
D
B
C
B
x
AB 2  AD( AC )
GEOMETRIC MEAN #2 -- Using the fact
that BDC  ABC, set up the
proportion:
right 
DC BC


whole BC
AC
C
B
Notice that all three triangles are similar to each other by AA. It is this
unique relationship that will create three different geometric mean
relationships.
left 
AD
AB


whole
AB
AC
o
o
A Left ∆D
o
x
A
B
C
Whole ∆
B
o
x
D
Right ∆
o
C
A
x
Whole ∆
C
BC 2  DC ( AC )
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ASSESSMENT
Using Similarity & Congruence to Solve & Prove
GEOMETRIC MEAN #3 -- Using the fact that ADB  BDC, we can
set up the proportion:
right  AD BD


left 
BD DC
BD2  AD( DC )
B
B
o
x
o
A Left ∆ D
G.SRT.5
x
Right ∆
D
The three geometric means are:
C
B
AB 2  AD( AC ) BC 2  DC ( AC ) BD2  AD( DC )
o
AB  AD( AC )
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BC  DC ( AC )
BD  AD( DC )
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A
x
D
C
2/8/2016
ASSESSMENT
Using Similarity & Congruence to Solve & Prove
G.SRT.5
CONCEPT 2 – Use congruence and similarity criteria for triangles to solve problems and to
prove relationships in geometric figures – Special Right Triangles.
There are two triangles that receive ‘special’ attention because of the specific geometric forms. These two triangles
both come from regular polygons. The side ratios and angle measures found in these ‘special’ triangles get used all
throughout mathematics and form the basis for ‘exact’ values for angles.
THE 45° - 45° - 90° TRIANGLE (RIGHT ISOSCELES)
The right isosceles is easily formed by drawing one of the diagonals
of a square. By the properties discovered in G.CO.11 it is known that
the diagonals of a square (or a Rhombus) will be angle bisectors.
Thus, the diagonal will divide the 90 angle into two 45angles.
45°
45°
Because the Right Isosceles is formed from the square,
45°
45°
two of the sides must be congruent and the base angles must be equal (45each). Look at the side relationships.
The legs of the right isosceles triangle will always be congruent to each
other because it is an isosceles triangle, and the hypotenuse will be
(leg ) ( 2)  hypotenuse or
leg 
x
x
2 times longer than either of the legs.
hypotenuse
2
x 2
THE 30° - 60° - 90° TRIANGLE
The 30- 60 - 90 triangle is formed when the perpendicular
bisector of a side of an equilateral triangle is drawn. The
perpendicular bisector divides that side into two equal parts while
also bisecting the opposite angle of 60 into two equal angles of 30.
60°
30° 30°
60°
Because all three sides were congruent to start with, the side that
was bisected is exactly half of the hypotenuse. This relationship is
quite easy to remember, so focus on the relationship these two sides
have with the other leg.
60°
30°
2x
60°
1
2
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60°
60°
30°
2x
60°
x
1
2
x
2/8/2016
ASSESSMENT
Using Similarity & Congruence to Solve & Prove
G.SRT.5
The relationships found in this ‘special’ triangle are:
 Short Leg is Half of the Hypotenuse

Long Leg is
3 times longer than the Short Leg
hypotenuse
short leg 
2
( short leg ) 2  hypotenuse or
30°
2x
30°
x 3 cm
Hypotenuse
60°
3  long leg
( short leg )
short leg 
or
long leg
3
Long
Leg
60°
x
Short Leg
Why does the size of these two triangles not alter these side relationships?
x
x
45°
30°
2x
x 3 cm
45°
60°
x 2
x
The answer is very simple…. SIMILARITY!
All 45 - 45 - 90 are similar (by AA) and so the sides will maintain these proportional relationships.
All 30 - 60 - 90 are similar (by AA) and so the sides will maintain these proportional relationships.
At the heart of special right triangles is similarity.
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ASSESSMENT
Using Similarity & Congruence to Solve & Prove
G.SRT.5
1. The geometric mean of 4 and 16 is:
A) 8
B) 10
C) 12
D) cannot be determined
2. If g is the geometric mean of a and b then which of the following is NOT true?
A)
a
b

g
g
B)
g
b

a
g
C) ab  g 2
D)
a
g

g
b
3. Which proportion is NOT true?
A)
AD AB

AB
AC
B)
B
BC CA

CD BC
C)
AD AB

DC BC
D)
AD BD

BD DC
o
A
4. The value for x is:
x
D
o
A) 7 2
B) 14
C) 14 2
C
14 2
D) 28
o
x
5. The value for x is:
30°
A) 12
B) 12 2
C) 12 3
D) 24
x
12
6. Which of these proportions represent a geometric mean?
a) G.M. or Not G.M.
4 6

6 9
b) G.M. or Not G.M.
2
8

8 32
c) G.M. or Not G.M.
8
10 2

25
10 2
d) G.M. or Not G.M.
1 10

5 50
7. Find the missing values. (If not a whole number, leave it in exact simplest radical or reduced fraction form)
a)
b)
x
z
y
12
4
x = ________ y = _________
z = ________
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c)
x
z
y
3
12
x = ________ y = _________
z = ________
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5
z
y
3
x
x = ________ y = _________
z = ________
2/8/2016
ASSESSMENT
Using Similarity & Congruence to Solve & Prove
G.SRT.5
8. Solve for the unknowns.
(E) = leave your answer in exact form (without rounding) - in some cases that will be a reduced radical.
a)
b)
c)
d)
o
x
12 3
60°
x
15
y
x
60°
x = __________ (E)
y = __________
y = __________
e)
f)
o
6 3
x = __________ (E)
x = __________ (E)
y = __________ (E)
6 2
h)
o
60°
x
y
o
60°
y
g)
x
15
x
o
y
x = __________ (E)
10
6
o
2 3
x
60°
y
x
x = __________ (E)
x = __________
x = __________
y = __________
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x = __________
y = __________ (E)
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ASSESSMENT
Using Similarity & Congruence to Solve & Prove
G.SRT.5
Answers:
1)
2)
3)
4)
5)
6)
A
A
C
B
C
a) G.M
b) G.M.
c) G.M.
d) Not G.M.
7) a) x = 4 6 , y = 4 2 , z = 4 3
b) x = 3 5 , y = 6, z = 6 5
1
2
c) x = 5 , y = 4, z = 6
3
3
8) a) x = 6 3 , y = 18
b) x = 15 3 , y = 30
c) x = 6 2
d) x = 5 3 , y = 10 3
e) x = 5 2
f) x = 6, y = 12
g) x = 12
h) x = 6, y = 4 3
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