In the Parallelogram, E is the midpoint of AC and EF is parallel to AB

Hi chetan2u, 2 quick questions:

Now area of ABX = area of AEX + area of AFX, as the height is same being between two same parallel lines and base too is equal.

Shouldn't it be BFX instead of AFX? Are we considering AB the base for these 2 triangles or AE and BF as bases?

Thank you!

Hi Crytiocanalyst, question;
how do I know these areas are divided into X, X, 2X and 2X is equal to AEX + BFX? Since No value is given other than EX=6.
Thank You!

Yes, it is BFX. Corrected the typo. Thanks

The ABX region can be further divide into 2 similar areas by drawing a perpendicular and we can make the resultant triangle and take their individual areas as X hence when we add we get 4x hope it clarifies

Sir, to divide the triangle into 2 equal parts we have to consider that X is in the middle of EF, isn't it? However, the question does not mention that. Is there something that I am missing?

Triangle ABX can be divided into 2 equal half and not the outer triangle the trianglr which is situated in the middle is divided , yes youre right for the triangles to be divide the the outer triangles it needs to be in between , however i never touched the outside triangle in the first place

The key to answering the question is to notice that the entire figure is a parallelogram and that EF is the parallel mid segment that divides the parallelogram into 2 symmetric/congruent areas.


Thus, we have parallelogram ABEF = Area of (1/2) (36) = 18


Rule: between 2 parallel lines, the perpendicular height will always be the same.

Call the area of triangle AEX = [AEX]

Area of triangle ABX = [ABX]

Area of triangle BFX = [BFX]


[AEX] + [ABX] + [BFX] = 18 (equation 1)


Choose base sides for each triangle:

EX = 6

Let XF = b

Since the opposite sides of a parallelogram are equal, AB = 6 + b


And all 3 triangles will have an equivalent perpendicular height drawn to the respective bases above ——> call it H = perpendicular distance between the 2 parallel lines AB and EF

[AEX] = (1/2) (6) (H) = 6H / 2

[BFX] = (1/2) (b) (H) = bH / 2


[ABX] = (1/2) (b + 6) (H) = (6H + bH) / 2


As you can see, the Area of triangle AEX + the Area of triangle BFX = the target triangle’s Area, triangle ABX


[AEX] + [BFX] = [ABX] (equation 2)

Combining this with (equation 1) above

[AEX] + [BFX] + [ABX] = 18


Substitute (equation 2) into (equation 1)


[ABX] + [ABX] = 18

(2) * [ABX] = 18

Area of triangle ABX = [ABX] = 9

D

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