Australian Maths Competition Area Question 
PQRS is a square. T and U are midpoints of the sides PS and PQ
respectively. TQ, SU and PR intersect at V.

This is a question from the 2009 Intermediate Division AMC paper.We are currently preparing for it and I would like some help on the process of solving this. I hope it will give me an idea of how to think when doing questions like this.
Question and multiple choice options:

 A: Construct the diagonal $QS$.Then, $\text{ar}(PQS) = \dfrac{1}{2}\Delta$ if $\Delta$ be the area of the square. Note that $V$ is the centroid of the isosceles triangle $PQS$ and recall that the medians divide any triangle into 6 triangles of equal area. Hence, $$\text{ar}(VSPQ) = 4 \times \dfrac{1}{6} \times \text{ar}(PQS) = \dfrac{1}{3} \Delta$$
$$\therefore \text{ar}(QVSR) = \dfrac{2}{3}\Delta $$
A: Notice that $\triangle PUV \sim \triangle VSR$ with ratio $\frac{1}{2}$. Also, note that $\triangle PVT \sim \triangle RVQ$ with the same ratio of $\frac{1}{2}$. Also, note that since $PT = PU$, $\angle TPV = \angle UPV = 45$ and $PV = PV$, by SAS congruency, $\triangle TPV \cong \triangle UPV$, and note that the areas of $PVT$ and $TVS$ and the areas of $PVU$ and $VUQ$ are all equivalent, because they have the same height and the same base. Let $[PVT] = [PVU] = [TVS] = [VUQ] = x$. Then, $[SVR] = [RVQ] = 4x$, and therefore $\frac{[SVQR]}{[PQRS]} = \frac{8x}{12x} = \frac{2}{3} \implies C$.
A: 
Then, [PQRS] = 3 + 3 + 6 = 12
[quad SRQV] = 12 - 4
Result follows.
