In one of my tutorial classes, when I was studdying in 9th class (I am in 10th now), our tutor gave us a problem saying it’s a difficult one, and to him, *it was incomplete*.

This is *that* problem:

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> **In a rectangle *ABCD*, point *B* is joined to the points *P* and *Q* on the sides *AD* and *DC* respectively. And *P* and *Q* are joined to form *three* distinct triangle; namely *PDQ or x*, *QCB or y*, *BAP or z* and *BPQ or f*.**
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> **Given the area of *x = 3m^2*, *y = 4m^2* and *z = 5m^2*, find the area of *f*.**
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> **Note:** all the information regarding this problem has been provided. The dimensions of any of the line segments are *not known*. This also means that the exact positions of *P* and *Q* on their respective line segments are also *not known*.
>  ![diagram](https://i.stack.imgur.com/24fhY.jpg)
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> 
> 

Because of the lack of information (See *Note*), my tutor declared it as an incomplete problem (he is just a high-school Science and Mathematics teacher, please don’t expect him to think beyond the limits of a typical high-school student, that is, in case he is wrong).

After a few hours of brainstorming, I came up with what I called an *ingenious method*. But then I felt that my method actually was not accurate.

**So, can anybody tell me if this problem is complete and solvable, if yes, is my method correct (see below), if no, what is the best/better method to solve it.**

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MY METHOD:
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> 
> As all the triangles with their areas given are on the edge of the rectangle, we know that they all are *right triangles*. So we can say that one of the arms of *the right angle* is *the altitude* and the other is *the base*.
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> 
> Now, we can use the formula *(1/2) \* base \* altitude* to calculate their areas. Lets take *a* as *altitude* and *b* as *base*.
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> 
> (Here’s the first part of *the trick*) We now try all possible \_a\_s and \_b\_s for the three triangles, where *a* and *b* are **natural numbers only**. (This is also where the major weakness of my method is. More numbers, i.e., any *real numbers* should have been tried, but that is infeasible). Note that *all possible values* are those values that form a triangle of that specified area.
> 
> 
> **For triangle *x*, we have the following dimensions:**
>  -***a = 1*, *b = 6***
>  -*a = 2*, *b = 3*
>  -*a = 3*, *b = 2*
>  -*a = 6*, *b = 1* #
> 
> 
> **For triangle *y*, we have the following dimensions:**
>  -*a = 1*, *b = 8*
>  -***a = 2*, *b = 4***
>  -*a = 4*, *b = 2*#
>  -*a = 8*, *b = 1*
> 
> 
> **For triangle *z*, we have the following dimensions:**
>  -***a = 1*, *b = 10***
>  -*a = 2*, *b = 5*
>  -*a = 5*, *b = 2*
>  -*a = 10*, *b = 1*#
> 
> 
> (Here’s the other part of the trick) Then, we select those specific dimensions of triangles, which when placed together form a rectangle.
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> 
> **The condition we require:**
> **When the triangles are placed together to form a *quadrilateral* the opposite sides of it must be equal.**
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> 
> If this *quadrilateral’s* opposite sides are equal, we know that it is a *parallelogram*. And as all of the triangles that form the edges are *right angled*, we have at least one of the angles of the *parallelogram* right angled\_, so we know that this *parallelogram* is a *rectangle*. 
> 
> 
> The correct set of dimensions (that we will be using here) are high-lighted in bold, while the once suffixed with a ‘#’ can also be used.
> 
> 
> So, in *triangle x* we take *PD* as altitude and *DQ* as base.
>  In *triangle y* we take *BC* as altitude and *CQ* as base.
>  In *triangle z* we take *PA* as altitude and *AB* as base.
> 
> 
> Henceforth, we now have:
>  *(PA + PD) = BC*
> *(DQ + CQ) = AB*
>  Therefore, the condition is matched.
> 
> 
> **WE NOW HAVE THE CORRECT DIMENSIONS OF THE THREE TRIANGLES**
> 
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> Now we can either find the area of *f* by finding the area of the rectangle and subtracting it by the areas of the three triangles, or by finding the three triangle’s *hypotenuse* using the *Pythagorean formula* and then apply the \_Herron’s formula \_on the hypotenuses. I prefer the former.
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> 
> >*After calculating, we have, \_f = 8m^2\_*
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Thank you =)