[PriOlym_20151130RTAC] Ratio with One Circle Overlapping Two

Question

Introduction

     This question is like this previous one, except it is of olympiad standard.  I illustrate the solution of this without algebra, by using Distinguised Ratio Units.  As before, I try to match parts to an equal number.  But here we have quite a mixture of different types of units.

Solution

Commentary

     Basically we make the triangle units to number 12 and do the same for the circle and square units.  It turns out that one triangle unit is the sum of one circle unit and square unit.  We deduce that 9 circle units (for the area of A) plus 6 circle units (for the area of B) is the same as 8 circle units and 8 circle units.  The reduction of circle units must be equally compensated by the increase in the circle units.  Thus one circle unit is the same as two square units.  From here, things become easy.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H11. Solve part of the problem

Suitable Levels

* Primary School Olympiad Mathematics

* Primary School Mathematics (challenge)

* other syllabuses that involve areas and ratios

* anyone who is game for a challenge

[Pri20151129RTAO] Equalising Ratio Units for The Overlap

Question

 

Introduction

     This is a primary school ratio problem that is quite a favourite among question setters, but poses headaches for pupils and parents.  The trouble is that the ratios use different base units and this makes it difficult to compare the ratios.  Can we avoid using algebra or trial and error?  

Strategy

     Note [H04, H09] that the difference in the areas between the rectangle and the square (including the shaded overlapping part) is exactly the same as the difference between them without the overlapping part.  With this crucial observation, we can proceed to try to equalise the ratio units [H10] of the aforementioned differences.  This can be done by multiplying to get to the Lowest Common Multiple, which, in this example is 6.  Henceforth we can be sure of using the same ratio units, because the same number of units are used to refer to the same quantity.

Solution

Summary

     Ratio problems are solved by making sure that we use the same type of units.

Heuristics Used

H02. Use a diagram / model        [ table ]

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

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve areas and ratios

* anyone who wants to learn

[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.

[Pri20150306PPP] A Perspective for Perplexing Parallelograms

Question

Comprehension

     Let us first be clear about what is given in the problem.  I have indicated the numerical values of the areas in the diagram below.

Planning

     First we construct  HK  parallel to  AB  passing through  E.  [ Heuristics H02 & H09 ]  This makes the problem easier to solve because there are congruent triangles and there are connecting ratios along the sides of the parallelogram as well as along the diagonal.  Namely  BF : FA = BE : ED = BK : KC.  We denote the  Area of DBKE by  x, and observe that it is the same as  Area of DBFE  by congruency (you can shift and rotate  DBKE  to get  DBFE).   Area of DAHE = Area of DAEF = 1 cm²  because of congruent triangles.  [ H04 ]

     Let us try to figure out  x [ H11 ].  Making further observations [ H04 ], note that  Area of DADB = Area of DCDB (big congruent triangles) and Area of DHDE = Area of DGDE  (congruent triangles).  See diagram below.

     Because of all these congruent triangles, the parallelogram AHEF and EGCK are forced to have the same areas (area measures), even though they may not be congruent:-

      Area EGCK = Area DCDB – Area DGDE – Area DBKE

                           = Area DADB – Area DHDE – Area DBFE 

                           = Area AHEF = 2 cm².

We conclude that  Area DBKE = Area DBFE = ¹³/₅ – 2 = ³/₅  [cm²].  [ H05 ]

Here is the important thing:-

If triangles/parallelograms have the same height, their area ratio equals their base ratio.

     [ This is different for similar triangles/parallelogram. Do not confuse these two situations. ]

     Area DBFE : Area DAEF = BF : FA = BE : ED = BK : KC  = 3 : 5.

Now we can quickly fill in the areas of the remaining pieces as follows:-

      Area ABCD = 2 ´ Area DABD = 2 ´ (³/₅ + 2 + ⁵/₃) = 8 ⁸/₁₅  [cm²]   Done!

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H11. Solve part of the problem

Thinking Back

     After solving a problem (especially a difficult one), it is always good to think back and recollect what we have learnt by solving the problem.  We have used the following important geometrical facts

#1   If two shapes are congruent (that means you can shift, rotate and/or reflect so        that they coincide), then their areas are equal. #2   If triangles/parallelograms have the same height, their areas’ ratio equals their        bases’ ratio. #3   Sometimes, two shapes can have the same area even if they are not congruent.

      Drawing a construction line splits the diagram into various parallelograms and triangles with the same height, for which we can compare the ratios.  Using congruent triangles, we realise that two of the smaller parallelograms have the same area even though they are not congruent.  With the right perspective, we can deal with these perplexing parallelograms.  Using comparison, we found the area of the small triangular piece  x, and from there we use fact #2 above to work out the rest.

[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.