An intuitive approach to solving a question of geometry

Question

In the square $ABCD$ of side $6$, points $P$ and $Q$ move along sides $AB$ and $CB$ so that $\overline{PB}=\overline{BQ}=x$. Consider the segment $OQ$ parallel to $AB$. Which of the following expressions describes the measure of the length of $\overline{OQ}$ as a function of $x$?

My approach: I named, for simplicity, $\overline{OQ}=y$. I have considered that $\text{area}(\triangle PBC)=3x$; $\text{area}(PBOQ)=\frac{(x+y)\cdot x}{2}$, $\text{area}(\triangle OQC)=\frac{y\cdot (6-x)}{2}$. Hence

$$\text{area}(\triangle OQC)=\frac{y\cdot (6-x)}{2}=3x-\frac{(x+y)\cdot x}{2}$$

Solved $y$ in function of $x$ we have:

$$y=f(x)=x-\frac 16 x^2.$$

Now the question is the following: given that my 18-year-old students will have to deal with simulations from the National Institute for the Evaluation of the Education and Training System, how will they be able to solve the problem without pen and paper as required comprising 41 questions in an hour? I would not have succeeded without to use pen and paper.

Is there an alternative proof instead of my approach?

Answer 1

Use similar triangles $\triangle COQ\cong \triangle CPB$, therefore $$\frac{\overline{OQ}}{\overline{PB}} = \frac{\overline{CQ}}{\overline{CB}} = \frac{6-x}{6}$$

So $$\overline{OQ} = \frac{x(6-x)}{6}$$

Answer 2

The triangles $\triangle PBC$ and $\triangle OQC$ are similar, which shows that $$\frac{y}{x}=\frac{6-x}{6}.$$ It follows that $$y=x-\frac16x^2.$$