geometry – Singapore Maths Tuition

Secondary Four O Level Maths Tuition (E Maths & A Maths Tuition) at Bishan starting in 2014!

Maths Group Tuition starting in 2014!

https://mathtuition88.com/group-tuition/

Secondary Four O Level Maths Tuition (E Maths & A Maths Tuition) at Bishan starting in 2014!

Location: Block 230 Bishan Street 23 #B1-35 S(570230)

Schedule: Monday 7pm-9pm

Thursday 7pm-9pm

(Perfect for students who have CCA in the afternoon, or students who want to keep their weekends free.)

Google Maps: http://goo.gl/maps/chjWB

Email: mathtuition88@gmail.com

Phone: 98348087

Mr Wu’s O Level Certificate (with A1 for both Maths). Mr Wu sincerely wishes his students to surpass him and achieve their fullest potential.

Despite being in the Gifted Education Programme (GEP), Mr Wu is just an ordinary Singaporean. His secret to academic success is hard work and the Maths Techniques he has discovered by himself while navigating through the education system.

Directions to Bishan Tuition Centre:

A) Via BISHAN MRT (NS17/CC15)

(10 minutes by foot OR 2 bus stops from Junction 8. From J8, please take bus numbers, 52, 54 or 410 from interchange. The centre is just after Catholic High School, just beside Clover By-The-Park condominium.

Other landmarks are: the bus stop which students alight is in front of Blk 283, where Cheers minimart and Prime supermarket are.)

It’s one street away from Raffles Institution Junior College (RIJC), previously known as Raffles Junior College (RJC). It’s also very convenient for students of Catholic Junior College (CJC), Anderson Junior College (AJC), Yishun Junior College (YJC) and Innova Junior College (IJC).

Other secondary schools located near Bishan are Catholic High School, Kuo Chuan Presbyterian Secondary School, and Raffles Institution (Secondary).

Tips on attempting Geometrical Proof questions (E Maths Tuition)

Tips on attempting Geometrical Proof questions (O Levels E Maths/A Maths)

1) Draw extended lines and additional lines. (using pencil)

Drawing extended lines, especially parallel lines, will enable you to see alternate angles much easier (look for the “Z” shape). Also, some of the more challenging questions can only be solved if you draw an extra line.

2) Use pencil to draw lines, not pen

Many students draw lines with pen on the diagram. If there is any error, it will be hard to remove it.

3) Rotate the page.

Sometimes, rotating the page around will give you a fresh impression of the question. This may help you “see” the way to answer the question.

4) Do not assume angles are right angles, or lines are straight, or lines are parallel unless the question says so, or you have proved it.

For a rigorous proof, we are not allowed to assume anything unless the question explicitly says so. Often, exam setters may set a trap regarding this, making the angle look like a right angle when it is not.

5) Look at the marks of the question

If it is a 1 mark question, look for a short way to solve the problem. If the method is too long, you may be on the wrong track.

6) Be familiar with the basic theorems

The basic theorems are your tools to solve the question! Being familiar with them will help you a lot in solving the problems.

Hope it helps! And all the best for your journey in learning Geometry! Hope you have fun.

“There is no royal road to Geometry.” – Euclid

Animation of a geometrical proof of Phytagoras... — Animation of a geometrical proof of Pythagoras theorem (Photo credit: Wikipedia)

O Level 2007 E Maths Paper 2 Q3 Solution (Geometry Question)

(a)
$\angle DAB=\angle DCB$
(opposite angles of parallelogram)

$\displaystyle\angle PAD=\frac{1}{2}\angle DAB=\frac{1}{2}\angle DCB=\angle RCB$
(shown)

(b)
$AD=BC$

$\angle PAD=\angle RCB$
(from part a)

$\angle ADC=\angle ABC$
(opposite angles of parallelogram)

$\displaystyle\angle ADP=\frac{1}{2}\angle ADC=\frac{1}{2} ABC=\angle RBC$

Thus, triangles ADP and CBR are congruent (ASA).

(c)(i)
$\angle ADC+\angle DAB=180^\circ$
$\angle DAP+\angle ADP=90^\circ$

Considering the triangle ADP,
$\angle DPA=180^\circ-90^\circ=90^\circ$
(shown)

(ii)
$\angle DAB+\angle ABC=180^\circ$

$\angle BAQ+\angle ABQ=90^\circ$

Considering the triangle ABQ,
$\angle PQR=180^\circ-90^\circ=90^\circ$
(shown)

Challenging Geometry E Maths Question — St Andrew’s Sec 3 Maths Tuition Question

Question:

ABCD is a rectangle. M and N are points on AB and DC respectively. MC and BN meet at X. M is the midpoint of AB.

(a) Prove that $\Delta CXN$ and $\Delta MXB$ are similar.

(b) Given that area of $\triangle CXN$ : area of $\triangle MXB$ =9:4, find the ratio of,

(i) DN: NC

(ii) area of rectangle ABCD: area of $\triangle XBC$ . (Challenging)

[Answer Key] (b) (i) 1:3

(ii) 20:3

Suggested Solutions:

(a)
$\angle MXB=\angle NXC$ (vert. opp. angles)

$\angle MBX = \angle XNC$ (alt. angles)

$\angle BMX = \angle XCN$ (alt. angles)

Therefore, $\Delta CXN$ and $\Delta MXB$ are similar (AAA).

(b) (i) $\displaystyle\frac{NC}{BM}=\sqrt{\frac{9}{4}}=\frac{3}{2}$

Let $BM=2u$ and $NC=3u$

Then $DC=2\times 2u=4u$

So $DN=4u-3u=u$

Thus, $DN:NC=1u:3u=1:3$

(ii)

We now have a shorter solution, thanks to a visitor to our site! (see comments below)

From part (a), since $\Delta CXN$ and $\Delta MXB$ are similar, we have $MX:XC=2:3$

This means that $MC:XC=5:3$

Thus $\triangle MBC:\triangle XBC=5:3$ (the two triangles share a common height)

Now, note that $\displaystyle\frac{\text{area of }ABCD}{\triangle MBC}=\frac{BC\times AB}{0.5 \times BC \times MB}=\frac{AB}{0.5MB}=\frac{2MB}{0.5MB}=4$

Hence area of $ABCD=4\times\triangle MBC$

We conclude that area of rectangle ABCD: area of $\triangle XBC=4(5):3=20:3$

Here is a longer solution, for those who are interested:

Let area of $\triangle XBC =S$

Let area of $\triangle MXB=4u$

Let area of $\triangle CXN=9u$

We have $\displaystyle\frac{S+9u}{S+4u}=\frac{3}{2}$ since $\triangle NCB$ and $\triangle CMB$ have the same base BC and their heights have ratio 3:2.

Cross-multiplying, we get $2S+18u=3S+12u$

So $\boxed{S=6u}$

$\displaystyle\frac{\triangle BCN}{\triangle BDC}=\frac{3}{4}$ since $\triangle BCN$ and $\triangle BDC$ have the same base BC and their heights have ratio 3:4.

Hence,

$\begin{array}{rcl} \triangle BDC &=& \frac{4}{3} \triangle BCN\\ &=& \frac{4}{3} (9u+6u)\\ &=& 20u \end{array}$

Thus, area of $ABCD=2 \triangle BDC=40u$

area of rectangle ABCD: area of $\triangle XBC$ =40:6=20:3