Problem http://puu.sh/rHIrw.png
$PQRS$ is a rectangle. $MS=NP=x$ units and the area of the triangle $MNQ$ is 38 units squared. Find the possible values of $x$.
The way I did it was substituting $MR$ as $12-x$ and $SN$ as $9-x$ and used that information to apply the pythagorean theorem to figure out the sides $MQ$ and $MN$. But I'm not sure if that'll help me. I know that certain triangles in a quadrilateral are right triangles but I don't know what tools should I use to prove that. In the end, all of my calculations are really just using the pythagorean theorem. My intuition is telling me that $MQ$ is the height but I don't trust myself unless I do a proof about it. Thus, I can't figure out a way on how to use the information regarding the area of the triangle $MNQ$.
The key to this problem is to compute the area of $\triangle MNQ$ indirectly by subtracting the other regions of rectangle $PQRS$. These other regions are $\triangle NPQ$, $\triangle MSN$ and $\triangle QRM$, which have areas $\frac12(12x),\frac12x(9-x),\frac12(9(12-x))$ respectively. The area of the whole rectangle is $9\cdot12=108$, so the area of $\triangle MNQ$ is $$108-\frac12(12x)-\frac12x(9-x)-\frac12(9(12-x))$$ $$=108-6x-\frac12(x(9-x)+9(12-x))$$ $$=108-6x-\frac12(9x-x^2+108-9x)$$ $$=108-6x+\frac12x^2-54$$ $$=\frac12x^2-6x+54$$ Since the area is equal to 38: $$\frac12x^2-6x+54=38$$ $$\frac12x^2-6x+16=0$$ $$x^2-12x+32=0$$ $$(x-4)(x-8)=0$$ $$x=4\text{ or }x=8$$ Since both these numbers are less than 9, they are the possible values for $x$.