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$(ADE)=(CDF)=\frac{(ABCD)}{4}$ $\frac{AE}{CD}=\frac{1}{2}=\frac{AH}{HC}$ $\Rightarrow AH=HG=GC$ $(BEF)=\frac{ABC}{4}$ $(GDH)=\frac{ACD}{3}$ $(GHFE)=(ABCD)-(ADE)-(CDF)-(BEF)-(DGH)=\frac{5\times(ABCD)}{24}=\boxed{5}$

Write a solution. If we draw the lines joining the midpoints of all 4 sides of the quadrilateral ABCD as above, point I will be the center, where all diagonals and midpoint lines cross. Four equal partitions of smaller quadrilaterals will then be formed, such as ⎕BEIF, and has an area of 24/4 = 6 cm2.

As a result, △EIF has an area of 6/2 = 3 cm2.

Now notice that EF//AC//KJ, and they all have even space from its adjacent line because EF and KJ are both formed from the midpoints of the big quadrilateral. Moreover, ⎕AEFC is identical to ⎕AKJC as they have all the same sides and angles.

Now notice that △EIF has the same base as △EDF but just different heights.

Nonetheless, if there were an imaginary line passing point D and parallel to KJ, that line would have equal space as KJ would have with AC because the diagonal would be divided into 4 lengths equally when passing these lines.

As a result, the ratio of the heights of △EIF and △EDF is equal to the ratio of the spaces between EF to AC and EF to the line passing point D.

In short, the space is tripled, and so the height of △EDF is 3 times longer than △EIF. Likewise, its area is also 3 times larger.

Therefore, the area of △EDF is 3*3 = 9 cm2.

Since △EDF and △HDG have parallel bases and share the same angle D, they are similar. Considering the parallel spacing, △HDG has 2/3 of the height of △EDF, so the base of △HDG will also be 2/3 of the base of △EDF.

In other words, the area of △HDG = (2/3)(2/3)*(9) = 4 cm2.

Finally, the shaded area = 9-4 = 5 cm2.

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Area of the shaded region GHEF is 5 cm^2

area of ∆ADE = area of ∆CDF = (area of parallelogam ABCD)/4

AE/CD = 1/2 = AH/HC

⇒ AH = HC = GC

area of ∆BEF = (area of ∆ABC)/4

area of ∆GDH = (area of ∆ACD)/3

Therefore, area of the shaded region GHEF = (area of parallelogam ABCD) - (area of ∆ADE) - ( area of ∆CDF) - (area of ∆BEF) - (area of ∆GDH) = ( 5* area of parallelogam ABCD)/24 = (5*24)/24 = 5 cm^2

area of ∆ADE = area of ∆CDF = (area of parallelogam ABCD)/4

AE/CD = 1/2 = AH/HC

⇒ AH = HC = GC

area of ∆BEF = (area of ∆ABC)/4

area of ∆GDH = (area of ∆ACD)/3

Therefore, area of the shaded region GHEF = (area of parallelogam ABCD) - (area of ∆ADE) - ( area of ∆CDF) - (area of ∆BEF) - (area of ∆GDH) = ( 5* area of parallelogam ABCD)/24 = (5*24)/24 = 5 cm^2

area of the shaded region GHEF is 5 cm^2

area of ∆ADE = area of ∆CDF = (area of parallelogam ABCD)/4

AE/CD = 1/2 = AH/HC

⇒ AH = HC = GC

area of ∆BEF = (area of ∆ABC)/4

area of ∆GDH = (area of ∆ACD)/3

Therefore, area of the shaded region GHEF = (area of parallelogam ABCD) - (area of ∆ADE) - ( area of ∆CDF) - (area of ∆BEF) - (area of ∆GDH) = ( 5* area of parallelogam ABCD)/24 = (5*24)/24 = 5 cm^2

To find the area of the shaded region the areas of the triangles: BEF, AEH, CFG and ACD will be deducted from the area of the parallelogram ABCD.

The area the triangle ACD is equal to 1/2 of the parallelogram since triangles ACD and ABC are congruent and so have equal areas.

The area of the triangle BEF is equal to 1/4 of the area of the triangle ABC. This is due to their similarity as doubling the size of the triangle BEF will give lengths BE and BF the same as AB and BC respectively and both triangles. Halving the edges result in 1/4 of the area of the original triangle.

To find the area of triangle CFG the line BD will be drawn. The the intersection between AC and BD (M) is half way along both lines since triangles BCM and ADM are both congruent due to AAS. Since M is halfway along BD then lines CM and DF are medians. This means that line CG is 2/3 CM and 1/3 AC. This means the triangle CFG has a perpendicular height 1/3 of ABC and a base 1/2 of ABC. This means triangle CGF is 1/6 the area of ABC and 1/12 area of parallelogram ABCD. In similar reasoning triangle AEH is 1/12 of ABCD.

These triangles leave 5/24 of the area of parallelogram of ABCD for the shaded area AFGH.

Therefore, the area of AFGH is 5.