## Thursday, 23 October 2014

### The Outstanding Team Award goes to....

Group 3 with 95 points

## Sunday, 19 October 2014

### Australian Mathematics Competition 2014

Dear S1-01

Here's the summary of your class' performance:

 High Distinction & Prudence Award Winner Keith Tang Yi Heng Distinction Lucas Obe Chiu Chen Ning Ho Matsuoka Evangelista Miyuki Sophia Credit Haw Jin Yu Clement Yap Wagle Prateek Shantanu Noel Lim Siddhartha Jaruhar Xavien Teo Ming Rui Biswakarma Jaishree Roy Tay Proficiency Johan Soo Durwa Walimbe Weisan Nishtha Gupta Bhojwani Ria Manoj Everi Yeo Joshua Wee Yu Shuen Kaung Htet Wai Yan Tang Jun Mei Shanice Tham Kai Heng Participation Kuan Yew Chong Wong Min Quan

## Tuesday, 7 October 2014

### Concluding Year 2014...

Dear S1-01

It has been a pleasure to be with you on your maiden learning journey in SST :)
The closure of this academic year would a good time for us to reflect and plan for the following year.

We would like to hear your experience as the feedback would be taken into account when we shape your learning experience, as well as your juniors'.

## Saturday, 27 September 2014

### 2-to-3 Quiz (Revision) Construction

Question 1

To construct the figure:
(i) Measure and draw line AB using ruler.

(ii) Measure and draw angle ABC using protractor.

(iii) Stretch the 2 arms of the compass to measure 5.2 cm and mark an arc on the line BC. Label it point C.

Here are 2 ways of drawing the parallel line DC
(iv) Use a protractor to draw angle BCD
OR
(iv) Use a set-square and ruler to draw line CD

(v) Stretch the 2 arms of the compass to measure 3.8 cm, place the sharp end at "C" and mark an arc on the line DC. Label it point D.

(vi) Join A and D to complete the figure.

Question 2

Three properties to take note of:

• Diagonals of the rhombus bisect each other
• Diagonals of the rhombus are perpendicular to each other
• Opposite angles are equal

### [Someone asked...] The difference between Approximation and Estimation...

Estimation involves calculation, whereas Approximation is about rounding off a number.
• The rounding off of a number can be carried out before or after the calculation.

Consider the following scenario:
The length of the door is 1.95 m and its height is 2.06 m (assuming that the dimensions given are exact).

(a) Find the area of the door and round the answer off to 2 significant figures.
The actual area of the door is 1.95 m x 2.06 m = 4.017 m^2.
Answer = 4.0 m^2 (to 2 sig fig) [Ans]

The step to round off a number (to present the final answer) is an approximation.
There is no mention of estimation in question (a), so we use the given values to compute the area first, then round off as required.

(b) Estimate the area of the door to 2 significant figures.

The purpose of doing 'estimate' or 'estimation' is to enable us carry out calculation quickly and when we don't need an exact value. A rough figure should give us a sense of the actual value. It should not be too far off from the actual value.

To estimate, we will round the numbers off to the required "degree of accuracy" (in this case, 2 significant figures) first before doing any calculation. In other words, we are using the approximated numbers to do calculation.

Area of the door ≈ 2.0 m x 2.0 m <<<<< This is the 'estimation' step when the approximated values are used for calculation
= 4.0 m^2 [Ans]

Since 1.95 m ≈ 2.0 m (2 sig fig) and  2.06 m ≈ 2.0 (2 sig fig)

Exception
When doing Estimation, we round off the values based on the degree of accuracy given.

However, there is an exception when dealing with roots because by rounding it off to 1 SF, 2 SF, etc. may not land itself nicely to be rooted.

To estimate square root of 26.77, it does not help us to figure out what number it's 'near' to if we round it off to 1 SF (i.e. 30) or 2 SF (i.e. 27).
However, we know that 25 (which is a perfect square) is quite close to it; so, we will estimate it to be square root of 25, and get the answer "5".
If you key square root 26.77 in the calculator, your answer should be quite close to 5.

To extend/ combine the above....
If the question asks: Estimate 150.5677 x "square root of 8.343" to 2 sig fig
Then we'll have 150.5677 x "square root of 8.343" ≈ 150 x "square root 9"
= 150 x 3
= 450 [Ans]

150.5677 when rounded off to 2 sig fig = 150
square root of 8.343 will be square root of 9 (which is the closest perfect square).

## Monday, 22 September 2014

### Revision (2-to-3 Quiz): Linear Graph Question 1 - It should be...

(a) Recall, on the graph paper, each 'big' square with the 'darkened' green link measures 2 cm x 2 cm.

(b) Substitute the given values of x and y to find the unknowns.

(c) Remember to check the domain for the plot.

(d) Things that must be present when plotting the graph:

• label x and y axes
• markings on the axes must be clearly indicated/ written (including the origin)
• label the line with the given equation
• plot the points with crosses clearly

(e) To read points, the working is represented by the dotted lines.

### Revision (2-to-3 Quiz): Linear Graph Question 1

Refer to Handout 2-to-3 Quiz (10) Question 1
For discussion...

## Sunday, 21 September 2014

### Content Pages - 2nd Semester

Content Page for the individual topics (2nd semester) are now available via the GoogleSite.

• Click HERE to view Summary [hardcopy will be given]

### [Someone asked...] When simplifying expressions... do we factorise?

When we are asked to simplify expressions, we would usually expand to work on the like terms. However, there are occasions when we need to look out for "patterns".

Here are some possible scenarios:

1. If no variables are involved in the denominator of any algebraic fractions, we can just expand to simply.

2. If there is variable in the denominator, we'll need to see how complex it is.

In the following example, you notice that the denominator is very simple - it has only a numerical value multiplied to "x".
• Now, we simply need to simplify the numerator like what we did (usually).
• When we reach Line 2, then check if there's any common factor amongst the terms in the numerator.
• If there is, 'take out' the common factor, just in case it can be 'cancelled' with the denominator.
• In this case, you notice that in Line 3, we have "2x" in the numerator, which we can reduce further with the denominator "6x", resulting in what we see in Line 4.

3. In the following, you notice that the denominator seems a bit 'more complex', and there seems to be a common factor between the terms.

• In addition, the numerator has 2 'brackets', hinting a possibility of 'reducing' with some terms in the denominator. So, do not expand the 'brackets' in the numerator yet.
• In Line 2, we 'take out' the common factor of the denominator.
• Now check if there's any common factor between the numerator and the denominator., You would notice (x+2) is the common factor. 'Cancel' them and we'll get Line 3.
• Now, simplify and the final answer is as shown in Line 4.

### [Someone asked...] How to form that INEQUALITY?

Someone asked: How to form the inequality of the following problem
(Fairfield Methodist Secondary End-Of-Year 2013 Paper 2)

Here's how you could approach the problem step-by-step:

### End-of-Year Revision - Suggested Solutions

Please refer to the GoogleSite for the suggested solution/ working for the revision papers.

## Friday, 19 September 2014

### Term 4 Week 2/3 Consultation

Dear S1-01

As you should have moved into the final stage of the exam preparation in Term 4 Week 2, instead of conducting remedial lessons, I've created several consultation slots so that you can see me to go areas that you are not sure, e.g. questions in the revision papers.

• 22 Sep (Monday) 2.30 pm to 4.30 pm
• 24 Sep (Wednesday) 1.45 pm to 2.15 pm
• 29 Sep (Monday) 11 am to 12 noon
• 30 Sep (Tuesday) 12 noon to 1 pm
You may give me a call (@419, outside the staff room) or send me an email alert.
Venue: Outside Staff Room

## Tuesday, 9 September 2014

### [Someone asked....] Number of sides of a polygon if the exterior angle is a reflex angle...?

How to find "The number of sides a regular polygon has if its exterior angle is ≥180º?

In the syllabus, we focus on convex polygons.
One of the properties that all convex polygons have would be the sum of Interior and Exterior angles is 180º.
Hence, the size of each exterior angle cannot be more than 180º.

Below are 2 examples of polygons with exterior angles being acute angle and obtuse angle.

## Thursday, 4 September 2014

### Support Programme during Sep Holidays

Hi Everybody

I'll be in the school (during the Sep Holidays) to run the Maths Support Programme for some of your classmates on
• 8 Sep 2014 (Monday)
• 9.30 am to 11.30 am
• Venue: Info Hub (Level 4/5)
• Attire: School uniform
You may turn up if you wish to...

What are we doing:
Attempt a Practice Paper from another school; consultation if you would like to clarify your doubts face-to-face.

If you intend to turn up, things to bring:
1. Long Ruler, Eraser & Sharp pencil - for graph plotting; Graph paper
2. Calculator
3. Writing materials

Do let me know if you are unable to turn up so that I would prepare enough materials for you.

Who's attending the support programme:

 7 WALIMBE DURWA NITIN 11 JOSHUA WEE YU SHUEN 12 KAUNG HTET WAI YAN 16 NOEL LIM

## Monday, 1 September 2014

### Pythagoras' Theorem: How would Romeo meet Juliet?

[Post activity discussion]

Some of you enquired the "complete" set of solution (after the class discussion) - i.e. the various possibilities that arise from various assumptions.

Here are the possibilities and assumptions that we need to be clear before solving the problem:

In scenario 2, some of you suggested using nets (something you learnt in primary school).
Here, it's good to draw the net and write the dimensions clearly to ensure the correct numbers are used for calculation.

Look, there are 2 possible 'nets' that could give the shortest path.
Do NOT assume the 2 nets give the same "diagonal" length.
[Do not apply "seeing is believing"]
Always compute the numbers to check.

The 3rd scenario was the one that we spent a significant amount of time to discuss in class - Refer to the "Romeo & Juliet" box that I brought to the class to help you visualise better.
Here, you would have to apply Pythagoras Theorem twice.

## Saturday, 30 August 2014

### S1-01 Group 3 Sub Group Number Pattern-Question 5

Our suggested solution :

Group Members: Jai, Nishtha and Keith

## Tuesday, 26 August 2014

### Number Pattern Activity - Group 1 Question 4

Group 1: (Sub-Group) Jin Yu, Roy, and Durwa

### Group Work: Number Pattern Tasks

Each group is to be re-organised into 2 sub-groups.

Each "sub-group" is given a task.

1. Take a photo of your suggested solution - make sure the image is clear & big enough.

2. Take a short video clip - no longer than 2 minutes, post it in Youtube - name it something similar to "S1-01-Group 5 Q4"

3. In your Maths blog, in the same blog:
• Title of Post: Number Pattern Activity - Group 5 Question 4
• Post a photograph of your suggested solution
• Embed the video clip just below the photograph
• Include the name of all the members
On top of that, you've been invited to the G+ Event.

### Update: Who have already responded...

For those who have not responded to the survey-quiz,
Click HERE to answer the Quiz.
 Submitted by Group Score Johan Soo 1 8 Joshua 1 8 Roy 1 7 Sophia 1 9 Wong Min Quan 1 8 Bhojwani Ria Manoj 2 5 Kaung Htet Wai Yan 2 8 Tham Kai Heng 2 5 Chiu Chen Ning 3 9 Jaishree 3 7 Lucas 3 7 Prateek 4 9 Yew Chong 4 9

### Transformation: From Circle to Cone

Some of you will find the following demonstration useful - to see how a 'circle' is transformed into a 'cone'.
Watch from 1:17 onwards

## Saturday, 23 August 2014

### Investigating: Volume of Cone & Cylinder

The volume of a cone fills up one-third of the volume of cylinder, i.e. we need three cones to fill up a cylinder.

### Investigating: Volume and Surface Area of Sphere

What's the relationship between the volume of a cylinder and a sphere

To examine the total surface area of a sphere

## Wednesday, 20 August 2014

Edit:

Original:

### [Homework 20140820] Pythagoras' Theorem - A Challenge

You are going to find the volume of the following figure.
Hint: Pythagoras Theorem

• Take a picture of your working and post it in the Maths blog.
• Subject title: Pythagoras' Theorem - A Challenge (by Your Name, Group no.)

## Tuesday, 19 August 2014

### [T3W8] Support Programme (Group 1)

 Geometrical Construction 10 JOHAN SOO T3W8 Support Prog 19 August 2014 (Tues) 2.15 pm to 3.15 pm S1-04 classroom 11 JOSHUA WEE YU SHUEN 12 KAUNG HTET WAI YAN 16 NOEL LIM 17 SIDDHARTHA JARUHAR 18 TAY YONG KIAT, ROY
Remember to bring along your construction set.

## Sunday, 17 August 2014

### Geometry with Donald!

The short clip beautifully connects various concepts that we've learnt and will be learning in the next couple of weeks.

Look out for the concepts that you are familiar with (e.g. Golden Ratio)

Enjoy your exploration with Donald Duck!

### [Someone asked....] Solving Algebraic Word Problems

Sometime ago, someone (from another class) asked how to solve the following word problem (from the Textbook).
Here's the step-to-step explanation on how to tackle this question.

• Click at the slides to proceed.
• The working in each slide addresses to a part of the problem (highlighted in red).

## Saturday, 16 August 2014

### Homework on Basic Geometry

Dear S1-01

A gentle reminder that you were given 2 handouts on Thursday.
Do remember to complete the assignments and submit them on coming Monday (18 August 2014).

## Tuesday, 12 August 2014

### Little Red Riding Hood needs help!

Diagram constructed by Nishtha

### Where to meet?

The location of the houses of three friends, Antonio, Benedict and Celine happened to form an equilateral triangle. They meet quite often to catch up with each other; however, none of them want to invite their friends home. To be fair, they decided to meet at a place that is most 'central' to all.

• What would you suggest?
• How would you do it if you have their houses marked on a map already?

Note: There is a typo-error in the above diagram, it should be "Benedict"

Diagrams constructed by Sophia

## Wednesday, 6 August 2014

Dear S1-01

### Submission of Assignments

Many of you have not submitted the following assignments (in the last 2 weeks).
While these topics would be tested by Thursday (Level Test), you still have to hand it as you would need to demonstrate your understanding and mastery at the end-of-year exam; and you need to know them well when you move on to other levels.

You must hand in these owed assignments by next Tuesday (12 August) by the last period; otherwise, you will have to stay back after school to complete and submit your work.

For those who failed to submit the assignments, you will stay back after school to complete & submit them (12 August; 2 pm to 3.30 pm @ Tutorial Room 1)

## Monday, 4 August 2014

### Alternative Assessment 2: Status of Submission

Up to 4 August 2014, 0030h - the following have already submitted their alternative assessments:

 Name in full Class Register number Bhojwani Ria Manoj S1-01 1 Everi Yeo Yi Yun S1-01 3 Ho Matsuoka Evangelista Miyuki Sophia S1-01 4 Nishtha Gupta S1-01 5 Tang Jun Mei, Shanice S1-01 6 Walimbe Durwa Nitin S1-01 7 Chiu Chen Ning S1-01 8 Haw Jin Yu S1-01 9 Joshua Wee S1-01 11 Keith tang S1-01 13 Kuan Yew Chong S1-01 14 Lucas Obe S1-01 15 Roy Tay S1-01 18 Xavien S1-01 19 Tham Kai Heng S1-01 20 Wagle Prateek Shantanu S1-01 21 WONG MIN QUAN S1-01 22 Wei-san Wee S1-01 24