Seeking Precise Solution for Triangle Area Problem

Question

I hope this message finds you in high spirits. I am writing to seek your expertise in solving a captivating geometry problem that I recently encountered in a competitive exam. Despite my best analytical efforts, the solution has proven elusive, with results oscillating between $48$ square units and $64$ square units. In my pursuit of a definitive resolution, I turn to this knowledgeable community to shed light on the accurate approach and solution.

Problem Overview:

In triangle $\triangle ABC$, points $E$ and $F$ divide side $BC$ into equal thirds. $M$ and $N$ are the midpoints of the sides $AB$ and $AC$ respectively. If the area of ​​quadrilateral $EFQP$ is 12 square units, find the area of $\triangle ABC$.

My Efforts:

I've tried various analytical approaches, but my results have been inconsistent. A more systematic strategy involving geometric properties, algebraic manipulation, or other mathematical tools seems necessary.

Seeking Collaboration:

I kindly request your expertise in tackling this problem. A collaborative analysis holds the key to a precise solution that accurately calculates the area of triangle $\triangle ABC$.

Expected Result:

Together, I believe we can solve this geometry puzzle and arrive at a comprehensive solution for the area of triangle $\triangle ABC$. This endeavour will enhance my grasp of geometry and contribute to the community's mathematical knowledge.

I'm sincerely thankful for your time and insights. Feel free to ask if you need further details.

Appreciate your help greatly!

Answer 1

Triangles $\triangle ABE , \triangle AEF , \triangle AFC $ have the same area, each one-third of $[ABC]$.

Segment $MN$ is parallel to $BC$. Therefore,

$\triangle AMP$ is similar to $\triangle $ABE$, and it follows from similarity that

$ MP = \frac{1}{2} BE $

Similarly, we obtain

$ PQ = \frac{1}{2} EF $

and we also get that $[APQ] = \frac{1}{4} [AEF] $

Hence $[PEFQ] = \frac{3}{4} [AEF] = 12$

From which, $[AEF] = 16 $

And finally $[ABC] = 3 [AEF] = 48 $

Answer 2

The Base of the 3 Quadrilaterals are Same , heights are Same.

Hence areas will be Same $12$.

Total area of the Base Section is $12+12+12=36$

The upper Part of the whole triangle must be $(1/2) \times (1/2) = (1/4)$ of the whole triangle [[ Check Similar triangles $\triangle ABC$ & $\triangle AMN$ ]]

In other words , upper Part is $1/3$ of the Base Section.
$36 \times 1/3 = 12$

Whole triangle is upper Part + Base Section , hence $12+36=48$

Additional Details :

When we check the triangle with the 2 Mid-Points , we have the triangle with $1/2$ height & $1/2$ Base :
https://en.wikipedia.org/wiki/Midpoint_theorem_(triangle)

Hence , area must be $1/4$ of whole triangle.

Midpoint for a triangle area question?