1
Secondary 4 Mathematics Lesson Study
Topic: Finding the ratio of the area of triangles
Teachers from Shatin Tsung Tsin secondary School:
Mr. Cheung Man Wai, Mr. Mak Shing Chit, Mr. Chang Wing Kang, Mr. Fung Kwok
Leung, Miss Chung So Sum, Mr Lau Chi Hin, Mr. Choi Wai Man
Members from Hong Kong Institute of Education:
Dr. Sze Chong Lap, Mr. Nelson Chu, Mr. Micheal Lee
This lesson study was carried out in the period of February – May 2004, during which
9 meetings (including 1 post-test conference) were held among six teachers and three
research team members in the school.
Background
Before the lesson study, students should have learnt the ways to find the area ratio of
triangles by using similar triangles.
Stage 1: Incubation of ideas
Choosing and defining the object of learning
From teachers’ experiences, students found it difficult to manage the questions related
to the area ratio of common height triangles and there is no textbook highlighting this
topic. Therefore, our focus was set on finding the area ratio by using the properties of
common height triangles and similar triangles.
Objectives of this lesson study are as follows.
(1) Identify the height of a triangle with respect to given base.
(2) Identify the common height of a pair of triangles.
(3) Finding similar triangles on a given figure.
(4) Finding triangles with common height.
(5) Finding area ratio of similar triangles.
(6) Finding ratio of sides of similar triangles from area ratio.
(7) Finding area ratio of common height triangles.
Stage 2: Pretest Setting and Result Analysis
A. Design of Pretest
A pretest was designed to gather information on students’ prior knowledge and
students’ understanding about the ratio of areas of triangles of common height.
A pilot test was given to a sample of six S.5 students. Based on their responses,
the questions were refined. The pretest was administered to classes 4A and 4B.
2
B. Results of Pretest
The result of the pretest are summarized as follows:
Qn. Ability testing
Percentage of correct answer or method
4A
4B
4A & 4B
1
Identifying the height of a triangle with
respect to given base
63.4%
58.3%
61.0%
2
Identifying the common height of a pair
of triangles
87.8%
80.6%
84.4%
3a. Finding area ratio of similar triangles
75.6%
55.6%
66.2%
3b. Finding ratio of sides of similar triangles
85.4%
63.9%
75.3%
from area ratio.
.
3c. Finding area ratio of common height
triangles
51.2%
50.0%
50.6%
4a. Finding similar triangles on a given figure
85.4%
88.9%
87.0%
4b Finding triangles with common height
59.8%
72.2%
65.6%
4c Finding area ratio of common height
triangles
41.5%
38.9%
40.3%
4d. Finding the ratio of the length by using
similar triangle
85.4%
75.0%
80.5%
4e. Finding area ratio of common height
triangles
36.6%
13.9%
26.0%
4f. Finding ratio of lengths of similar
triangles
92.7%
63.9%
79.2%
4g. Finding area ratio of similar triangles.
56.1%
30.6%
44.2%
5a Finding area ratio of common height
triangles
73.2%
58.3%
66.2%
5b Finding triangles with common height
70.7%
63.9%
67.5%
5c Finding area ratio of similar triangles
70.7%
66.7%
68.8%
5d Finding ratio of sides of similar triangles
from area ratio.
92.7%
83.3%
88.3%
5e Finding area ratio of common height
triangles.
46.3%
41.7%
44.2%
6
9.8%
2.8%
6.5%
Problem solving:
Finding the area ratio by using the
common height and similar triangles.
3
C. Analysis of the results of the pretest
1.) From the results of question 1, it was found that some students drew the height
with respect to the horizontal base instead of the line AB as required.
2.) Question 3c was a typical question of finding the area ratio by using the common
height of triangles. The results show around 50% students failed to do so.
3.) For question 4c and 4e, only around 40.3% and 26% of students could answer the
question correctly respectively. This shows students find it difficult to find the
area ratio by identifying the triangles with common height especially when the
bases are not horizontal.
4.) For question 4g, the results could reflect some students might forget the
relationship between the area ratio and the corresponding sides of similar
triangles.
5.) The results of questions 5e and 6 show that students find it even harder to solve
the mixed type problems.
D. Conclusions drawn from the analysis of the pretest results (V1)
1.) Some students might not know the relationship between the area ratio of common
height triangles and their corresponding sides.
2.) Most students could not find the area ratio of common height triangles because
they were not able to identify the common height of triangles especially when the
base was not horizontal.
E. Determination of the critical feature
From the results of the pretest, it was found that many students did not know the
relationship between the area ratio and the corresponding bases in the common height
triangles.
From the research of the past CE papers, there are three basic approaches in
solving problems related to the area ratio of triangles. They are as follows.
1.) Considering the common angle of triangles and applying the formula of
Area 
1
ab sin C ;
2
2.) Considering similar triangles and applying
A1
l
 ( 1 )2 ;
A2
l2
4
3.) Considering common height triangles and applying
A1 b1
 .
A2 b2
After discussion, the group of teachers determined the critical features were 1) and 2).
F. Variation of teaching strategy (V2)
This topic is not discussed in common textbooks. The following teaching strategies
were discussed.
A
l
1.) Use HKCEE questions to show the techniques including 1  ( 1 ) 2 and
A2
l2
A1 b1
in solving this kind of problems.

A2 b2
2.) Explain how the formula Area 
1
ab sin C can be used to solve different
2
problems on ratio of areas of triangles.
3.) Ask students to find common height triangles on a figure. Then ask them to draw
the common height of the common height triangles.
Stage 3: Implementation of the Research Lesson
A. Planning of the lesson study
One double lesson (80 minutes) would be conducted to go through the above critical
features. The planning is as follows.
Activity
Time required Remarks
Height of triangle
5 min
Draw the height of the triangle
regarding to a specific side as
base
Revision on the area ratio of similar
triangles
10 min
Examples
5 min
Classwork (Practice)
10 min
Check students’ understanding
Introduce the concept of finding the
area ratio of common height triangle
10 min
Introduce and apply the
A
b
formula 1  1
A2 b2
Examples
5 min
Classwork (Practice)
10 min
Applying
A1
l
 ( 1 )2
A2
l2
Check students’ understanding
5
Activity
Time required Remarks
Examples on harder problem
combined with finding the area ratio
of similar and common height
triangles
10 min
Problem solving
technique in solving problems
related to the area ratio of
similar and common height
triangles.
Classwork
10 min
Check students’ understanding
Discussion of the classwork
5 min
B. Use of Variation (V3)
Learning activity
Revise how to find the
area ratio of similar
triangle
Variation
Background (Invariant)
Critical feature
Common
height
triangles and
similar
triangles
Area ratio of two similar
triangles is equal to the
square of their
corresponding sides.
Date
Event
Class
Teacher
25/3
Pilot test
5B
LCN
19/3
Pretest
4A/4B
CMW/LCN
22/3
Tryout lesson
4D
FKL
23/3
Lesson study
4B
LCN
25/3
Lesson study
4A
CWM
1/4
Post-test
4A/4B
CWM/LCN
Show how to find the
area ratio by identifying
the common height
triangle and taking the
To identify and
use the property
of common
height triangles
as the critical
Area ratio of common
height triangles is equal to feature to solve
the area ratio
the ratio of their
problem.
corresponding bases.
ratio of their bases.
B. Schedule
A tryout lesson was given by FKL in 4D. Here are the observations of the tryout
lesson:
a) Most students could draw the common heights of two triangles.
A
l
b) Most students knew the formula 1  ( 1 ) 2 for similar triangles.
A2
l2
c) Most students found no difficulty on finding ratios of sides of similar triangles.
d) Some students got confusion between similar triangles and common height
triangles.
6
e) Some students did not know how to find the ratio of bases of common height
triangles.
A
l
f) Many students wrongly applied 1  ( 1 ) 2 to common height triangles.
A2
l2
g) Many students found difficulty on solving mixed type problems.
h) There was not enough time to finish the worksheet.
After the tryout lesson, we made the following changes to the lesson plan.
a) The time for the students to draw the heights of triangles should be shorter.
A
l
b) The time for revision of the formula 1  ( 1 ) 2 should be shorter.
A2
l2
c) The difference between similar triangles and common height triangles should be
emphasized.
A
l
A
b
d) The difference and time to apply the formulae 1  ( 1 ) 2 and 1  1 should
A2
l2
A2 b2
be emphasized.
e) Mixed type problems should be discussed more and in detail.
After the 4B lesson, it was suggested to revise the area ratio of similar triangles first
as the introduction. Students should then be asked to draw the height of triangles and
then draw the common height of triangles. The focus of the lesson would shift to the
area ratio of common height triangles. The flow of the lesson could then be more
smooth. Also, full size copies of worksheet should be given to students so that there
would be enough room for them to write down their steps and answers.
After the implementation of the lesson plan on 4A and 4B, we have the following
conclusions.
a) Students were interested in the recreational problems.
b) Most students understood the difference between similar triangles and common
height triangles.
c) The critical feature was the common height. Students could find the ratio of the
areas if they knew where the common height was. Therefore, it was important to
ask the students to draw the common heights.
d) Students need more training on mixed type problems so that they can break down
the problem into smaller part and separate the two concepts themselves.
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Stage 4: Evaluation
A. Result of Post-test
The teachers administered the post-test, which was parallel to the pretest, to the
students after the research lessons. Since some of the students knew the numerical
answers of the pre-test, we changed some numerical values and the way of
questioning. The results of the pre-test and post-test are summarized as follows.
Qn Ability testing
.
Percentage of correct answer or method
4A
Pre
1 Identifying the height of a triangle
4B
Post
Pre
4A & 4B
Post
Pre
Post
63.4% 68.3% 58.3% 81.1% 61.0% 74.4%
with respect to given base
2 Identifying the common height of a
pair of triangles
87.8% 95.1% 80.6% 97.3% 84.4% 96.2%
3a. Finding area ratio of similar triangles 75.6% 90.2% 55.6% 81.1% 66.2% 85.9%
3b. Finding ratio of sides of similar
triangles from area ratio.
85.4% 95.1% 63.9% 91.9% 75.3% 93.6%
.
3c. Finding area ratio of common height
triangles
51.2% 97.6% 50.0% 83.8% 50.6% 91.0%
4a. Finding similar triangles on a given
85.4% 100% 88.9% 91.9% 87.0% 96.2%
figure
4b Finding triangles with common height 59.8% 80.5% 72.2% 67.6% 65.6% 74.4%
4c Finding area ratio of common height
triangles
41.5% 80.5% 38.9% 67.6% 40.3% 74.4%
4d. Finding the ratio of the length by
using similar triangle
85.4% 90.3% 75.0% 81.1% 80.5% 85.9%
4e. Finding area ratio of common height
triangles
36.6% 75.6% 13.9% 45.9% 26.0% 61.5%
4f. Finding ratio of lengths of similar
92.7% 90.3% 63.9% 75.7% 79.2% 83.3%
triangles
4g. Finding area ratio of similar triangles. 56.1% 63.4% 30.6% 62.2% 44.2% 62.8%
5a Finding area ratio of common height
triangles
73.2% 80.5% 58.3% 70.3% 66.2% 75.6%
5b Finding triangles with common height 70.7% 97.6% 63.9% 94.6% 67.5% 96.2%
5c Finding area ratio of similar triangles 70.7% 100% 66.7% 94.6% 68.8% 97.4%
5d Finding ratio of sides of similar
triangles from area ratio.
92.7% 97.6% 83.3% 91.9% 88.3% 94.9%
5e Finding area ratio of common height
46.3% 95.1% 41.7% 73.05 44.2% 84.6%
triangles.
8
Qn Ability testing
.
Percentage of correct answer or method
4A
Pre
6 Problem solving:
Finding the area ratio by using the
common height and similar triangles.
4B
Post
Pre
4A & 4B
Post
Pre
Post
9.8% 56.1% 2.8% 29.7% 6.5% 43.6%
7. Harder Problem solving:
Finding the area ratio by using the
common height and similar triangles.
41.5%
13.5%
28.2%
Here are some observations about the post-test.
a) Nearly all students knew how to find the area ratio of similar triangles after the
lesson (Q3a, 4g, 5c).
b) The number of students knowing how to find the ratio of sides of similar triangles
from the area ratio (Q3b, 4d, 4f, 5g) had increased after the lesson.
c) The number of students knowing how to find the area ratio of common height
triangles (Q3c, 4c, 4e, 5e) had a large increase compared with the pretest. This
A
b
indicated that most students learnt the formula 1  1 for common height
A2 b2
triangles.
d) Very few students answered Q6 correctly before the lesson and about 43% of them
answered it correctly after the lesson. This indicated that many students had
difficulty on handling mixed type problem.
e) An extra mixed type problem (Q7) was added to the post-test. About 28% of
students answer it correctly. The result was satisfactory.
f) Cross-tabulation
Considering the typical question of area ratio of common height triangle (Q3c and
5e), there were 50.6% and 44.2% of students could get the right answers in the
Q3c and Q5e in the pretest respectively. There was a high percentage of students
getting the correct answers (91.0% of 3c and 84.6% of 5e) after the research
lesson. This showed that student learnt the objectives as we expected.
B. Reflection
There are several reasons why the topic was chosen. First, many students find this
topic very difficult. Second, this topic was not taught formally in previous years.
Finally this topic has not been discussed in common textbooks. Most teachers taught
this topic by showing some past HKCEE questions and many students reflected that
this topic is very difficult. Therefore we chose this topic and hope that the students
may learn how to solve this kind of problems.
9
After the lesson studies, we confirm that the critical feature of this topic is the
common height. If students can draw the common height of a pair of common height
triangles, they can find the ratio of their area. Many students find it difficult because
they cannot find the common height. Before the lesson study, many students wrongly
A
l
apply the formula 1  ( 1 ) 2 to common height triangles. After teacher’s
A2
l2
emphasizing the difference between the two formulae, the performances of the
students have been improved very much.
C. Teacher’s professional growth
These lesson studies give us an opportunity to share our ideas and methods in
teaching and learning in this specific topic. When we discussed on how to teach this
topic, we found that there are many different approaches and it is difficult to find
which one is the best. After the pretest and the tryout lesson, we are able to find the
weakness of students and the critical feature. We found that students could not solve
this kind of problems because they could not find the common height. Therefore we
emphasized on how to find the common height of a pair height triangles. We also
A
l
A
b
found that students were confused by the two formulae 1  ( 1 ) 2 and 1  1 .
A2
l2
A2 b2
Therefore we also emphasized on the difference between them.
It is agreed that this kind of lesson study is helpful and valuable to be continued for
both students and teachers if there is sufficient resources. By combining different
point of views of teachers, the critical feature can be found much easier. Teachers may
also learn from each other. The most valuable output of the lesson study is not the
particular lesson plan. By continuous practice of lesson studies and term works, both
students and teachers will be benefited.
Appendix:
1. Pretest/Post-test
2. Pretest/Post-test with solution
3. Worksheet
4. Coding
5. Sketchpad and Powerpoint files (teaching aids)
10
Shatin Tsung Tsin Secondary School
S.4 Mathematics Pre-test
Name: ______________________ Class: ______ ( _____ ) Date: _______________
1. Draw the heights with bases AB for the following triangles.
B
B
A
B
A
A
2. Draw the common heights of ABC and ACD in the following figures.
A
1
B
C
B
C
2
D
D
A
3.
a)
R
In the figure PS // QR , PS  4 and QR  6 .
S
area of PST : area of RQT
T
4
6
5
= _______________________
P
Q
b)
Given that ABC is similar to DEF ,
area of ABC : area of DEF  1 : 4 .
F
C
A
AB : DE
B
D
6
E
c)
= ______________________
A
area of ABC : area of ACD
4
5
= _______________________
B
2
C
3
D
7
11
4.
C
4
D
E
6
A
8
B
In the figure, it is given that AB  8 , AD  6 and DC  4 .
a) Write down all pairs of similar triangles.
______________________________________________________________
______________________________________________________________
b) Write down all pairs of non-overlapping (不重疊) triangles which have common
3
height.
______________________________________________________________
______________________________________________________________
c) Find area of ADE : area of DEC .
4
______________________________________________________________
d) Find CE : EB .
7
______________________________________________________________
e) Find area of ABE : area of AEC .
______________________________________________________________
f) Find CE : CB .
7
______________________________________________________________
g) Find area of DEC : area of ABC .
______________________________________________________________
5
12
5.
R
S
In the figure PS : QR  2 : 3 .
T
P
Q
a) Write down all pairs of similar triangles.
3
______________________________________________________________
b) Write down all pairs of non-overlapping triangles which have common height.
4
______________________________________________________________
c) area of PST : area of RQT = ________________________________
5
d) ST : TQ = ____________________________________________________
6
e) area of PST : area of PTQ = __________________________________
7
6.
D
C
E
In the figure, ABCD is a parallelogram.
Given that
area of CEF : area of ADE  1 : 4 .
F
A
B
Find area of CEF : area of ABCD .
______________________________________________________________
______________________________________________________________
______________________________________________________________
13
Shatin Tsung Tsin Secondary School S.4 Mathematics Pre-test
Name: ______________________ Class: ______ ( _____ ) Date: _______________
3. Draw the heights with bases AB for the following triangles.
B
B
A
B
A
A
4. Draw the common heights of ABC and ACD in the following figures.
A
1
B
C
B
C
2
D
D
A
3.
a)
R
In the figure PS // QR , PS  4 and QR  6 .
S
area of PST : area of RQT
T
4
6
5
= 4:9
P
Q
b)
Given that ABC is similar to DEF ,
area of ABC : area of DEF  1 : 4 .
F
C
A
AB : DE
B
D
6
E
= 1:2
A
c)
4
5
area of ABC : area of ACD
7
B
2
C
3
D
14
= 2:3
4.
C
4
D
E
6
A
8
B
In the figure, it is given that AB  8 , AD  6 and DC  4 .
h) Write down all pairs of similar triangles.
DEC and ABC
3
i) Write down all pairs of non-overlapping (不重疊) triangles which have common
height.
A D E and DEC
A B E and AEC
4
j) Find area of ADE : area of DEC .
3:2
k) Find CE : EB .
7
2:3
l) Find area of ABE : area of AEC .
3:2
m) Find CE : CB .
7
2:5
n) Find area of DEC : area of ABC .
5
15
4:25
5.
R
S
In the figure PS : QR  2 : 3 .
T
P
Q
f) Write down all pairs of similar triangles.
3
P S T and RQT
g) Write down all pairs of non-overlapping triangles which have common height.
P S T and PTQ ,
PQT and QTR
5
h) area of PST : area of RQT = 4:9
i)
ST : TQ =
6
2:3
7
j) area of PST : area of PTQ = 2:3
6.
D
C
E
In the figure, ABCD is a parallelogram.
Given that
area of CEF : area of ADE  1 : 4 .
F
A
4
B
Find area of CEF : area of ABCD .
EF : DE  1 : 2
area of CEF : area of CED  1 : 2
area of CEF : area of A C D 1 : 2  4  1 : 6
area of CEF : area of ABCD  1 : 12
16
Worksheet
C
A
C
C
B
B
A
B
A
Draw the height if we
Draw the height if we regard Draw the height if we
regard AB as base.
AC as base.
regard BC as base.
Theorem 1:
For any two similar figures,
area of ABC  l1 
 
area of PQR  l 2 
2
Example
1. In the figure, BC// DE, AC= 3 cm and CE= 4
cm.
Find
D
B
area of ABC

area of ADE
A
3 cm
C 4 cm
E
Class work
2. In the figure, PSQ, QXR and RYP are
straight lines.
If the area of PQR is 225 cm 2 , find the
P
area of the parallelogram SXRY.
S
Q
20 cm X
Y
30 cm
R
17
For triangles with common height,
Theorem 2:
ABD and ADC with common height,
area of ABD

area of ADC
Example
3. Given that BC : DC  5 : 1.
A
area of ABD

area of ADC
B
C
D
4. Given that AD = 3 cm and CD = 1 cm.
A
area of ABD

area of BDC
3 cm
D
1 cm
C
B
Class Work
5. (Modified HKCEE 1996) In the figure,
find area of PRX : area of QRX .
R
5
4
Ans:______________________
P
X
Q
18
6. In the figure, QX : XR  5 : 6 and
P
PY : YR  5 : 3 . If the area of PQR is 44 cm 2 ,
calculate the area of
(a) PXR
Y
Q
R
X
(b) RXY
Example
7. (Modified 1999 Q54) In the figure PQRS is a
rectangle. M is a midpoint of QR. PR and MS
intersect at N. Find area of NRS : PQMN .
P
S
N
Q
R
M
Class Work:
8. (Modified 1994 ) In the figure, PX:XQ = 1: 2,
PY:YR = 3:2. Find area of QXY : area of PQR .
Q
X
P
9. (Modified 1997 Q53.) In the figure, PQRS is a
rectangle. RSX is a straight line and PX// QS. If the
area of PQRS is 24 and Y is a point on QR such that
QY :YR = 3:1, find the area of SXY .
R
Y
X
P
Q
S
Y
R
19
10. (Modified 1996 Q35) In the figure, if
area of RSX
1
 ,
area of QRX 3
Find
S
R
area of RSX
.
area of trapezium PQRS
X
P
Q
Homework
11. In the figure. PQ// SR. PR and QS intersect at
X. Area of PXQ and that of RXS are 16 and 25
P
respectively. Find the area of the trapezium PQRS.
Q
X
S
R
12. In the figure, PQ// RS. Given the area of
PXS is 18 cm 2 and QX: XS =2:3.
Find the area of the trapezium PQRS.
S
P
X
Q
R