**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 D*BKE*by*x*, and observe that it is the same as Area of D*BFE*by congruency (you can shift and rotate D*BKE*to get D*BFE*). Area of D*AHE*= Area of D*AEF*= 1 cm^{2}because of congruent triangles. [ H04 ]
Let us try to figure out

*x*[ H11 ]. Making further observations [ H04 ], note that Area of D*ADB*= Area of D*CDB*(big congruent triangles) and Area of D*HDE*= Area of D*GDE*(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 D*CDB*– Area D*GDE*– Area D*BKE*
= Area D

*ADB*– Area D*HDE*– Area D*BFE*
= Area

*AHEF*= 2 cm^{2}.
We conclude that Area D

*BKE*= Area D*BFE*=^{13}/_{5}– 2 =^{3}/_{5}[cm^{2}]. [ 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 D

*BFE*: Area D*AEF*=*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

#1 If two shapes are congruent (that means
you can shift, rotate and/or reflect sothat they coincide), then their areas are equal. #2 If triangles/parallelograms have the same
height, their areas’ ratio equals theirbases’ 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.
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