[Pri20150504PER] Pythagoras Precludes Perimeter Problem?

Question

Introduction

     Just like this problem, this primary school problem involving lengths seems to require Pythagoras’ Theorem, which is not taught until secondary school.  But with some clever observations, we can solve it by looking at it in another way.

Observations

     First,  note that being diagonals of the rectangle BOEC,  BC  is the same as  OE,  which is the same as  13 cm, the radius of the quadrant.  Secondly, we know that the sum of  BO  and  OC  is 17 cm,  being half of the perimeter of the rectangle BOEC.  We do not have to know the individual lengths  BO  and  OC,  and we do not need Pythagoras’ Theorem.  Since  AO  and  OD  are radii of  13 cm  each,  their total length is  26 cm.  We can just subtract  17 cm  from this to get the total of  AB  and  CD.  We do not need to know the individual lengths of  AB  and  CD.  Once we understand these points, we are ready to solve the problem.

Solution

    Perimeter / cm   = (AB + CD) + BC + arc AED

                              = [(OA + OD) – (BO + OC)] + OE + ¼ ´ 2 ´ p  ´ 13

                              = [  13  +  13   –     ½ ´ 34  ] +  13  + ½ ´ 3.14 ´ 13

                              = 42.41

Ans:  The perimeter of the shaded region is  42.4  cm.

Remarks

     Your answer can never be more precise than the precision you use for your calculation.  When we use  3  significant figures for the value of  p,  our answer is at best correct to  3  significant figures.

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 perimeters and lengths

[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.

[Pri20150306MOD] A Telescope Too Far?

Question

Introduction

     This is a nasty multiple-choice question to be set for a primary 6 (~ grade 6) pupil for a test.  It could also be a time trap as the pupil might spend a lot of time to no avail just to try to score that miserable mark.  With advanced knowledge, we can solve this using telescoping sums (a.k.a. the method of differences), where many terms cancel and the sum can be shortened, much like a telescope.

     However, this solution looks like over-kill.  Is there a solution that is more accesible to a primary 6 pupil?  Read on!

A Simpler Solution?

     We observe that a denominator of  5  is common.  In fact, since every term has an even factor, we have a common factor of 2 in the denominator also.  That means every term in the sum can be expressed as ¹/₁₀ of something [ Heuristic H09 ].  We can quickly work out the first few partial sums and express each of them as  ¹/₁₀ of something.  [ Heuristic H04 ]

     Tabulating the results [ H02 & H03 ], we observe that each partial sum is  ¹/₁₀ of something slightly less than ½.  So we try to express them as ½ minus something.  How?  We can subtract to find out.  [ H05 ]  For example,  if  ⁵/₁₂ = ¹/₂ – ???   then   ??? = ¹/₂ – ⁵/₁₂ = ¹/₁₂.

     Later on, we notice   [ H04 again ] another pattern: the factors in the denominators of the subtracted quantity matches the last two denominators of the last term.  If we considered the full sum, then what would those factors be?  (See the part highlighted in yellow).  These would be 20 and 21, as per the last term of the series.  Using this pattern, we work out the required sum as follows

That is the answer!

Heuristics Used

H02. Use a diagram / model

H03. Make a systematic list

H04. Look for pattern(s)

H05. Work backwards
H09. Restate the problem in another way