[Enrich20170101SQT] Calculating Square Roots by Hand

Introduction

          Happy New Year to our readers!  I wish this year will be a fruitful one for everybody.

          Today, I will illustrate how to calculate square roots by hand, using  54 756  as an example.  It is similar to long division, but has some modifications.

Solution

          Starting from the right, pair up the digits.

          2×2 = 4  is the nearest perfect square to  5.  Subtract and bring down the next two digits, giving  147.

          Double the digit  2  to get  4.  Think:  ? × 4?  gives  147  or nearest possible value.  We have 3×43 = 129.

          Subtracting and bringing down the next two digits gives  1856.  Replicate the digit  4  on the left and double the digit  3,  giving  46.

          Now think:  ? × 46?  gives  1856  or nearest possible value.  It turns out that  4 × 464 gives exactly  1856.  We are done!  The square root of  54 756  is  234.

How does it work?

          This relies on the algebraic identity  (10a + b)² = 100a² + 20ab + b², the right-hand expression is equal to   100a² + (20a + b)b.  For example, at stage 4, we have  a = 23,  b = 4  and  (20a + b) = 464.

          Did you learn something today?

[OlymLSec_20160118PPPC] A Square Proof by Contradiction

Question

Explanation

     If  a + b = 11,  then  2ab = (a + b)² – (a² + b²) = 121 – 100 = 21.  But  2ab  is an even number, whereas  21  is odd.  This is a contradiction.  So (B) is impossible.  ©

Remarks

     Short and sweet isn’t it?  This uses the square-of-sum identity   (a + b)² = a² + 2ab + b².  I used the tactic of assuming the answer is correct  [H08]  and showing that this leads to something nonsensical [H05].  So the original assumption must be wrong.  This is called “proof by contradiction” or reductio ad absurdum (in Latin).

     By the way, the correct answer option is (E) from the Pythagorean Triplet   8² + 6² = 10²  with  {a, b} = {8, 6}.  The question seems to be taken from some Kangaroo mathematics competition.

Heuristics Used

H05. Work backwards

H08. Make suppositions

H09. Restate the problem in another way

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* Lower Secondary Mathematics competition

* GCE ‘O’ Level “Elementary” Mathematics (challenge)

* other syllabuses that involve whole numbers and Pythagorean triplets

* any precocious or independent learner who loves a challenge

[Pri20151117MSAS] MCQ tactic for the Area of a Hollow Square

Question

Solution 1

     Width of the smaller square  WX = (156 ¸ 4) cm   = 39 cm

     Width of the larger square     AB = (39 + 2´8) cm = 55 cm

     Area = (55² – 39²) cm² = 1504 cm² 

     Ans: (1)

Solution 2

     This is a Multiple Choice Question (MCQ).  Observe that the shaded area is an even number, because it is 8 cm width all around.  Since 1504 is the only even number among the options, (1) is the correct choice.

Remark

     No tedious calculation is needed!

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels
* Primary School Mathematics
* other syllabuses that involve areas and perimeters
* anyone who loves his/her brain tickled

[AM_20150621TGCARF] Sum of sin and cos the cost of tan?

Question

Introduction

     This is a bonus Maths II (rough equivalent of Additional Mathematics syllabus) question from a test by Hwa Chong Institution (HCI) this year 2015.  Independent schools in Singapore like HCI are free to set their own curricula, but they usually end up covering slightly more than the mainstream curriculum, since their students also take the national examinations.  For their internal tests and exams, they can set bonus questions.  These are harder but optional questions that students can attempt and if they are successful, the bonus marks can be added to their normal marks.  It is also OK not to attempt the bonus questions.  That gives students the choice and opportunity to stretch their minds, but they are not penalised if they are unable to solve the bonus questions.     In this article, I present two approaches to tackling this question.  Let us review some important formulas first.

Some Useful Formulas

Solution 1  (via the Pythagorean Identity)

  

Solution 2  (via R-formula)

Reflections / Extension

     Here is another HCI question on trigonometry that involves sine, cosine and tangent.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

* other syllabuses that involve trigonometry

[AM_20150621TGSCQT] sin and cos Embroiled in some Quadratic

Question

Introduction

     This is a bonus question from another version of a test set by Hwa Chong Institution this year (2015).  It turns out that different students take the test at different dates, and the school took the trouble to set different versions of the test.  They have the manpower resources to do that!

     It is good to know what topics each question involves.  In this example, students’ knowledge of quadratic theory and trigonometry are being tested in a combined fashion.  Let us first review the relevant material.

Reminders

Solution

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

* other syllabuses that involve trigonometry

[S2_20150523XFDS] Numbers that can be Difference of Squares

Question

Introduction

     This is likely an primary mathematics olympiad-type of question, but lower secondary pupils can also try this.  It involves deeper thinking.  But where do we begin?  Sometimes it is good to begin from the beginning, and then follow your nose. 

Reminders

Solution

     Suppose  N  is a whole number such that  1 < N < 1000  and  N  can be expressed as

                                        N = a² – b²  = (a – b)(a + b)

a difference of squares.  So  N  can be split as a product of two factors  (a + b)  and  (a – b).  Observe that     (a + b) – (a – b) = 2b,       which is an even number.

     The difference between the two factors is an even number.  This can only mean that the two factors are  both odd  or  both even.  You cannot have one of them odd and the other even, because when you subtract them, you would get an odd number.  We now have three cases:-

     Case 1a:  N  is even but not divisible by 4.

     Case 1b:  N  is divisible by 4 (and, of course, is even)

     Case 2:    N  is odd  i.e. both  (a + b)  and  (a – b)  are odd

Ans:  750

Remarks

     In the foregoing, it is possible for  b  to be zero.  0 happens to be a perfect square, because  0² = 0.  However, we need not worry about this, because the above algebra is general enough to cover the case where  b  is  0.
   We have solved the problem using logic, even-vs-odd analysis and the three important algebraic identities under reminders (highlighted in orange).  We also used the special cases (highlighted in light blue) and made observations based on them.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H11. Solve part of the problem

H12* Think of a related problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels
* Primary School Mathematics Olympiad
* Secondary 2 Mathematics » grade 8 (expansion and factorisation)
* anyone who is game for a challenge in algebra and number theory

[S2 Expository] Square-of-Difference Identity for Algebra

     An algebraic identity is an equation that is true for all values of the variables involved.  If we substituted any set of values to the Left Hand Side (LHS) and the same values to the Right Hand Side (RHS), the equation will be true i.e. the LHS will always be equal to the RHS.  The square-of-difference identity

                                       

is one of the three identities that students have to learn in secondary two.  Many students have difficulty remembering this, and they mix this up with the other identity, which involves  a² – b².  However  (a – b)²  is not the same as  a² – b².  They do not understand why the above formula is true, because almost nobody explains it.  Perhaps a few teachers explain the identity for  (a + b)².  But if the  ‘+’  is changed to a  ‘–’  this is a little trickier.  Let me try to explain the formula visually, and with colours to boot, for perhaps the first time in history.

     We start (on the left) with a square of side  a,  whose area is  a².  This is shown in green in the diagram.  We partition each side of the square into  a – b  and  b.  Our goal is to get an area of   (a – b)².  Let us flip the strip of width  b  on the right of the square.  This strip has area  ab  and is shown in pink in the middle square.  This is the same as saying we are subtracting one copy of  ab.  Note on the bottom of the square, there is another strip of area  ab  (shown outlined in orange).  If we subtracted that, we would have subtracted  2ab  (see the square on the right), and we would seem to get  (a – b)².   But then the little square of area  b²  (indicated by a darker green) would have been subtracted twice.  So we need to add  b²  back, so as to restore balance in the universe. 

     You can imagine doing this with a square of area  a²  made of layer of sand.  We remove strips of area  ab  two times – from the right and from the bottom.  Then we patch up the  b²  hole by adding back a layer of sand.  We finally end up with a layer of sand of area  (a – b)².  This illustrates why  a² – 2ab + b² = (a – b)².

     Isn’t this kewl?

Suitable Levels

* Lower Secondary Mathematics

* other syllabuses that involve algebra, expansion and factorisation

[Pri20150408YYS] Yin-Yang Semicircles?

Question

Introduction

     The diagram looks a bit like the Yin and Yang symbol, doesn’t it?  This problem can be solved easily using the correct insights.  I present two solutions: the first one is by direct calculation (in terms of p), and the second solution uses the powerful concept of ratio of similar figures.

     Whichever method is used, first we must make observations.  Can you see that there are three types of semicircles (small, medium and large)?  [H02, H04]

Let  S = area of small semicircle,  M = area of medium-sized semicircle,  and  L = area of large semicircle.

Note that  area of A  = area of C  = L – M + S,  and  area of B = 2´(M – S).  [H10, H11]

Solution 1 (by direct calculation)

     L = ½p(3)² = ^9p/₂.   M = ½p(2)² = ^4p/₂.   S = ½p(1)² = ^p/₂.

     area of A  = area of C  = ^9p/₂ – ^4p/₂ + ^p/₂  =  3p

     area of B = 2´(^4p/₂ – ^p/₂) =  3p

\ area of A : area of B : area of C  =  1 : 1 : 1.  (The areas are all the same)

Many pupils feel more comfortable using concrete approximations like  p  » ²²/₇  or  p  » 3.14,  but this tends to obscure relationships between entities, and makes the calculations messier.

Solution 2 (using similar shapes)

     An powerful idea is that the ratio of areas of similar shapes is the square of the ratios of their lengths.  When a figure is enlarged by a factor of (say) 5, we get a similar figure and the area becomes  5² = 25 times as large.  So  M = (2)²S = 4S  because the radius of the medium-sized semicircle is twice that of the small semicircle.  Likewise,  L = (3)²S = 9S.
     area of A  = area of C  = 9S – 4S + S  =  6S

     area of B = 2´(4S – S) =  6S

\ area of A : area of B : area of C  =  1 : 1 : 1.  (The areas are all the same)

Commentary

     The second solution is neater because we do not need to deal with fractions or with p.  The  ratio of areas of similar shapes is the square of the ratios of their lengths.  This concept is in fact required knowledge in secondary school mathematics, including GCE ‘O’ level “Elementary” Mathematics.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels
* GCE ‘O’ Level “Elementary” Mathematics (“similar figures”)
* Primary School Mathematics (“areas”)
* other syllabuses that involve areas, ratios and or similar shapes.

[Pri20150429PPP] Pythagoras for Primary Pupils?

Question

Introduction

     This primary (elementary) school mathematics examination question created quite a stir among some parent support groups on Facebook.  The issue is that the height of the triangle seems to have been omitted.

Pythagoras’ Theorem

     Some participants who know secondary school mathematics were quick to suggest the use of Pythagoras’ Theorem to find the height of the triangle, which works out to be  12 cm.  This leads to the answer (2) 114 cm²,  which is correct.  The problem is that pupils are not taught Pythagoras’ Theorem until secondary school, and so it would seem an unfair test for the pupils.  So the discussion turned to thinking of various methods by which a primary school pupil may uncover the answer without resorting to advanced knowledge.

Elimination and Educated Guessing

     Mr Teo Kai Meng, a tutor who regularly participates in the support groups, offered some insightful observations.  Assuming that the height measurement is a whole number of centimetres, only options (2) and (4) need to be considered as they were divisible by  19, which the area had to be under the said assumption.  [ Another tutor, Melissa Song had a similar idea by observing that since the triangles DKLN and DKMN  have the same height, the ratio of their areas is the ratio of their bases LN : MN = 19 : 16. ]  We can ignore choices (1) and (3).  As we know,  area = ½ ´ base ´ height .  Since teachers like to catch students for being careless in forgetting to multiply by ½  (or dividing by 2), it is quite likely that option (4) was set up as a booby trap.  Thus one may intelligently surmise that option (2) should be the answer.

     Mr K L Chua, a tutor who calls himself “Mathematics Specialist”, used a similar reasoning.  He worked backwards from each of the four choices to get the heights and chose the most plausible answer.  Of the two whole-number answers, (4) was eliminated and (2) was chosen since from the diagram the height should roughly be near to  16 cm  even though the diagram was not drawn to scale.

Scale Drawing

     Another participant suggested doing a scale drawing to estimate the height.  Indeed this can be done, and is a good tactic too, since this could be done quickly with a ruler and pencil.

A Visual Solution

     Assuming no knowledge of Pythagoras’ Theorem, it is possible to construct a visual solution.  First we take four copies (indicated in orange/light-orange) of the right-angled triangle  DLNK  and arrange them to surround a square of side  37  cm (indicated in green),  which is the same as the longest side (hypotenuse) of the said triangle.     This green area is  1 369 cm².  Now we rearrange the triangular pieces as indicated by the red arrows.  The areas of the orangey triangles do not change when you shift them.  Neither will the green area change, since the total area everything in the containing square (orangey plus green areas) remains the same.

With pairs of right-angled triangles joined together along their longest sides, we now obtain two green squares, the larger of which has side  35 cm.  This gives an area of  1 225 cm².  The total area of the two green squares is the same as the area of the large green sqaure before the shifting, namely  1 369 cm².  Hence the area of the small square is  (1 369 – 1 225) cm² = 144 cm².

From here we quickly deduce that the unknown height is  12 cm,  12 being the square root of  144.  Hence we conclude that the area of  DKLM  is  114 cm².

Remarks

     I have shown that it is theoretically possible for primary school pupils without knowledge of Pythagoras’ Theorem to derive the answer in an exact manner.  By the way, the method of shifting triangular pieces as indicated above can be generalised to give a proof of Pythagoras’ Theorem.

     Some parents expressed fear that this is another one of those Cheryl-like or olympiad type of problems.  Is there a conspiracy by the school teachers to purposely set difficult questions and make life difficult for pupils, disadvantaging those who cannot afford private tutors?  In this case, could this just have been an oversight on the part of the teacher who set the question?

     Entrepreneur John Low Jiayong and tutor John Lim sourced for and managed to obtain faithful copies of the original question.  It turned out that some school-paper vendors had inadvertently erased the 12 cm measurement.  Thus in the original question, the height  KN  was given as 12 cm, and the measurement of  37 cm for the hypothenuse  LK  was purposely given as extraneous information to distract students.

     Some parents observed that this is after all just a multiple-choice question that carries a credit of only one mark.  If this were an exam situation with the  12 cm omitted, it would be best to either sacrifice the  1  mark and move on, or to use tactics like elimination, educated guessing or estimation with scale drawing.
     Anyway, it had been quite a fruitful discussion, as adults (parents and tutors) attacked this problem from many angles in a purposeful way.  Many parties put in concerted effort and contributed in an engaged matter.  It would be good if this type of rich discussion were enacted in our classrooms everyday among pupils and teachers, perhaps enabled by social technology.  For then, pupils would deeply learn and perhaps would not be so stressed out, nor be needing so much extra external help.  Private tutors could then move on to focus on value-added  mentorship  for 21^st Century Learning, instead picking up the tab where school teachers have left off.

Heuristics Used / that can be Used

H01. Act it out (e.g. scale drawing)

         (as a class learning activity, pupils can use scissors to cut out paper triangles

          and physically move them around)
H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H07. Use guess and check

H08. Make suppositions

H09. Restate the problem in another way

Suitable Levels
* Primary School Mathematics
* Lower Secondary Mathematics
* any precocious / gifted pupil who wants to learn
* any person of any age interested in creative problem solving

[Pri20150306TTT] The Tricky Triangle

Question

Introduction

     This is another one of those tricky primary school mathematics questions involving areas.  A perfunctory glance at the area seems to suggest there are four pieces.  Later you might realise that you can think of it as two quarter-circles with two little 45°-45°-90° isosceles triangles removed. 

Plan

     Our plan will be to first find the areas of the two quarter circles and then to subtract the areas of the isosceles triangles.  This is our usual divide-and-conquer strategy [ Heuristics H10 & H11 ].  Notice that the two quarter-circles can be rearranged [H09] into a semi-circle with radius 10 cm.  Simple enough.

What about the two exised triangles?  Notice that the longest sides  (the sloping sides) of the triangles (highlighted in green) are each equal to the radius  10 cm  of the quarter-circles, simply because they, by touching the arcs, are themselves also radii of the quarter-circles.

However, the problem seems to be that we do not know the base and the height of each triangle.  Examiners for Singapore Primary School mathematics like to set this sort of questions involving areas of isosceles right-angled triangles, in which you are given only the length of the hypotenuse (the longest side).  How to tackle this kind of situation?  By using our imagination!

Imagine that the two triangles are brought together.  This forms a larger right-angled isosceles triangle.  However, now you realise it is half of a 10 cm by 10 cm square.  You can also imagine turning the triangle around until one of the 10 cm sides is horizontal.  Treating this as the base, the height of the triangle is 10 cm.  Either way, you are able to solve it and get the same answer.

     All that is left now is to subtract this from your area of the semi-circle found earlier.

Solution

   Shaded Area [in cm²]

= Area of two quarter-circles – area of two triangles

= Area of semi-circle – area of combined triangle

= ½ ´ p ´ (10) ²  – ½ ´ 10 ´ 10

= 50p  – 50

Ans: Shaded area = 107.08 cm².

Heuristics Used

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Commentary

     We solved this problem by breaking it down into smaller problems.  Since areas are unchanged when you shift them, or turn them, or reflect them, we are able to arrange the two quarter-circle pieces into one semi-circle.  We can also combine the two triangles into a larger triangle for which we know the base and the height.  By breaking down the problem and transmuting these smaller problems into equivalent problems, our task becomes much simpler, allowing us to get the solution quickly.

Please refer to this similar problem.