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.
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[ 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:-
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
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.
#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.
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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.
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