[Pri20160223FPDM] Pernicious Portion Problem? Shift Happens!

Problem / Question

Strategy

     This seems to be a confounding question on decimals.  What shall we do with the triangles?  Is there a short cut?

     Yes!  What you can do is to imagine putting the two triangles together to form a rectangle.  And then the solution becomes easy!  This is because the area is unchanged and hence the proportion of the shaded area is unchanged, is the same as before.  We can make use of fractions and convert it to a decimal.

Solution

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve fractions and decimals

* any learner who is interested

[OlymPri20150526] The Smallest Angle in a Special Trapezium

Question

Introduction

     This looks challenging because we do not seem to be given much information.  For example, we do not know the individual angles of the trapezium (American: trapezoid).  Does this mean that this puzzle cannot be solved?  Are we trapped by this trapezium?  What is the secret key that unlocks the problem?

Solution

     First sketch trapezium  ABCD.  Introduce point  F,  the mid-point of  AD.  All contructed lines and points are shown in grey (American: gray).  Then  DF = FA = BC. 

Now draw a line parallel to  BC  passing through  A  and intersecting  DC  at  E, say.  Note that ÐAED = ÐBCD  (corresponding angles).  Then  ÐADE + ÐAED = ÐADC + ÐBCD = 120°.  Hence the remaining angle  in  DDAE,  ÐDAE = 180° – 120° = 60°.  We can deduce this, even though we do not (currently) know the individual angles  ÐADE  and  ÐAED.  This is the key step.

From here, things get easier.  DAFE  is an isosceles triangle with  ÐAFE = ÐAEF = (180° – 60°) ¸ 2 = 60°.  So  DAFE  is in fact an equilateral triangle.  That means  FE = FA = FD.  So  DFDE  is an isosceles triangle.

Note that  ÐAFE  is an exterior angle of the triangle  DFDE.  If you know that exterior angle of a triangle is the sum of the interior opposite angles,  then from  ÐAFE = ÐFED + ÐFDE,  we easily see that  ÐADE = ÐFDE = 60° ¸ 2 = 30°.  If you do not know the theorem about exterior angles, you can still quickly work out that  ÐDFE = 120°,  and then use  ÐADE = (180° – 120°) ¸ 2  and arrive at the same conclusion.  ©

Remarks

     We can now see that actually  ÐBCD  = ÐAED  = 90°,  but we do not need to rely on that (or on accurate drawing) to deduce the answer.  From solving this question, we learn that even though we do not know the individual angles, by construction and using the sum of angles in a triangle, it is possible to solve for an important angle, namely  ÐAFE.  The rest of the solving uses isosceles triangles and equilateral triangles, which are part of the common repertoire of tactics.

HeuristicsUsed

H02. Use a diagram / model

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

H12* Think of a related problem   (Draw construction lines)

Suitable Levels
* Primary School Mathematics

* other syllabuses that involve area of triangles and circles

[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.