Comments on: Miss Loi’s Unfortunate Confluence of Factors http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors Sassy O Level Maths Tuition, Questions & Tips from Singapore's Favourite Private Tutor Sun, 05 Feb 2012 17:08:39 +0000 hourly 1 By: Miss Loi http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-38141 Miss Loi Fri, 02 Jul 2010 03:39:25 +0000 http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-38141 Some of the readers here are a bit 黑心 (<i>chao tar</i> heart) one you dunno <i>meh</i>? Some of the readers here are a bit 黑心 (chao tar heart) one you dunno meh?

]]> By: qwerty1106 http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-38119 qwerty1106 Sat, 26 Jun 2010 16:49:53 +0000 http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-38119 lol... turns out someone wanna see u chao da XD lol... turns out someone wanna see u chao da XD

]]> By: Miss Loi http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-23971 Miss Loi Tue, 07 Apr 2009 16:31:19 +0000 http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-23971 Almost a year has passed, and the thick smoke that had previously obscured her vision has finally cleared to reveal a <i>chao tar</i> Miss Loi emerging from the ruins of the fitting room. She rubbed her eyes at <b>mathslover</b>'s <a href="#comment-23869" rel="nofollow">express method</a> and lamented, at the same time, the tunnel-vision nature of her <a href="#comment-13512" rel="nofollow">original 'safe' approach</a> where she, like many stressed students in an exam, ended up having to prove that Δ<var>ABC</var> & Δ<var>PBS</var> are similar first. But if only she had stopped, taken a deep breath and a step backwards, she would've seen <b>mathslover</b>'s <em>line of symmetry</em> slicing through the diagram, and used this to quickly obtain <var>AC</var> even earlier in Part 2 as follows: <ol> <li>Δ<var>PBS</var> & Δ<var>QDR</var> are isosceles ⇒ <var>DM'</var> ⊥ <var>QR</var> & <var>M'</var> is mid-point of <var>QR</var> </li> <li>Δ<var>PBS</var> ≡ Δ<var>QDR</var> & <var>ABCD</var> is a square & <var>PQRS</var> is a rectangle ⇒ <var>MBDM'</var> is a straight line <span class="fineprint">(your line of symmetry)</span> (=90 cm) </li> </ol> ⇒ <var>BD</var> = 90 − 2(40) cm = 10 cm Since <var>AC</var> & <var>BD</var> are diagonals of square <var>ABCD</var>, <var>AC</var> = <var>BD</var> = 10 cm <img src="http://www.exampaper.com.sg/blog/wp-content/uploads/2009/04/similarity-congruency-answer-diagram-3.gif" alt="Answer Diagram" /> And since the diagonals <var>AC</var> & <var>BD</var> divide the square into four right-angled triangles with height = base = 10/2 = 5 cm, we can simply get the area via 4 × (1/2)(5)(5) = 50 cm<sup>2</sup> It must have been the smoke she breathed in ... but <span class="big">WHY DID IT TAKE ONE YEAR FOR SOMEONE TO RAISE THIS?!</span> Almost a year has passed, and the thick smoke that had previously obscured her vision has finally cleared to reveal a chao tar Miss Loi emerging from the ruins of the fitting room.

She rubbed her eyes at mathslover's express method and lamented, at the same time, the tunnel-vision nature of her original 'safe' approach where she, like many stressed students in an exam, ended up having to prove that ΔABC & ΔPBS are similar first.

But if only she had stopped, taken a deep breath and a step backwards, she would've seen mathslover's line of symmetry slicing through the diagram, and used this to quickly obtain AC even earlier in Part 2 as follows:

  1. ΔPBS & ΔQDR are isosceles
    ⇒ DM' ⊥ QR & M' is mid-point of QR
  2. ΔPBS ≡ ΔQDR & ABCD is a square & PQRS is a rectangle
    ⇒ MBDM' is a straight line (your line of symmetry) (=90 cm)

⇒ BD = 90 − 2(40) cm = 10 cm

Since AC & BD are diagonals of square ABCD, AC = BD = 10 cm

Answer Diagram

And since the diagonals AC & BD divide the square into four right-angled triangles with height = base = 10/2 = 5 cm, we can simply get the area via 4 × (1/2)(5)(5) = 50 cm2

It must have been the smoke she breathed in ... but
WHY DID IT TAKE ONE YEAR FOR SOMEONE TO RAISE THIS?!

]]> By: mathslover http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-23869 mathslover Sun, 05 Apr 2009 13:20:52 +0000 http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-23869 For part 3, can we use this approach? Since from part 1 we proved ΔPBS ≡ QDR, and given that ABCD is a square, and both points B and D are common, doesn't that imply that the line AC is the line of symmetry of that diagram? Hence by letting X be the centre of the square ABCD, we can conclude that BX = (90/2)cm - 40cm = 5cm. Without go through another round of proving similarity... For part 3, can we use this approach?

Since from part 1 we proved ΔPBS ≡ QDR, and given that ABCD is a square, and both points B and D are common, doesn't that imply that the line AC is the line of symmetry of that diagram?

Hence by letting X be the centre of the square ABCD, we can conclude that BX = (90/2)cm - 40cm = 5cm.

Without go through another round of proving similarity...

]]> By: Miss Loi http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-13636 Miss Loi Fri, 19 Sep 2008 16:18:23 +0000 http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-13636 <b>Kiddo:</b> To be honest, as a <i>chao tar</i> Miss Loi was squeezing her *ahem* 30 cm<sup>2</sup> figure through the 50 cm<sup>2</sup> window, she was sort of expecting some students to protest: <blockquote>Miss Loi! Cannot! The textbook says that the ratios for similar triangles are only for the sides! They never say their heights are of the same ratio too!</blockquote> For this, Miss Loi's answer would then be: <blockquote>If Δ<var>PBS</var> and Δ<var>ABC</var> are similar, shouldn't Δ<var>PBM</var> and Δ<var>ABX</var> be similar too?</blockquote> Kiddo: To be honest, as a chao tar Miss Loi was squeezing her *ahem* 30 cm2 figure through the 50 cm2 window, she was sort of expecting some students to protest:

Miss Loi! Cannot! The textbook says that the ratios for similar triangles are only for the sides! They never say their heights are of the same ratio too!

For this, Miss Loi's answer would then be:

If ΔPBS and ΔABC are similar, shouldn't ΔPBM and ΔABX be similar too?

]]> By: kiddo http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-13629 kiddo Fri, 19 Sep 2008 14:14:07 +0000 http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-13629 Thanks Miss Loi for thy clarification. I've just learnt something obvious which may not be explicitly stated in textbooks. From your working for 2nd part of part 2 ie. finding length of AC, notice that ratio of corresponding heights (BX & BM) of similar triangles is same as ratio of corresponding sides (AC & SP). This knowledge can be applied when solving questions involving heights of similar triangles. Thanks Miss Loi for thy clarification. I've just learnt something obvious which may not be explicitly stated in textbooks.

From your working for 2nd part of part 2 ie. finding length of AC, notice that ratio of corresponding heights (BX & BM) of similar triangles is same as ratio of corresponding sides (AC & SP). This knowledge can be applied when solving questions involving heights of similar triangles.

]]> By: kiddo http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-13512 kiddo Wed, 17 Sep 2008 04:28:17 +0000 http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-13512 Miss Loi won't reveal answer if we don't attempt so.. <div class="highlight"> * A <i>chao tar</i> Miss Loi rises from the ashes ... to present the following diagram: * <img src="http://www.exampaper.com.sg/blog/wp-content/uploads/2008/09/similarity-congruency-answer-diagram-2.gif" alt="Similarity Congruency Answer Diagram II" /> </div> 2.Last hint given, angle SPB = 45° Angle PMB = 90° (?) So angle PBM = 180 - 45 - 90 = 45° (sum of angles in triangle), so we've isosceles triangle PBM Since M is midpoint of PS, PM = MS = 80/2 = 40cm By isosceles triangle, BM = PM = 40cm <div class="highlight"> Yes, when answering longish multi-part <a href="/tag/similarity" rel="nofollow">similarity</a> & <a href="/tag/congruency" rel="nofollow">congruency</a> question like this one, it's always a good idea to use whatever proofs you've already established in earlier parts. So from <a href="/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-8236" rel="nofollow">Part 1</a>, we have: ∠<var>ABC</var> = 90° & ∠<var>PSB</var> = 45° ⇒ ∠<var>SPB</var> = 45° (sum of angles of Δ<var>PSB</var>) ⇒ Δ<var>PSB</var> is isosceles (∠<var>SPB</var> = ∠<var>PSB</var> = 45°) Since Δ<var>PSB</var> is isosceles and given <var>M</var> is mid-point of <var>PS</var> ⇒ <var>BM</var> bisects ∠<var>PBS</var> ⇒ ∠<var>PBM</var> = 90°/2 = 45° ⇒ Δ<var>PBM</var> is also isosceles (∠<var>MPB</var> = ∠<var>PBM</var> = 45°) ⇒ <var>PM</var> = <var>BM</var> = 40 cm </div> *Extend PC to RS & let intersection point be T. Angle PTS = 45°(corresponding angles,BM//TS?) Triangle PTS is isosceles since angle SPT=PTS so TS = PS = 80cm RT = RS - TS = 90 - 80 = 10cm DB = RT = 10cm (?) so AC = 10cm as diagonals of square are equal in length.* * * is crap. Also didn't use BM at all in working.XD <div class="highlight"> * <i>Deng Deng Deng Deng</i> ... * When the magic word <em>Hence</em> is uttered in a <a href="/questions/a-maths/differentiation-integration-a-hence-question" rel="nofollow">Hence Question</a>, thou MUST follow the Holy Word and refrain from resorting to your pagan methods, even though your final answer is correct! So in order to use <var>BM</var> to find <var>AC</var>, you'll first have to proof that Δ<var>PBS</var> and Δ<var>ABC</var> are similar: We already know that ∠<var>PBS</var> = ∠<var>ABC</var> = 90° Since <var>ABCD</var> is a square, ⇒ <var>AC</var> bisects ∠<var>BAD</var> & ∠<var>BCD</var> ⇒ ∠<var>BAD</var> = ∠<var>BCD</var> = 90°/2 = 45° ⇒ ∠<var>BAC</var> = ∠<var>PSB</var> & ∠<var>BCA</var> = ∠<var>SPB</var> (all angles equal! - highlighted in red in diagram) ⇒ Δ<var>PBS</var> and Δ<var>ABC</var> are similar (<strong>AAA</strong>) Armed with this revelation, we have the following ratios: <m>AC/SP = BX/BM</m> (where <var>X</var> is the midpoint of <var>AC</var>) And since we've also proven in <a href="/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-8236" rel="nofollow">Part 1</a> that Δ<var>PBS</var> ≡ Δ<var>QDR</var>, and given once again that <var>ABCD</var> is a square, can you see from the diagram that <var>BX</var> = 5cm? So ... <m>AC/SP = BX/BM</m> <m>AC/80 = 5/40</m> ⇒ <var>AC</var> = 10cm ;) </div> 3.Area of square ABCD = 2 x (½ x 10 x 5) = 50cm<sup>2</sup> (Reason:diagonals of square bisect each other at right angles.) <div class="highlight"> For Part 3, there's no holy <em>Hence</em> word to command what approach you should take, so your answer is fine :) Just to highlight another approach should you be required to use <var>AC</var> from Part 2: Since <var>ABCD</var> is a square, let <var>AB</var> = <var>BC</var> = <var>x</var> Using Pythagoras' Theorem, <var>AB</var><sup>2</sup> + <var>BC</var><sup>2</sup> = <var>AC</var><sup>2</sup> ⇒ <var>x</var><sup>2</sup> + <var>x</var><sup>2</sup> = 10<sup>2</sup> (using value of <var>AC</var> obtained in Part 2) ⇒ 2<var>x</var><sup>2</sup> = 100 ⇒ Area of <var>ABCD</var> = <var>x</var><sup>2</sup> = 50 cm<sup>2</sup> ;) </div> Awaiting enlightenment... Miss Loi won't reveal answer if we don't attempt so..

* A chao tar Miss Loi rises from the ashes ... to present the following diagram: *

Similarity Congruency Answer Diagram II

2.Last hint given, angle SPB = 45°
Angle PMB = 90° (?)
So angle PBM = 180 - 45 - 90 = 45°
(sum of angles in triangle),
so we've isosceles triangle PBM
Since M is midpoint of PS, PM = MS = 80/2 = 40cm
By isosceles triangle, BM = PM = 40cm

Yes, when answering longish multi-part similarity & congruency question like this one, it's always a good idea to use whatever proofs you've already established in earlier parts.

So from Part 1, we have:

∠ABC = 90° & ∠PSB = 45°
⇒ ∠SPB = 45° (sum of angles of ΔPSB)
⇒ ΔPSB is isosceles (∠SPB = ∠PSB = 45°)

Since ΔPSB is isosceles and given M is mid-point of PS
⇒ BM bisects ∠PBS ⇒ ∠PBM = 90°/2 = 45°
⇒ ΔPBM is also isosceles (∠MPB = ∠PBM = 45°)
⇒ PM = BM = 40 cm

*Extend PC to RS & let intersection point be T.
Angle PTS = 45°(corresponding angles,BM//TS?)
Triangle PTS is isosceles since angle SPT=PTS
so TS = PS = 80cm
RT = RS - TS = 90 - 80 = 10cm
DB = RT = 10cm (?)
so AC = 10cm as diagonals of square are equal
in length.*

* * is crap. Also didn't use BM at all in working.XD

* Deng Deng Deng Deng ... *

When the magic word Hence is uttered in a Hence Question, thou MUST follow the Holy Word and refrain from resorting to your pagan methods, even though your final answer is correct!

So in order to use BM to find AC, you'll first have to proof that ΔPBS and ΔABC are similar:

We already know that ∠PBS = ∠ABC = 90°
Since ABCD is a square,
⇒ AC bisects ∠BAD & ∠BCD
⇒ ∠BAD = ∠BCD = 90°/2 = 45°
⇒ ∠BAC = ∠PSB & ∠BCA = ∠SPB (all angles equal! - highlighted in red in diagram)
⇒ ΔPBS and ΔABC are similar (AAA)

Armed with this revelation, we have the following ratios: AC/SP = BX/BM (where X is the midpoint of AC)

And since we've also proven in Part 1 that ΔPBS ≡ ΔQDR, and given once again that ABCD is a square, can you see from the diagram that BX = 5cm?

So ... AC/SP = BX/BM
AC/80 = 5/40 ⇒ AC = 10cm ;)

3.Area of square ABCD = 2 x (½ x 10 x 5) = 50cm2
(Reason:diagonals of square bisect each other at
right angles.)

For Part 3, there's no holy Hence word to command what approach you should take, so your answer is fine :)

Just to highlight another approach should you be required to use AC from Part 2:

Since ABCD is a square, let AB = BC = x

Using Pythagoras' Theorem,
AB2 + BC2 = AC2
⇒ x2 + x2 = 102 (using value of AC obtained in Part 2)
⇒ 2x2 = 100
⇒ Area of ABCD = x2 = 50 cm2 ;)

Awaiting enlightenment...

]]> By: clarion-x http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-9398 clarion-x Mon, 09 Jun 2008 06:31:33 +0000 http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-9398 Paiseh lah... No time... XD Paiseh lah... No time...
XD

]]> By: Miss Loi http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-9396 Miss Loi Mon, 09 Jun 2008 05:06:37 +0000 http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-9396 Hello <b>pooboi</b>! How come nobody informed Miss Loi that she made a guest appearance in last year's O Level paper?! In any case, the first part of congruency questions normally requires you to <em>prove</em> (if it carries > 2 marks), or simply <em>identify/state</em> the congruent triangles (if it carries 1-2 marks). Speaking of which, parts 2-3 of this question are still unsolved. <b>Clarion</b> you do homework do halfway!!! Tsk tsk. Hello pooboi! How come nobody informed Miss Loi that she made a guest appearance in last year's O Level paper?!

In any case, the first part of congruency questions normally requires you to prove (if it carries > 2 marks), or simply identify/state the congruent triangles (if it carries 1-2 marks).

Speaking of which, parts 2-3 of this question are still unsolved. Clarion you do homework do halfway!!! Tsk tsk.

]]> By: pooboi http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-9372 pooboi Sun, 08 Jun 2008 03:10:14 +0000 http://www.exampaper.com.sg/questions/e-maths/miss-lois-unfortunate-confluence-of-factors#comment-9372 wah! isn't part a similar to last year's GCE 'O' question?! hahaha! wah! isn't part a similar to last year's GCE 'O' question?! hahaha!

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