Introduction to Unit Lesson Plan
By Amber Fickes
For my unit lesson plan, I was assigned to the topic of similarities in a geometry class.
This topic includes: Ratios and Proportions, Similar Polygons, Proving Triangles Are Similar,
Similarity in Right Triangles, Proportions in Triangles, and Perimeters and Areas of Similar
Figures. I had the opportunity to gain access to the actual geometry textbook used in the local
high school: Prentice Hall Mathematics Geometry 2004 Textbook. I also had access to the
textbook’s website, where I took some questions for my quizzes. In my unit lesson plan I
included six lessons, where I took the approach of inquiry and scaffolding in each lesson. In
some of the lessons, when I was teaching about how to prove similar triangles, I had them do
some instructional activities so they would have some idea about the information before I even
taught anything. I actually found these activities in the textbook, which I found very interesting
because it showed me they really took the time to help teach the students. I did these teaching
methods so I could help get my students to participate in their own learning. They are the future
so they should care about how and what they learn. I put the standards for each lesson from the
list of NC Standards. I do use technology throughout my unit. I want to be able to use geometry
software so the students can look at proportions within triangles before we ever start the lesson.
This way they can have a hands-on-experience and create their own knowledge/conclusions. I
talk to my students in each lesson. I may ask them leading questions so we can dive into the
lesson that way. When I put in my lesson that we talk as a class about the answers I have them
explain to me what their answers are and why. I assess my students a variety of ways so I can
meet the needs of each child. I do problems in class with them so they will know how to do the
homework. We have quizzes which are all multiple choice because that’s how the end-of-course
exam will be. I have a project based on similarities so the students can be creative with their
knowledge. There are construct your own concepts in some lessons. Finally there is a test so I
can really see what students understand, and it is also multiple choice. If I have students with
special needs I will address them individually. If a student needs extra time with a test or quiz I
will give to him/her, and if they need to be in a room alone then that will done. If I have some
students who excel at math I will have harder problems for them to do if they request them. I
don’t want to force anything on the students. One of the things I put into my unit lesson plan that
I feel is really important is a review day. A teacher should be ahead enough to allow a review
day where students can ask questions. I made up a key concepts page and wrote down some
practice problems. I will do all that I am able to do to help my students achieve their goals: in
and outside the classroom.
Ratios and Proportions
1. Prior Knowledge: A student will need to know what a ratio is or the idea of a ratio. If the
student knows the ratio is then he should know how to simplify it. Though I will use
inquiry based questioning for the students to come up with their original definitions for
some new terminology.
2. Objectives: NC Standard 1.02 Use length, area, and volume of geometric figures to solve
problems. Using ratios and proportions.
3. Motivation: I would use motivation of real life problems to get the students interested in
ratios and how they relate to a whole lot to the real world. An example is finding
dimensions of a bedroom for an architect. I will pose a problem like that in the lesson.
This information can be helpful in solving areas of triangles and proofs of triangles.
4. Misconceptions: Some students may think that simply putting two different factions with
an equal sign makes it a proportion.
5. Instructional Activities:
a. First pose the question: What is a ratio? I want you to write your thoughts about a
ratio and we will discuss it as a class. Also come up with other names for the
word ratio that are mean the same (3 minutes).
b. Then Ill ask several students to read their “definitions” of ratios. Ill let one student
to talk at a time. Then that student will call on someone else and ask if he/she
agrees and if he/she has anything to add to it (5 minutes).
c. Ill then put up my already completed overhead sheet with my definition of a ratio:
A ratio is a comparison of 2 quantities or numbers. We can write this as a/b or as
a:b. In English this says “a to b” (3 minutes).
d. Ill ask them to find two numbers that are the same in fraction form such as one
that is reduced from the other. (3 minutes)
e. We will write these on the board. Ill ask them if are these entire equal and we will
make sure that they are. Then ill state that they have just proven the definition of a
proportion: a statement that two ratios are equal. They can be written in the form
of a/b=c/d or a:b=c:d. In English this says “a is to b as c is to d” (3 minutes).
f.
I will show them some properties of proportions and I want them to explain them
to me instead of the other way around. a/b=c/d is equivalent/equal to
i. ad=bc (This one is the cross-product: multiplying both sides of a/b=c/d by
bd results in this. I want them to come up with their own term.)
ii. b/a=d/c
iii. a/c=b/d
iv. a+b/b=c+d/d (10 minutes)
g. Ill ask them to think about what a scale is, and we will talk about what our
perceptions of it are and how it is used. Ill show them a definition of a scale,
which compares each length in a drawing to the actual length. It can be seen as
1in to 100 mi or 1in=12ft. We will use proportions to find real dimensions.
6. Leads to: This will lead to finding empty lengths for shapes when they are similar by
setting up proportions and solving for the variable.
7. Assessment: Ill will show them some examples now at the end of the class to practice so
they can be prepared for their homework.
i. Write two proportions to m/4=n/11. Ill ask the students to pair up and
come up with two proportions on their own and we will share with the
class. (6 minutes for making them up and sharing).
ii. Solve for the given variable. (y+3)/8=y/4. I would like them to do it first
on their own or in groups, and then we will go over it.
1. 4(y+3)=8y
2. 4y+12=8y
3. 12=4y
4. 12/4=y
5. 3=y
iii. Find the dimensions of the bedroom in the scale drawing on the board.
Use them to find the real dimensions of the room. We will all do this one
together as a class. Our scale is 1in=16ft
1. We will have a student come measure the drawing with a ruler and
announce to us what the dimensions are. They happen to be 7/8in
by 5/8in.
2. The scale can be written in fraction form of 1/16. Find the length
by setting up a proportion: 1/16= (7/8)/x.
3. x=16(7/8)
4. x=14
5. To find the width we set up the proportion 1/16= (5/8)/y.
6. y=16(5/8)
7. y=10
8. The dimensions are 14ft by 10 ft.
8. Homework: I will give them a worksheet of a few problems that I will come by and see if
they attempted the problems. We will then go over any questions that people have the
next day.
a. The base of the pyramid at the right is a square whose sides measure 0.555m. The
intent was for the sides to measure 555m. What is the ratio of the length of a base
side in the small pyramid to the length of a base side in the intended pyramid?
i. 0.555m/555m. This can be reduced to 0.001m
b. a/b=1/2 Find the missing pieces of these equivalences.
i. 2a=
ii. 2/1=
iii. (a+1)/1=
iv. a/1=
1. 2a=1b
2. 2/1=b/a
3. (a+1)/1=(b+4)/4
4. a/1=b/2
c. Solve the proportion
i. (x+3)/3=(10+4)/4
1. 4(x+3)=3(10+4)
2. 4x+12=30+12
3. 4x+12=42
4. 4x=42/12
5. 4x=
=7/3
6. x=7/3 *1/4
7. x=7/12
d. Theree will be a picture inserteed. The scalee for this mapp of Louisiaana is 1in=455mi.
On thee map, the distance
d
from
m Lake Charlles to Baton Rouge is abbout 3 1/8 in.
Aboutt how far apart are the tw
wo cities?
1. 1in=45 mi and from Lake Charles
C
to Baaton Rouge it’s about 3 1/8in.
2. 1in//45mi= (25/8in)/x. Solvee the proporrtion
3. (1125/8 in*mi) =xin.
4. (1125/8)mi=x.
5. So it is 1125/8 miles from Lake
L
Charles and Baton Rouge.
Similar Polygons
1. Prior Knowledge: You will need to be able to solve proportions. Know what
congruent means.
2. Objectives: 2.03 Apply proportions, definitions, and theorems of two-dimensional
figures to solve problems and write proofs: a) Triangles, b) Quadrilaterals, c)
Other polygons, d) Circles. Similar polygons. Applying similar polygons.
3. Motivation: I want to show them how this knowledge applies to the outside world.
Using this information you can find measures when using enlarged or reduced
images. Also similar polygons are used in art such as in the golden ratio. I will
briefly show them how this ratio can be found in the outside world so their
interests are piqued.
4. Misconceptions: Some students may think that just because two shapes are the
same then they are similar. However, this isn’t the case. They need to be
proportional with their measurements.
5. Instructional Activities/Assessment activities that we will go over after they
attempt it:
a. I want to start out with showing them two parallelograms with one bigger
than the other. I want them to try to solve this problem without me giving
them any knowledge. They need to construct their knowledge first. The
first parallelogram will be ABCD with the angle A being 53 degrees. The
smaller parallelogram will be EFGH with the angle F being 127 degrees. I
will ask them if those two parallelograms are similar. If they are then what
is the degree of angle E? (5 minutes)
b. I will let them discuss this in groups of 4. We will come back as a class
and I will have one person from each group tell what they came up with as
the answer. If a group comes up with yes they are similar, then I will
explain to them why this answer is correct. (3 minutes)
c. Two polygons are similar if 1) corresponding angles are congruent and 2)
corresponding sides are proportional. Then I will show them that the sides
of corresponding sides make up a similarity ratio. This ratio is similar to
the proportions we solved earlier. (3 minutes)
d. Now we will talk about if those parallelograms from earlier. Using those
definitions about similar polygons we can figure out what the
measurement of angle E is. So I ask them what angle from parallelogram
ABCD corresponds with angle E. We will discuss how angle A
corresponds with angle E. (5 minutes)
e. Now I want them to fill in this proportion: AB/EF=AD/x. We will discuss
how the side AB corresponds to EF. Thus what does AD correspond to? It
corresponds to EH. (4 minutes).
f. Now we will take this information to the next level. We will determine if
two triangles ABC and DEF are similar and if they are then we will write
a similarity statement and give a similarity ratio.
We will discuss how these two triangles are similar because of there are
three pairs of congruent angles. Also, there are three proportional sides of
the triangles. To prove this we can set up proportions: AB/FE=15/20=3/4,
BC/ED=12/16=3/4, AC/FD=18/24=3/4. All of these proportions are the
same so thus those triangles are similar. Our similarity statement is
triangle ABC ~ triangle DEF. The similarity ratio is ¾ or 3:4. (8 minutes)
g. Now we can use algebra with similar figures. I will have them do this in
the same groups that they worked in before. I will go around and check
peoples work and select a student to put it up on the board.
Find the value of x given those two trapezoids LMNO and QRST.
After they have done the problem, I pick a student and we will discuss
his/her work on the board.
1. First write a proportion of corresponding sides.
LM/QR=ON/TS.
2. Next substitute the numbers for the sides. 5/6=2/x.
3. Now we must cross-multiply like we did in lesson 1.
5*x=2*6. 5x=12.
4. Now we have to get x by itself by dividing by 5 on both
sides. 5x/5=12/5. X=2.4 (8 minutes).
h. Now for our real world application. Suppose we have an older person with
vision problems, and there was a magnification system that would help
that person read better. The person places a document under the
magnifying lens so to see it. It appears on the video screen. The screen
pictured is 16 in wide by 12 in tall. The actual text is 6 in wide by 4 in tall.
What is the largest complete video image possible for this block of text? I
want the students to try this out on their own first, and then we discuss
their work and I will comment on it. I will tell them before they begin to
think about this problem as solving proportions for similar figures.
1. We can picture this as two rectangles.
2. The smaller rectangle has a width of 6 in which can be magnified
to 16 in. Let the bigger rectangle have x in tall.
3. We can set up a proportion. 6/16=4/x.
4. Now we can solve for x by cross-multiplying. 6*x=4*16. 6x=64
5. We have to get x by itself. So we have to divide by 6 on both sides.
6x/6=64/6. X=32/3 in.
6. The largest image possible on the screen is 16 in by 32/3 in. (8
mins)
i. I want to talk to them briefly about this related to art. The golden rectangle
is a rectangle that can be divided into a square and a rectangle that is
similar to the original rectangle. The golden ratio is 1.618 to 1 which is the
length and width of any golden rectangle. It apparently is very appeasing
to the human eye, and is found in architecture and art since ancient times.
Leonardo da Vinci was very intrigued with it and even wrote about it in
The Divine Proportion. This information will not be tested, but it is just
another trivia fact about math (5 mins).
6. Leads to: Proving triangles are similar through theorems and postulates. Using these ideas
with right triangles and other polygons.
7. Homework: I will give them several problems similar to class work so they can work on them
at home to practice. I will check that they did the homework not for accuracy. We will go over
the answers and any questions the following day.
a.
Complete the following congruence and proportion statements.
1. Angle D=____ the answer is angle B
2. AB/ED=AC/_____ the answer is EF
b. Is the polygon similar? If so write a similarity ratio and similarity
statement. If they are not explain why.
The answer is:
AB/FE=3.5/2.5=1.4.BC/ED=6.3/4.5=1.4.AC/FD=8.4/6=1.4.
They all equal 1.4 or 7/5. So yes those triangles are similar. A
similarity statement is triangle ABC ~ triangle FED. A
similarity ratio is 7/5 or 7:5.
c. The polygons are similar. Find the value of each variable.
The answer is:
1. First we need to set up proportions. AB/EF=5/10.
AC/EG=z/15. CD/GH=y/35. BD/FH=10/x.
2. We can solve for x. AB/EF=5/10. BD/FH=10/x. 5/10=10/x.
We cross-multiply. 5x=100. We divide by 5. X=25.
3. We can solve for y. AB/EF=5/10. CD/GH=y/35.
5/10=y/35. We cross multiply. 175=10y. We divide.
Y=17.5
4. We can solve for z. AB/EF=5/10. AC/EG=z/15. 5/10=z/15.
We cross-multiply. 75=10z. We divide. Z=10.5
d. A map has dimensions 9in by 15in. You want to reduce the map so that it
will fit on a 4in by 6in index card. What are the dimensions of the largest
possible complete map that you can fit on the index card?
The answer is: The map is 15 in. in length and it must be reduced down to
6 in. So we can set up a proportion. 15in/6in=9in/x where we need to solve
what the smallest width could be for the map. We solve for x by crossmultiplying. 15x=54in. We divide. X=18/5 in. So the dimensions the map can
be on in the index card is 4in by 18/5 in.
e. (This problem involves the golden rectangle and golden ratio. Use
proportions to solve for length to width. Remember the golden ratio for
length and width is 1.618:1). A switch plate for a standard wall switch has
the shape of a golden rectangle. The longer side of the switch plate is
about 120 mm. How long is the shorter side? Round your answer to the
nearest millimeter.
This is the answer: First you need to write a proportion. You can use L for
the longer side. 120/L=1.618/1. We can cross multiply. 120=1.618L. We
can divide. L=74.
Proving Triangles Are Similar
1. Prior Knowledge: Knowing how to use Side-Side-Side and Side-Angle-Side theorems to
prove triangle congruence. Knowing how to use Angle-Side-Angle and Angle-AngleSide theorems to prove triangle congruence.
2. Objectives: 2.03 Apply properties, definitions, and theorems of two-dimensional figures
to solve problems and write proofs: a) Triangles. The AA postulate and the SAS and SSS
theorems. Applying AA, SAS, and SSS similarity.
3. Motivation: I like to apply this subject to the real world. We can prove triangles are
similar in architecture such as houses. Also with determining heights of different objects
in nature.
4. Misconceptions: Two triangles have the exact same degrees in the triangles then the two
triangles are similar. They actually need to have the same degrees in the same spots
within the two triangles. Some people may think that if you can use one theorem you can
use them all. However, you have to meet the requirements for each theorem and usually
those requirements are not all in the one triangle.
5. Instructional Activities:
a. I want the students to do an introduction activity in groups of 4 to get them
thinking about triangles and similarity. It involves triangles with two pairs of
congruent angles. They first draw two triangles of different sizes, each with a 60
and 90 degree angles. Then measure the sides of each triangle to the nearest
millimeter. The find the ratio of the lengths of each pair of corresponding sides.
They will answer the following questions: 1.What conclusion can you make about
the two triangles? 2. Complete this conjecture: If two angles of one triangle are
congruent to two angles of another triangle, then the triangles are _?__. This
activity will help them remember the different theorems they learned earlier. It
will lead to the angle-angle similarity postulate. We will talk about this activity
after the groups are done. (15 minutes)
b. I will tell them how that activity proves the Angle-Angle Similarity (AA ~)
Postulate. This states “If two angles of one triangle are congruent to two angles of
another triangle, then the triangles are similar.” (3 minutes)
c. We will then discuss some of the other theorems we learned earlier in the year
and how they might apply to proving triangle similarity. I ask if they remember
Side-Angle-Side theorem. I tell them the definition again: if an angle of one
triangle is congruent to an angle of a second triangle, and the sides including the
two angles are proportional, then the triangles are similar. (5 minutes)
d. Next we will start talking about the Side-Side-Side Similarity (SSS ~) theorem.
We come up with a definition: if the corresponding sides of two triangles are
proportional, then the triangles are similar. (5 minutes).
e. I talk about how we can use similar triangles and measurements to find distances
that are difficult to measure directly. This is called indirect measurement. We can
use similar triangles that are formed by certain figures and their shadows to find
difficult heights (3 mins).
6. Assessment: We are now going to do some problems as a class. The students will attempt
these problems, and then we go over the answers.
a. Explain why the triangles must be similar. Write a similarity statement.
Angle QRP is congruent to angle XYZ because they are right angles.
QR/XY=3/4 and PR/ZY=6/8=3/4. Therefore, triangle QRP ~ triangle XYZ by the
SAS ~ theorem.
b. Explain why the triangles are similar. Write a similarity statement. Then find DE.
Angle ABC is congruent to angle EBD because vertical angles are congruent.
(We will go back over vertical angles). AB/EB=12/18=2/3 and CB/DB=16/24=2/3. Therefore,
triangle ABC ~ triangle EBD by the SAS theorem. CA/DE=2/3. 10/DE=2/3. 2DE=30. DE=15.
c. Ramon places a mirror on the ground 40.5 ft from the base of a geyser. He walks
backwards until he can see the top of the geyser in the middle of the mirror. At
that point, Ramon’s eyes are 6ft above the ground and he is 7ft from the image in
the mirror. Use similar triangles to find the height of the geyser.
Triangle HTV ~ triangle JSV because of the AA ~ postulate. The angle V on both
triangles are the same so this postulate works. HT/JS=TV/SV the corresponding
sides of ~ triangles are proportional. 6/x=7/40.5. 243=7x. 34.7=x. So the geyser is
about 35 ft high.
7. Leads to: Using theorems in finding the similarity in right triangles and other polygons.
8. Homework: The students will do these problems at home. I will check to make sure they
attempted the problems. I don’t check for accuracy but for work. We will go over the
answers and any questions they have the next day.
a. Can you conclude the triangles are similar? If so, write a similarity statement and
name the postulate or theorem you used. If not, explain.
i.
ii.
iii.
The answers are: i. Yes triangle ABC ~ triangle DEF. The theorem is SSS ~. ii. Yes triangle SRT
~ triangle XYZ. The theorem is SAS ~. iii. Yes the triangle ABC ~ triangle EDF. The postulate
is AA~.
b. Explain why the triangles are similar. Then find the value of x.
i.
ii.
The answers are: i. The angles given are congruent. Also there are vertical angles which are
congruent. 14/24=7/12. 7/12=x/22. 154=12x. x=77/6. ii. The left bottom angle from the smaller
triangle is congruent to the left bottom angle on the bigger triangle. Thus AA ~. 5/15=1/3.
1/3=4/x. x=12.
c. Explain why the triangles are similar. Then find the distance represented by x.
i. Kyle is at the zoo. He places a mirror on the ground 20 ft from a giraffe
behind a gate. He walks backwards until he can see the top of the giraffe
in the middle of the mirror. At that point, Kyle’s eyes are 6ft 3in above the
ground and he is 4 ft from the image in the mirror. Use similar triangles to
find the height of the giraffe.
The answer is: We need to convert everything to inches first. 6f 3in=75in.
4ft=48in. 20ft=240in. The triangles are similar because AA ~ with the
angles connected by the triangles. 75in/x=48in/240in. 18000in^2=48xin.
X=375in. Now we can convert this back to feet. X=31.25 ft.
Similarity in Right Triangles
1. Prior Knowledge: You will need to know how to solve proportions. Know what a right
triangle is. Know what the hypotenuse and the altitude are.
2. Objectives: 2.03 Apply proportions, definitions, and theorems of two-dimensional figures
to solve problems and write proofs: a) Triangles. Using similarity in right triangles
3. Motivation: Again I want to use this information in the real world. We can use similarity
in right triangles with finding how far someone is from a place when knowing the
distance between two places.
4. Misconceptions: Some people may confuse what a hypotenuse is in not so obvious right
triangles.
5. Instructional Activities:
a. I want to do another investigation activity to get the students thinking about right
triangles that will transition into the theorems we will learn. They will draw one
diagonal on a rectangular sheet of paper. Cut the paper on the diagonal to make
two congruent right triangles. In one of the triangles, use paper folding to locate
the altitude to the hypotenuse. Cut the triangle along the altitude to make two
smaller right triangles. Label the angles of the three triangles. Compare the angles
of the three triangles by placing the angles on top of one another. Answer these
questions: 1. Which angles have the same measure as angle 1? 2. Which angles
have the same measure as angle 2? 3. Which angles have the same measure of
angle 3? 4. Based on the results, what is true about the three triangles? 5. Use the
diagram at the right to complete the similarity statement triangle RST ~ triangle ?
~ triangle ? (15 minutes)
b. We discuss these results from the activity. We have just shown how the
hypotenuse yields three similar triangles. I show them how this proves this
theorem: 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. (3 minutes)
c. I ask them to remember what a proportion looks like and let one student write it
up on the board. (a/b=c/d). Well I explain to them for any two positive numbers a
and b, the geometric mean of a and b is the positive number x such that a/x=x/b.
(x=square root of ab). This is used in very important corollaries in the previous
theorem. (3 minutes)
d. I explain to them corollary 1: the length of the altitude to the hypotenuse of a right
triangle is the geometric means of the lengths of the segments of the hypotenuse.
(5 minutes)
e. The second corollary is: 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. (5 minutes)
6. Assessment: I will have the students so problems in groups and then we will go over the
answers as a class.
a. Find the geometric mean of 4 and 18.
The answer is: Write a proportion. 4/x=x/18. X^2=72. X=square root 72. X=6*square
root of 2. The geometric mean of 4 and 18 is 6*square root of 2.
b. We are going to apply the corollaries. Solve for x and y.
The answer:
Use the corollary 2 to solve for x:
Use the corollary 1 to solve for y:
Write a proportion. 4/x=x/(4+5)
Write a proportion. 4/y=y/5
Cross product property. X^2=36
Cross product property. Y^2=20
Take the square root. X=6
Take the square root. Y= 2*square root of 5
c. Now we apply this to the real world. The 300-m path to the information center
and the 400-m path to the canoe rental dock meet at a right angle at the parking
lot. Marla walks straight from the parking lot to the lake as shown. How far is
Marla from the information center?
The answer is: The altitude CD is perpendicular segment from point C to side AB,
is the shortest path to the lake. Triangle ABC is a right triangle. Using
Pythagorean triples, AB=500 m (300^2 + 400^2=25000. Square root of 25000 is
500). To find AD, apply corollary 2. AD/AC=AC/AB. AD/300=300/500.
AD=300/500 *300. AD=180. Marla is 180 m from the information center.
7. Leads to: Being able to solve the proportions in these corollaries and theorem helps find
proportions in triangles.
8. Homework: They will do these few problems at home. I will check them tomorrow to see
if they did the work. I check for work not accuracy. We will go over the answers and
answer any questions about them.
a. Find the geometric mean of each pair of numbers.
i. 4 and 10
The answer is: i. 4/x=x/10. X^2=40. X=square root of 40. X=2*square root of 10.
b. Solve for x.
i.
ii.
The answers are: i. You have to use corollary 1 to solve for x. To have the number
for the left hand segment on the bottom we subtract 40 from 50 which is 10. So
the proportion is 10/x=x/40. X^2=400. X=20. ii. You have to use corollary 2 to
solve for x. So the proportion is 3/x=x/(3+9). X^2=36. X=6.
c. Study the plan. A service station will be built on the highway, and a road will
connect it with Cray. How far from Blare should the service station be located so
that the proposed road will be perpendicular to the highway? How long will the
new road be?
The answer is: We need to know how long the highway should be. It is 50. To
find from Blare to the service station we use corollary 2. BS/40=40/50. BS=32.
So Blare should be located 32 m from the service station. Now to find the service
station length we should use the corollary 1. Well to find length from Alba to the
service station we need to subtract BS from 50 which is 18. So the proportion is
32/SS=SS/18. SS=24.
Exploring Proportions in Triangles Before Lesson
This activity uses geometry software and allows students to able to explore proportions within
triangles. We will do the activity the day before the lesson. We will discuss their conjectures at
the end of the lesson and then talk about them the next day about how they relate to the lesson.
Construct:
•
Use geometry software. Draw triangle ABC and construct point D on segment AB.
•
Construct a line through D parallel to segment AC.
•
Construct the intersection E of the parallel line with segment BC.
Investigate:
•
Measure segment BD, DA, BE, and EC.
•
Calculate the ratios BD/DA and BE/EC.
•
Manipulate triangle ABC and observe the ratios BD/DA and BE/EC (Save your
observations)
Construct:
•
Use geometry software. Draw triangle ABC. Construct the bisector of angle A. Construct
point D, the intersection of the bisector and segment CB.
Investigate:
•
Measure segment AC, AB, CD, and DB.
•
Calculate the ratios AC/AB and CD/DB.
•
Manipulate triangle ABC and observe the ratios AC/AB and CD/DB. (save your
observations).
Exercises:
1. Suppose a line parallel to one side of a triangle intersects the other two sides. Make a
conjecture about the four segments formed.
2. The bisector of an angle of a triangle divides the opposite side into two segments. Make a
conjecture about the two segments and then the other two sides of the triangle.
Extend:
1. Construct a line of AB parallel to line CD. Then construct line AC and line BD.
2. Construct point E on line AC.
3. Construct line EF parallel to line AB with point F the intersection of line EF and line BD.
4. Measure segment AC, CE, BD, and DF.
5. Calculate ratios AC/CE and BD/DF.
6. Manipulate the locations of A and B and observe the ratios of AC/CE and BD/DF.
7. Suppose three parallel lines intersect two transversals. Make a conjecture about the
segments of the transversals.
8. Suppose that four of more parallel lines intersect two transversals. Make a conjecture
about the segments of the transversals.
Proportions in Triangles
1. Prior knowledge: You need to know how you can tell triangles are similar and what
theorems to use. You need to be able to solve proportions.
2. Objective: Apply proportions, definitions, and theorems of two-dimensional figures to
solve problems and write proofs: a) Triangles. Using Side-splitter theorem. Using the
triangle-angle-bisector theorem.
3. Motivation: The real world application can relate to making the sails that people use on
the water. By using the theorems we will learn we can actually find different lengths of
the sail.
4. Misconception: People may confuse what a transversal and a parallel line are, and how
they relate to each other.
5. Instructional activity:
a. I want the students to reflect as a class about what they had done yesterday with
the activity and what they conjectured about the two three different pictures they
made. I explain to them that they were using proportions in triangles and
manipulating them with different triangles and parallel lines (7 mins).
b. I show them the side-splitter theorem. It states: if a line is parallel to one side of a
triangle and intersects the other two sides, then it divides those sides
proportionally. (10 mins)
c. I show them the proof so they can see how this theorem works.
d. I ask them to remember what the parallel lines and transversals the students drew
yesterday looked like and what they conjectured. I tell them there is a corollary to
side-splitter theorem that talks about this. Corollary: If three parallel lines
intersect two transversals, then the segments intercepted on the transversals are
proportional. a/b=c/d (10 mins).
e. I explain to them finally how the side-splitter theorem can be used to prove the
triangle-angle-bisector theorem. It states: if a ray bisects an angle of a triangle,
then it divides the opposite side into two segments that are proportional to the
other sides of the triangle. I will show them how this proof works. (10 mins)
6. Assessment: I will have the students attempt these problems in groups and then we will
go over the answers as a class.
a. Solve for x.
The answer is: Use the side-splitter theorem. TS/SR=TU/UV. Now substitute.
x/16=5/10/. Cross multiply. 10x=80. Solve for x by dividing. X=8.
b. Sail makers sometimes use a computer to create a pattern for a sail. After they cut
out the panels of the sail, they sew them together to form the sail. The edges of
the panels in the sail at the right are parallel. Find the lengths x and y.
The answer is: Use the side-splitter theorem. 2/x=1.7/1.7. Cross multiply. 3.4=1.7x.
Solve for x by dividing. X=2ft. To solve for y use the corollary to the side-splitter
theorem. 3/2=y/1.7. Cross-multiply. 5.1=2y. Now solve for y by dividing. Y=2.55ft.
c. Find the value of x.
The answer is: You need to use the triangle-bisector theorem. PS/SR=PQ/RQ.
x/6=8/5. Cross multiply. 5x=48. Now solve for x by dividing. X=9.6.
7. Leads to: Learning how to make and use proofs can lead to building more complicated
proofs in other areas of geometry and other math subjects.
8. Homework: I will give them a few homework problems to do on their own. I will check
their work tomorrow to see if they completed them. I don’t check for accuracy just for
completion. We will go over the answers and any questions afterwards.
a. Solve for x and or y.
i.
ii.
iii.
The answers are:
i. For this problem we have to use the side-splitter theorem to solve for x. This
is when a line parallel to one side of the triangle and intersects the other two
sides, then divides those sides proportionally. So our proportion is
2x/(x+4)=12/9. Now we cross-multiply. 24x=9x+36. We subtract 9x. 15x=36.
Now we divide by 15. x=2.4
ii. For this problem we have to use both the side-splitter theorem and the
corollary to that theorem. To solve for x we use the side splitter theorem. So
our proportion is x/35=20/31. We cross-multiply. 31x=700. We divide by 31.
x=22.58 (round to the nearest hundredth). Now to solve for y we must use the
corollary. It states if three parallel lines intersect two transversals then the
segments intercepted on the transversals are proportional. So our proportion is
20/31=21.5/y. Now we cross multiply. 666.5=20y. Now we divide by 20.
y=33.33.
iii.For this problem we have to use the triangle-angle-bisector theorem. This
states that if a ray bisects an angle of a triangle then it divides the opposite
side into two segments that are proportional to the other side of the triangle.
So our proportion is x/6-x=6/4 (we use 6-x because the whole bottom side is
6, but we only want the second half so its 6-x). We cross-multiply. 36-6x=4x.
We add 6x to both sides. 36=10x. We divide by 10. x=3.6
b. In Washington, D.C., 17th, 18th, 19th, and 20th Streets are parallel streets that
intersect Pennsylvania Avenue and I Street. How long to the nearest foot is
Pennsylvania Avenue to 19th Street and 18th Street? How long to the nearest foot
is Pennsylvania Avenue between 18th Street and 17th Street?
The answer is: You need to use the corollary to the side-splitter theorem. So our
proportion for the first part of question is 445ft/x=400ft/520ft. We cross-multiply.
231400ft^2=400xft. Now we divide by 400ft. x=579ft. The second part of the
problem we use the same thing except we are solving for a different area. Our
proportion is 579ft/y=520ft/620ft. We cross multiply. 358980ft^2=520xft. Now we
divide by 520ft. x=690ft.
Perimeters and Areas of Similar Figures
1. Prior Knowledge: You will need to know what and how to find a perimeter
and area of a figure. You will need to remember what similar figures are.
2. Objectives: 2.03 Apply properties, definitions, and theorems of twodimensional figures to solve problems and write proofs: a) Triangles, b)
Quadrilaterals, c) Other polygons. Finding Perimeters and Areas of Similar
Figures.
3. Motivation: Our real world connection can be using the information to make
plots of land available to the community for gardening.
4. Misconceptions: Some students may think that just because the figures are the
same shape we can call them similar and thus find their area/perimeter.
5. Instructional Activities:
a. For this last lesson, I want the students to do an activity to help getting
them thinking about perimeters/area with similar figures. They can
come up with some ideas first about the subject before I teach them the
lesson (15 minutes)
b. Perimeters and Areas of Similar Rectangles: On a piece of grid paper,
draw a 3-unit by 4-unit rectangle. Draw three different rectangles, each
similar to the original rectangle. Label them I, II, and III. Use your
drawings to complete a chart like this.
Rectangle
Perimeter
Area
Original
I
II
III
Use the information from the first chart to complete a chart like this.
Rectangle
I to original
II to original
Similarity Ratio
Ratio of Perimeters
Ratios of Areas
III to original
How do the ratios of perimeters and the ratios of areas compare with the
similarity ratios?
c. We will talk about their results as a class, and I will use this as a
transition into the material. I will explain to them the perimeters and
areas of similar figures with similarity ratios. If the similarity ratio of
two similar figures is a/b, then : 1. the ratio of their perimeters is a/b
and 2. the ratio of their areas is a^2/b^2. (15 mins)
6. Assessment: I will let them try to figure out the problem themselves in groups and we
will discuss them as a class afterwards. They should remember to use ratios and the
definition I just provided.
a. The trapezoids are similar. The ratio of the lengths of corresponding sides is 6/9
or 2/3. Find the ratio (smaller to larger) of the perimeters. Find the ratio (smaller
to larger) of the areas.
The answer is: The ratio of perimeters is the same as the ratio of corresponding sides,
which is 2/3. The ratio of areas is the square of the ratio of corresponding sides,
which is 2^2/3^2=4/9.
b. The area of the smaller pentagon is about 27.5 cm^2. Find the area A of the larger
regular pentagon.
The answer is: All regular pentagons are similar. Here is the ratio of the lengths of
corresponding sides is 4/10 or 2/5. The ratio of the areas if 2^2/5^2=4/25. So to find
the area of the larger regular pentagon we set up a proportion.
4cm^2/25cm^2=27.5cm^2/A. We cross-multiply. 4Acm^2=687.5cm^4. We divide by
4 cm^2. A=171.875 cm^2.
c. During the summer, a group of high school students used a plot of city land and
harvested 13 bushels of vegetables that the gave to a food pantry. Their project
was so successful that next summer the city will let them use a larger similar plot
of land. In the new plot, each dimension is 2.5 times the corresponding dimension
of the original plot. How many bushels can they expect to harvest next year?
The answer: The ratio of the dimension is 2.5:1. So the ratio of the areas is
(2.5)^2:1^2, or 6.25:1. With 6.25 as much land next year, the students can expect to
harvest (6.25)(13) or about 81 bushels.
d. The areas of two similar triangles are 50cm^2 and 98cm^2. What is the similarity
ratio? What is the ratio of their perimeters?
The answer is: Find the similarity ratio a:b. a^2/b^2=50cm^2/98cm^2. This can be
simplified to a^2/b^2=25cm^2/49cm^2. To find a/b we have to take the square roots
of both sides. a/b=5/7. The ratio of the perimeters is the same as the similarity ratios:
5:7.
7. Leads to: You can find areas of different polygons besides the basic triangles and
quadrilaterals.
8. Homework: The students will do the problems at home. I will check the work the
following day. I only check for completion not accuracy. We will go over the answers
and any questions they may have.
a. The figures in each pair are similar. Compare the first figure to the second. Give
the ratio of the perimeters and the ratio of the areas.
i.
The answer is: the similarity ratio is 16/24=2/3. The perimeter ratio is 2/3
and the area ratio is 2^2/3^2=4/9.
b. The figures in each pair are similar. The area of one figure is given. Find the area
of the other figure to the nearest whole number.
i.
The answer is: The similarity ratio is 12/18=2/3. The area of larger triangle
is 121m^2. So we set up a proportion is find the other area.
2m/3m=A/121m^2. We cross multiply. 242m^3=3Am. We divide by 3m.
A=81m^2.
c. An embroidered placemat costs $2.95. An embroidered tablecloth is similar to the
placement, but four times as long and six times as wide. How much would you
expect to pay for the tablecloth?
The answer is: the ratio is 4:6. So the area is 4^2:6^2=16:36. We need to set up a
proportion to figure out how much the new one will cost. 16/36=2.95/x. We cross
multiply. 16x=106.2. We divide by 16. x=$6.64. So the tablecloth will cost $6.64.
d. Find the similarity ratio and the ratio of perimeters for each pair of similar figures.
i. Two triangles with areas of 75m^2 and 12m^2.
The answer is: We set up a proportion of areas. A^2/b^2=12m^2/75m^2.
We can simplify this proportion. A^2/b^2=4^2/25^2. We can take the
square root of both sides to find the similarity ratio. a/b=2/5. the perimeter
ratio is 2/5 as well.
Unit Lesson Plan Quiz 1
This quiz will cover the first three lessons: Ratios and Proportions, Similar Polygons, and
Proving Triangles are Similar. I will use multiple choice because there will be multiple choice
questions on their end of course exam.
1. If g/h=6/5, which equation is NOT true?
a. g/6=h/5
b. gh=30
c. h/g=5/6
d. (g+h)/h=11/5
e. 5g=6h
2. Are the polygons similar? If they are, identify the correct similarity statement and ratio.
a. Yes, ΔABC
ΔFED with a similarity ratio of 21 : 11.
b. Yes, ΔABC
ΔDEF with a similarity ratio of 2 : 3.
c. The figures are not similar.
d. Yes, ΔABC
ΔDEF with a similarity ratio of 3 : 2.
e. Yes, ΔABC
ΔFED with a similarity ratio of 11:21.
3. State whether ΔADB
similar. ΔCDB, and if so, identify the theorem that proves the triangles
a. yes, AA~
b. yes, SAS ~
c. yes, SSS~
d. yes, side-splitter theorem
e. no they are not similar
4. Nine pounds of grass seed will cover 110 square feet. Let p represent the number of
pounds of seed needed to cover 680 square feet. Which proportion can you use to find p?
a. 110/9=p/680
b. 110/680=p/9
c. 9/110=p/680
d. 9/110=680/p
e. 110/p=9/680
5. The polygons are similar, but not necessarily drawn to scale. Find the values of x and y.
a. x=13, y=5
b. x=13, y=6
c. x=14, y=5
d. x=14, y=6
e. x=13, y=14
6. Write a similarity statement for the two triangles.
a. ΔTVU
ΔWXY
b. ΔTUV
ΔWYX
c. ΔVUT
ΔWXY
d. None of the above.
e. ΔTUV
ΔWXY
7. The Community Recreation Center is developing plans for a new sports facility.
Community members can submit suggestions for the new facility, along with basic scale
drawings of their ideas. Lupe wants to include a new 65- by 100-meter soccer field in the
athletic center. The scale that is used is 1/16 in. = 1 m. What is largest scale drawing he
can submit?
a. 8.5 in. by 11 in.
b. 10.5 in. by 13 in.
c. 6 in. by 12 in.
d. 3 in. by 7 in.
e. 5 in. by 10 in.
8. You want to draw an enlargement of a design that is painted on a 2 in by 4 in card. You
plan to draw on a 10 in by 26 in piece of paper. What are the dimensions of the largest
complete enlargement you can draw?
a. 10 in. by 19 in.
b. 15 in. by 26 in.
c. 10 in. by 20 in.
d. 20 in. by 10 in.
e. 17 in. by 26 in.
9. In sunlight, a cactus casts a 10 ft. shadow. At the same time a person 6 ft tall casts a 5 ft
shadow. Use similar triangles to find the height of the cactus.
a. 10 ft
b. 16 ft
c. 9 ft
d. 11 ft
e. 12 ft
10. ΔUVW
ΔXYZ. What is m Z?
a. 72°
b. 108°
c. 240°
d. 252°
e. 36°
Answers to the Unit Lesson Plan Quiz
1. B
2. D
3. B
4. C
5. A
6. E
7. A
8. C
9. E
10. A
Unit Lesson Plan Quiz 2
This quiz will cover the material learned in the last three sections. Those sections are:
Similarity in Right Triangles, Proportions in Triangles, and Perimeters and Areas in Similar
Figures. I will use multiple-choice so to help prepare them for the kinds of questions on the end
of course exam.
1. Solve for a and b.
a. a= 9, b= ¾
b. a= 9, b= 15
c. a= ¾, b= 15
d. a= 15, b= ¾
e. a= 15, b= 9
2. Find x to the nearest tenth.
a. 9.5
b. 8.4
c. 8.1
d. 9.0
e. 8.6
3. Find the
t ratio of thhe perimeterr of the largeer rectangle to
t the perim
meter of the
smalleer rectangle.
a. 9/7
b. 5/3
c. 7/9
d. 13/11
e. 3/5
t value of x.
4. Find the
a. 2
b.
c. 2
d.
19 e. 19 2
t values off x and y givven that
5. Find the
a. x= 28.5, y= 7
||
.
b. x= 7, y= 28.5
c. x= 9, y= 28.5
d. x= 14, y= 19
e. x= 19, y= 14
6. Solve for x.
a. 2
b. 9
c. 6
d. 8
e. 12
7. The area of regular octagon is 50 cm^2. What is the area of a regular octagon with
sides three times as large?
a. 150 cm^2
b. 450 cm^2
c. 7500 cm^2
d. 1500 cm^2
e. 515 cm^2
8. The areas of two similar triangles are 98 cm^2 and 128 cm^2. What is the
similarity ratio? What is the ratio of their perimeters?
a. 8/7; 7 : 8
b. 7/8; 8 : 7
c. 7/9; 7 : 9
d. 9/7; 7 : 9
e. 7/8; 7 : 8
9. Find the geometric mean of 75 and 3.
a. 25
b. 29
c. 39
d. 15
e. 9
10. The extendable ramp shown below is used to move crates of fruit to loading docks
of different heights. When the horizontal distance AB is 4 feet, the height of the
loading dock, BC, is 3 feet. What is the height of the loading dock, DE?
a. 12 ft
b. 7 ft
c. 6 ft
d. 11 ft
e. 9 ft
Answers to the Unit Lesson Plan Quiz 2
1. B
2. E
3. C
4. A
5. B
6. D
7. B
8. E
9. D
10. C
Project for Unit Lesson Plan
I am going to do a “Similarity is Fun” project that I found at this link:
http://tweb.lisd.net/lanne_bond/Microsoft%20Word%20-%20Similarity%20Project%20PAP%20Fall%202007.pdf. I think it is a fun way for students to take any object they can think of to
shrink and enlarge it so they can enhance their learning of ratios/similarity. Here are the project
details that I would give the students. I did tweak the project from the website.
“Similarity Can Be Fun!!”
• The objective of this project is to create an enlarged or scaled model of a similar or
original object using the ideas of similarity and proportion you have learned from
studying similar polygons.
• Model: Take an object small in size and construct a model that is completely
proportional, and bigger in size. Or, you may make a smaller model of an object that is
very large in size. Pay close attention to detail. Every side of your model should be
proportional to the original object. Accuracy is extremely important. You must either:
include the original object with your project, or take a photograph of the original object
(only if it is too large to attach) and attach it to your project. You are required to include
the form attached.
1. Put your report in a folder.
2. Include Title page and Contents page.
3. Include a drawing of the original object, with all of the dimensions marked clearly.
4. Include a drawing of your new model with all of the new dimensions marked
clearly.
5. On the attached form, you must show all of the calculations you used to find
the new dimensions of your model. Include the scale factor and show all
appropriate work. Be very neat and accurate.
•
•
•
Examples: Small to big- Take a Snickers bar and “blow it up” to 3 feet in length while
keeping the volume, width, and height proportional. Big to small Measure every part of a
car and make a scale model that could fit in your hand.
Your project will be graded on the following basis:
1. 50 points Accuracy of Scale/Proportions/Reasonableness of Scale
form/Calculations
2. 40 points Model – neatness and creativity
3. 10 points Following Directions (including a picture, scale, etc.)
Turn in sketches of before and after. Report measurements of both object such as length,
width, height, or radius. Include surface area and volume of both objects. Finally include
the similarity ratio, scale used, ratio of surface area, and ratio of volumes.
Rubric for Unit Lesson Plan Project
Name: __________________________________________
Similarity Project
Accuracy of Scale/Proportions/Reasonableness of Scale - Form
_________________________________________________________
_________________________________________________________
_________________________________________________________
50 points ____________
Neatness and Originality/Creativity - Model
_________________________________________________________
_________________________________________________________
_________________________________________________________
40 points ____________
Following Directions (including picture, scale etc)
_________________________________________________________
_________________________________________________________
_________________________________________________________
10 points ____________
Final Grade ____________
Review Sheet for Unit Lesson Plan Test
Key Concepts
•
Proportion-statement that two ratios are equal. a/b=c/d or a:b=c:d
•
Properties of Proportions: a/b=c/d is equivalent to
o ad=bc
o b/a=d/c
o a/c=b/d
o a+b/b=c+d/d
•
Scale compares each length in the drawing to the actual length.
•
Two polygons are similar is 1) corresponding angles are congruent and 2)
corresponding sides are proportional
•
Similarity ratio is the ratio of the lengths of the corresponding sides
•
Proving Triangles Are Similar
o Angle-Angle Similarity (AA~ Postulate): If two angles of one triangle are
congruent to two angles of another triangle, then the triangles are similar.
o Side-Angle-Side Similarity (SAS~ Theorem): If an angle of one triangle is
congruent to an angle of a second triangle, and the sides including the two
angles are proportional, then the triangles are similar.
o Side-Side-Side Similarity (SSS~ Theorem): If the corresponding sides of two
triangles are proportional, then the triangles are similar.
•
Similarity in Right Triangles:
o Theorem 8-3: The altitude to the hypotenuse of a right triangle divides the
triangle into two triangles that are similar to the original triangles and to
each other.
o Geometric mean: for any two positive numbers a and b, this mean of a and b
is the positive number x such that a/x=x/b.
o Corollary 1 to Theorem 8-3: The length of the altitude to the hypotenuse of a
right triangle is the geometric mean of the lengths of the segments of the
hypotenuse.
o Corollary 2 to Theorem 8-3: 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.
•
Proportions in Triangles:
o Side-Splitter Theorem: If a line is parallel to one side of a triangle and
intersects the other two sides, then it divides those sides proportionally.
o Corollary to Side-Splitter Theorem: If three parallel lines intersect two
transversals, then the segments intercepted on the transversals are
proportional. a/b=c/d
o Triangles-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.
•
Perimeters and Areas of Similar Figures: If the similarity ratio of two similar
figures is a/b, then
o The ratio of their perimeters is a/b and
o The ratio of their areas is a^2/b^2
Review Examples
1. If a/b=3/2 tell whether each equation must be true and explain why.
a. 3b=2a
b. 2/3=b/a
c. 2b=3a
d. a/3=b/2
2. The triangle is similar. Find the similarity ratio of the first to the second.
3. The polygons are similar. Find the value of each variable.
4. Are the triangles similar? If so, write the similarity statement and name the postulate or
theorem that you used. If not, explain.
a.
b.
5. A crate is 2 ft high and casts a 2 ft shadow. At the same time, an apple tree casts an 18ft
shadow. How tall is the tree?
6. Find the values of the variables. When an answer is not a whole number leave it in
simplest radical form.
7. Find the value of x.
a.
b.
8. For each pair of similar figures, find the ratio of the area of the first figure to the area of
the second.
9. The similarity ratio of two similar triangles is 3:7. The area of the smaller triangles is 36
cm^2. What is the area of the larger triangle?
Answers to Review Sheet for Unit Lesson Plan Test
1. A. This is true because this is the property of ad=ba. B. This is true because it is the
property of b/a=d/c. C. This is not true. It should be a/c=b/d. D. This is true because it is
the property a/c=b/d.
2. The similarity ratio is 2/3.
3. X equals 1.
4. A. These triangles are similar because of the Side-Angle-Side Theorem. B. These
triangles are not similar because the sides do not have the same similarity ratio or fulfill
any of the theorems.
5. The tree is 18ft tall.
6. X equals 2 21 . Y equals 4 3 . Z equals 4 7 .
7. A. X equals 7.5. B. X equals 5.5.
8. The ratio of the area of the first figure to the second figure is ¼.
9. The area of the larger triangle is 196 cm^2.
Uniit Lesson Plaan Test
1. The polygons beloow are simillar, but not necessarily
n
d
drawn
to scalle. Find the
valuess of x and y.
a. X=14, y=55
b. X=14, y=66
c. X=13, y=55
d. X=5, y=133
e. X=13, y=66
2. ΔUVW
W ‫ ׽‬ΔXYZ.. What is m‫ס‬
‫ס‬Z?
a. 125
b. 72
c. 252
d. 36
e. 108
3. Theree is a law thaat the ratio off the width too length for the Americaan flag shoulld be
10 : 19. Which dim
mensions aree NOT in the correct ratio?
a. 20 : 44 ft.
b. 20 : 38 in.
c. 30 : 57 ft.
d. 50 : 95 ft.
e. 100 : 190 ft.
wo rectanglees are similarr.
4. The tw
Which
h is a correctt proportion between corrresponding sides?
a. 4/14 = l/100
b. 14/10 = l/44
c. 4/10 = l/4
d. 14/4 = l/100
e. 14/4 = 10//l
t values off x and y givven that
5. Find the
a. X=5, y=199.5
b. X=19.5, y=
=4
c. X=8, y=133
d. X=13, y=88
e. X=4, y=199.5
6. If =
, which eqquation mustt be true?
a. h/g = 5/6
b. gh= 6 x 5
||
.
c. g/h = 6/5
d. h/6 = g/5
e. 5h = 6g
a
to thhe hypotenusse of a right
7. The raatio of the seegments intoo which the altitude
triang
gle divides thhe hypotenusse is 9 : 4. What
W is the leength of the altitude?
a
a. 3
b. 42
c. 36
d. 6
e. Can not bee determinedd
8. Find OM
O if
bisects ‫ס‬NLM
M, LM = 14, NO = 3, andd LN = 4. Roound your
answeer to the nearrest hundreddth, if necesssary.
a. 0.86
b. 12.27
c. 10.5
d. 18.67
e. 11.01
9. Solve for a and b.
=15/4
a. a=15/2, b=
b. a=3/4, b=99/2
c. a=9/2, b=3/4
d. a=9/2, b=15/2
e. a=3/4, b=15/2
10. For the pair of similar figures, give (a) the ratio of the perimeters and (b) the ratio
of the areas of the first figure to the second one.
a. a. 9/16 , b. ¾
b. a. 4/3, b. 16/9
c. a. 4/3 , b. 4/3
d. a. 16/9, b. 16/9
e. a. 16/9, b. 4/3
11. The extendable ramp shown below is used to move crates of fruit to loading docks
of different heights. When the horizontal distance AB is 4 feet, the height of the
loading dock, BC, is 3 feet. What is the height of the loading dock, DE?
a. 5 ft.
b. 11 ft.
c. 9 ft.
d. 7 ft.
e. 6 ft.
12. The areas of two similar triangles are 49 cm2 and 16 cm2. What is the ratio of the
corresponding side lengths? Of the perimeters?
a. 7 : 4, 7 : 4
b. 7 : 16, 7 : 4
c. 49 : 16, 49 : 4
d. 49 : 16, 16 : 49
e. 7 : 4, 7 : 16
13. Solve for x. If needed, round to the nearest hundredth.
a. 14.35
b. 13.13
c. 20.12
d. 10.25
e. 13.31
14. The triangles are similar but not drawn to scale. Write a similarity statement and
name the postulate or theorem that is used to prove similarity.
a. 4 : 5; Side-Angle-Side Theorem
b. 5 : 4, Angle-Side-Angle Theorem
c. 4 : 5, Angle-Angle Postulate
d. 5 : 4, Side-Angle-Side Theorem
e. None of the above
15. The triangles are similar. Find DE.
a. 30
b. 40
c. 50
d. 10
e. 70
16. An art class is painting a rectangular mural for a community festival. The students
planned the mural with a diagram that is 90 in. long and 26 in. high. The mural 14
ft. high. Find its length. Round to the nearest tenth.
a. 50.2 ft.
b. 48.5 ft.
c. 30.3 ft.
d. 25.5 ft.
e. 49 ft.
17. You want to make a scale drawing of your bedroom to help you arrange your
furniture. You decide on a scale of 5in. = 3 ft. Your bedroom is a 15 ft.-by-18 ft.
rectangle. What should be its dimensions in your scale drawing?
a. 30 in. by 25 in.
b. 20 in. by 25 in.
c. 25 in. by 30 in.
d. 30 in. by 35 in.
e. 25 in. by 20 in.
18. The area of the smaller regular triangle is about 30 cm^2. Find the area A of the
larger regular triangle. Round to the nearest hundredth.
a. 103.46
b. 102.37
c. 110.53
d. 103.47
e. 102.36
19. To estimate the height of a totem pole, Jorge uses a small square of plastic. He
holds the square up to his eyes and walks backward from the pole. He stops when
the bottom of the pole lines up with the bottom edge of the square and the top of
the pole lines up with the top edge of the square. Jorge’s eye level is about 2 m
from the ground. He is about 3 m from the pole. Estimate the height of the totem
pole.
a. 2 m
b. 5 m
c. 6 m
d. 3 m
e. 4 m
20. Why are these triangles similar. State which theorem or postulate makes them
similar.
a. Side-Side-Side Theorem
b. Angle-Angle Postulate
c. Side-Angle-Side Theorem
d. Side-Splitter Theorem
Answer Key to the Unit Lesson Plan Test
1. C
2. B
3. A
4. D
5. E
6. A
7. E
8. C
9. D
10. B
11. E
12. A
13. B
14. D
15. A
16. B
17. C
18. D
19. E
20. B
Conclusion to Unit Lesson Plan
This semester I have grown tremendously through my tutoring experience in this class. I
learned a lot from Ms. Garren and the way she teaches her Advanced Functions and Modeling
class. I like the innovative ways she gets the students attention by being real with them. She is
very friendly with all of her students, and knows small pieces of information about them that
makes her class personal. I find this to be a great ability to have when you are a teacher. You
need to be able to reach your students first on a personal level before you can ever reach them on
an academic level. I love how this semester I became closer to my students on this personal level
in such a quick amount of time. I really enjoy going to tutoring while in previous semesters I
didn’t have that feeling because there was no closeness amongst my students and I. My host
teacher also gave me really good tips each day about what to do in certain situations. I felt like I
could really talk to her about whatever was on mind concerning teaching. I definitely want to
work with her again in the future; maybe even do my student teaching with her. All in all, I think
semester of tutoring was probably the best I ever had, and I will always remember it through my
teaching career ahead.
Work Cited
Laurie E. Bass, R. I. (2004). Prentice Hall Mathematics Geometry. Upper Saddle River: Pearson Education Inc. www.pshchool.com