Geometry
Unit 3A Day 7
2 Dec 2016
8 - 9:30
Geometry Agenda
Bulletin
Geometric Mean and right triangles
Similarity and Dilation Solving
Problems-(Homework) Corrections
Practice test - quick look over
Homework - Prepare for test on Tuesday
Types of Means or Averages
Average test scores.
If you got 95%, 59%, 77% and 89% on 4 tests in
geometry what would your average score be?
Types of Means or Averages
Average test scores.
If you got 95%, 59%, 77% and 89% on 4 tests in
geometry what would your mean score be?
95% +59% +77% + 89% = 322 /4 = 80.5 %
4
Sum of all the values = mean (arithmetic mean)
Number of values
Types of Means or Averages
Arithmetic Mean
For an arithmetic sequence there is a common
difference between the consecutive terms
2, 4, 6, 8, 10
2+2 = 4
4+2 = 6
6+2 =8 etc.
Types of Means or Averages
Arithmetic Mean
For an arithmetic sequence there is a common
difference between the consecutive terms
2, 4, 6, 8, 10
The arithmetic mean of 6 and 2 = 6 + 2 = 4
2
4 is the value that is the same distance from both
2 and 6.
Types of Means or Averages
Geometric Mean
For a geometric sequence there is a common
ratio between the consecutive terms
4, 40, 400, 4000,
4 x 10 = 40
40 x 10 = 400
400 x 10 = 4000 etc
Types of Means or Averages
Geometric Mean
For a geometric sequence there is a common
ratio between the consecutive terms
4, 40, 400, 4000,
OR we can show the common ratio between
consecutive terms as
4 = 40
40
400
40 is the geometric mean of the terms 4 and 400
Geometric Mean
What if we didn’t have the whole series?
Only given 4, 400 and asked to find their
geometric mean, let’s call that x
Find geometric mean
4, x?, 400
4 = 40
40
400
Now we can write
4 =
x
x
400
Find geometric mean
4, x?, 400
What if we didn’t have the whole series?
Only given 4, 400 and asked to find their
geometric mean, let’s call that x
We can write
4 =
x
x
400
Then
4(400) = x2
Find geometric mean
4, x?, 400
What if we didn’t have the whole series?
Only given 4, 400 and asked to find their
geometric mean, let’s call that x
We can write
4 =
x
x
400
Then
4(400) = x2
1600 = x2 √1600 = √x2
Find geometric mean
4, x?, 400
What if we didn’t have the whole series?
Only given 4, 400 and asked to find their
geometric mean, let’s call that x
We can write
4 =
x
x
400
Then
4(400) = x2
1600 = x2 √1600 = √x2
x = 40
Geometric mean, x, for any 2 values, a and b
a =
x
x
b
Then
ab = x2
And x = √ab
Geometric mean, x, for any 2 values, a and b
a =
x
x
b
Then
ab = x2
And x = √ab
Find the geometric mean of
a) 4 and 9
b) 5 and 20
c) 1 and 25
Geometric mean of a) 4 and 9
4 =
x
x
9
Then
Geometric mean of a) 4 and 9
4 =
x
x
9
Then
4(9) = x2
And x =
Geometric mean of a) 4 and 9
4 =
x
x
9
Then
4(9) = x2
And x = √36 = 6
Geometric mean of b) 4 and 20
5 =
x
x
20
Geometric mean of b) 4 and 20
5 =
x
x
20
Then
5(20) = x2
And x = √100 = 10
Geometric mean of c) 1 and 25
Geometric mean of c) 1 and 25
1
x
Then
=
x
25
1(25) = x2
And x = √25 = 5
For a and b the geometric mean, x,
ab = x2 is true.
Geometric Means and Right triangles
Label both triangles on your paper.
A
A
bb
C
C
C
c
a
a
B
B
Now label the 2 inner triangles on the bottom
triangle
A
A z
X
XX
bb xx
C
C
C
y
c
a
a
B
B
Cut out the two large triangles
ABC, they should fit on top of each other
exactly, they are congruent.
Now cut the bottom triangle along the “altitude”
- the line drawn from C to X, that is a right
angles to the hypotenuse.
A
A z X
XX
c
y
bb xx
C
C
C
a
a
B
B
Arrange the 3 triangles on the desk so
you can see they are similar triangles.
C
b
x
A
z
X
B
C
a
y
x
X
The 2 smaller ones are easy to spot
△AXC
△CXB by AA similarity postulate
What rigid transformation is needed
for the largest triangle?
A
b
C
b
c
c
a
a
B
Reflection across AC (rewrite the
labels on the other side of the paper).
B
A
c
c
a
a
bb
b
c
c
C
a
a
Reflection across AC (rewrite the
labels on the other side of the paper).
B
A
c
c
b
a
a
C
Arrange the 3 triangles on the desk so
you can see they are similar triangles.
C
b
x
A
z
X
B
Now we can see
△AXC
△CXB
by AA
C
a
y
x
cX
c
△ACB
B
a
a
A
b
Notice that
z
x
=
x
y
2
x = yz
z
x
x is the geometric
mean of the
measures of the 2
segments of the
B
hypotenuse
x
y
y
A
z
x
Altitudes of right triangles &
Geometric Mean
The measures of the altitude drawn from the
vertex of the right angle of a right triangle to its
hypotenuse is the geometric mean between the
measures of the two segments of the
hypotenuse.
Unit 3A Similarity and Dilation p. 39
1. Two triangles have a scale factor of 3:4. If
one side of the small triangle is 9in long, how
long is the same side of the larger triangle?
3(SF) = 4
SF = 4/3
4/3 (9) = 12 inches
4
3
9
?
Unit 3A Similarity and Dilation p. 39
2. Two triangles have a ratio of similarity of 3:8.
The perimeter of the small triangle is 54 in.
What is the perimeter of the larger triangle?
3
8
3(SF) = 8
SF = 8/3
8/3 (54in) = 144 in
Unit 3A Similarity and Dilation p. 39
3. Two right triangles are similar.
46
8
6
30
What is the length of the
hypotenuse of the small
triangle?
Unit 3A Similarity and Dilation p. 39
3. Two right triangles are similar.
9.2
46
8
6
30
What is the length of the
hypotenuse of the small
triangle?
6 x 5 = 30
The scale factor is 5
We need to find ⅕ of 46.
Unit 3A Similarity and Dilation p. 39
4. Square I has an area of 27 in2
x
x
What is the area of
square II?
⅓x
⅓x
Unit 3A Similarity and Dilation p. 39
4. Square I has an area of 27 in2
x
x
What is the area of square II?
⅓x
X2 = 27in2
⅓x x = √27 = √3(9) = 3√3 in
Area Sq. II = (⅓ x) (⅓ x) = (⅓
3√3in)(⅓ 3√3in) =
(√3in)(√3in) = 3 in2
Unit 3A Similarity and Dilation p. 39
5. The shadow of a boy 7 feet tall, is 12 feet
long. The shadow of a tree at the same time is
14 feet long. How tall is the tree?
Similar triangles
7
=x
12
14
x = 14(7) / 12 = 49 / 6 = 8 ⅙ feet
Unit 3A Similarity and Dilation p. 39
6. The sides of a triangle are 4, 10 and 5. Find
the length of the longest side of a similar
triangle whose shortest side is 12.
10
4
= 10
4 x 3 = 12
4
x
12
5
12
x
10 x 3 = 30
4x = 10(12)
x = 120 = 30
4
Unit 3A Similarity and Dilation p. 39
7. Triangle ABC is similar to triangle RST
AB = 8
18
8
BC = 18
=
S
B
40
RS
ST = 40.
18
8
40
?
Find RS
C
A
R
T
Unit 3A Similarity and Dilation p. 39
7. Triangle ABC is similar to triangle RST
AB = 8
18
8
BC = 18
=
S
B
40
RS
ST = 40.
18
8
40
?
Find RS
C
A
RS = 40(8)
R
T
18
RS = 17.8
Unit 3A Similarity and Dilation p. 39
8. The sides of a triangle are in the ratio 2:4:8
and the perimeter is 56cm.
Find the length of the sides.
2x + 4x + 8x = 56cm
14x = 56cm
x=4
Unit 3A Similarity and Dilation p. 39
8. The sides of a triangle are in the ratio 2:4:8
and the perimeter is 56cm.
Find the length of the sides.
2x + 4x + 8x = 56cm
14x = 56cm
x = 4cm
Sides are:
2(4cm) = 8cm
4(4cm) = 16cm
8(4cm) = 32cm
Unit 3A Similarity and Dilation p. 39
9. Determine whether these triangles are
similar. If they are, justify why.
Yes, by SSS
similarity criteria
24
Triangles are in
proportion 1:3
30
9
10
27
Small sides x 3 =
larger sides
8
10. Find x, AB and BC
2x + 4 = x + 3
7
4
4(2x + 4) = 7(x + 3)
8x + 16 = 7x + 21
8x - 7x + 16 - 16 = 7x - 7x + 21 -16
x=5
10. Find x, AB and BC
x=5
AB = 2x + 4 = 2(5) + 4 = 14
BC = x + 3 = 5 + 3 = 8
11.
17 = x +17
10
20
17
Or
17
= 10
17+x
20
10
x
20
Z
Or notice 10 x 2
= 20
Must be same as
17.
12. Are these triangles similar?
M
6
N
Q
5
O
10
12
P
6 x 2 = 12
And 5 x 2 = 10
MN proportional
to NP and QN is
proportional to
NO
12. Are these triangles similar?
M
6
N
Q
5
O
10
12
P
△MNQ
△PNO by SAS
similarity theorem
6 x 2 = 12
And 5 x 2 = 10
MN proportional
to NP and QN is
proportional to
NO
When lines
cross, vertical
angles are
congruent.
13. Find x, BC and AC
5 = x +4
1
2
C
D
10 = x + 4
5
6
x+4
E
1
B
x=6
2
A
BC = x + 4 = 10
AC = 10 +2 = 12
14. Find x given the the triangles are
similar.
6m
4m
x
=
2m
6m
6m (4m) = x
2m
12m = x
2m
4m
x
15. Find m (notation is confusing here)
m = 4km
4km
6km 2km
m = 4km(6km) 2km 4km
2km
m = 12km
2km
6km
4km
m
15. Find m (notation is confusing here)
m = 4km
4km
6km 2km
m = 4km(6km) 2km 4km
2km
m = 12km
2km
6km
4km
m
16. SW = 9, ST = 18, SV = 20, TU = 10
9
x
2
=
18
S
18
9
W
V
U
T
16. SW = 9, ST = 18, SV = 20, TU = 10
9
x
2
=
18
S
18
9
W
V
9
T
U
16. SW = 9, ST = 18, SV = 20, TU = 10
20
x
2
=
40
S
20
9
W
40
V
20
9
T
U
16. SW = 9, ST = 18, SV = 20, TU = 10
WV
x
2
=
10
S
a) SU = 40
20
b) WV = 5
9
c)
WT
=
9
5
V
W
d) WU = 20
20
9
T
U
10
17. A student claims BE and CD are
parallel by look alone.
Let us find evidence to refute his claim.
A
12
6
B
E
9
3
C
D
10
If BE and CD are parallel, then the
sides would be divided proportionally.
But in this diagram
6 = 12
A
3
9
12
6
B
E
9
3
C
Since the sides of the
triangle are not
divided proportionally
by BE then it cannot
be parallel to CD.
D
10
18. Two farmers each have triangular
plots of land.
a) Are the plots similar?
6180
52°
58°
x
52°
7436
70°
8240
18. Two farmers each have triangular
plots of land.
a) Are the plots similar? Yes, by AA
similarity.
6180
52°
x
58° 70°
52°
8240
70°
18. Two farmers each have triangular
plots of land.
b) What is approx. value of x?
6180
52°
x
58° 70°
7436
52°
70°
Careful here - side x
corresponds to the
7436 ft side of other
field
8240
18. Two farmers each have triangular
plots of land.
b) What is approx. value of x?
6180
52°
x
58° 70°
7436
52°
70°
6180 = x
8240
7436
x = (6180)(7436) / 8240
8240
x approx. 5577ft
Homework
Complete corrections using class slides for the
homework from last time.
Finish Solving Problems worksheet p. 39 - 40 i.e.
questions 11 - 18
Complete 2 proofs out of the 6 on pg 41
Change of plans: We will work on geometric
means and review for 3A test on Friday and do
the Unit 3A test on Tuesday.
Tear out and turn in your 2 proofs
On page 41.
Practice test - let’s look over without writing
anything on the test.
Unit 3A test. Similar triangles and
Dilations
1 & 2 Angle Angle similarity
Corresponding sides are proportional
AB: DE
AC:DF
Corresponding angles are congruent.
Unit 3A test. Similar triangles and
Dilations
3 & 4 Angle Angle similarity
But Corresponding sides are not easy to see
with this diagram - redraw as 2 separate
triangles and look carefully to find which sides
correspond to each other..
Unit 3A test. Similar triangles and
Dilations
5 and 6 ) Triangle proportionality theorem: If a
line parallel to one side of a triangle intersects
the other two sides, then if divides those sides
proportionally.
Unit 3A test. Similar triangles and
Dilations
7 & 8) For parallel lines cut by a transversal,
alternate interior angles are congruent.
When lines cross, vertical angles are congruent.
AA similarity
So corresponding sides are proportional - check
you’ve got the sides matched up correctly.
Unit 3A test. Similar triangles and
Dilations
9 & 10
Pythagorean theorem
Draw the two triangles separately
Actually, the altitude (12) is the geometric mean
of 16 and y. Find y first then use Pythagorean
since you’ll know hypotenuse (16+y) and one leg
of the large right triangle.
Unit 3A test. Similar triangles and
Dilations
11. Think about it - try to come up with an
isosceles triangle with one 60 degree angle
where the triangles wouldn’t be similar.
12& 13). For dilations, numbers greater than 1
give enlargements and numbers less then 1
(fractions) are reductions. Negative signs - move
shape as well.
Unit 3A test. Similar triangles and
Dilations
14 and 15) Just make sure you have
corresponding vertices in the same order for each
triangle and you know how to draw a similarity
sign. Make sure the number you multiply the
sides by maps the large triangle onto the smaller
one in this case.
16. Dilations - Tell me now...
Homework - Study for 3A test
Find similar problems you’ve already done and
write them out neatly.
Then try doing the practice without looking at any
notes.
I will post answer key, Sun night in google
classroom.
Homework - Study for 3A test
CSF tutoring is on Mondays.
I can do enrichment, but have Science Bowl club
meeting as well.