Q16 Geometry (Similar Triangle & Area ratios)

[Jefferson]

Please help me with the final part, iv, of this question. I'm struggling to make sense of many of the steps they took, (including the 1:3 ratio or 3x area).
Below are attachments to question and provided solutions.

If possible, is there any clearer approach/worked solution?
Thank you!

[RuiAce]

I did it "my own way", but I think it does somewhat coincide with their approach.

The intuition behind my approach is that I kept using \( Area=\frac12 a b \sin C\) throughout parts ii) and iii). For ii) I chose to show that \( \frac{Area_{\triangle CAS}}{Area_{\triangle BQS}}=4 \) and \( \frac{Area_{\triangle BSA}}{Area_{\triangle QSC}}=1 \). These were all done using the proportional sides from part i):
\[ \frac{CS}{BS}=\frac{AS}{QS}=\frac{CA}{BQ}=2. \]
Let \(A_1 = Area_{\triangle BQS}\) and \(A_2 = Area_{\triangle BSA}\). Observe that the question claims the area of the rectangle is \(12 A_1\). This means that somehow \(A_2\) needs to have left the puzzle.

This leads to the question: Can we express \(A_2\) in terms of \(A_1\)?
\begin{align*}
\frac{A_2}{A_1} &= \frac{Area_{\triangle BSA}}{Area_{\triangle BQS}}\\
&= \frac{\frac12 \cdot BS \cdot SA \cdot \sin\angle BSA}{\frac12 \cdot BS \cdot QS \cdot \sin\angle BSQ}\\
&= \frac{AS}{QS} \cdot \frac{\sin \angle BSA}{\sin \angle BSQ}\\
&= 2\cdot 1\\
&= 2\\
\therefore A_2 &= 2A_1.
\end{align*}
In my method, the tricky bit was to see that these sines also cancel out. This is a consequence of the second quadrant property: \( \boxed{\sin (180^\circ - \theta) = \sin\theta }\). Here, we know that \( 180^\circ - \angle BSQ = \angle BSA \) since \(ASQ\) is a straight line, so indeed we have \( \sin \angle BSQ = \sin \angle BSA \).

Now that we've found that \(A_2 = 2A_1\), we can consider all the areas we have so far. Using parts ii) and iii), the area of trapezium \(BQAC\) is \( A_1 + 4A_1 + 2A_1 + 2A_1 = 9A_1\). So if we want the area of the rectangle to be \(12A_1\), fingers crossed that \(Area_{\triangle APB} = 3A_1\).

This is the only time in the entire question where I just used \( \frac12 bh \). I only did it because I wanted to avoid the angles.
\begin{align*}
Area_{\triangle APB} &= \frac12 \cdot PB \cdot PA\\
&= \frac12 \cdot BQ\cdot QC\\
&= Area_{\triangle BQC}
\end{align*}
(Note: In fact, \(\triangle APB\) and \(\triangle BQC\) are congruent. The loose explanation is because they're identical triangles from symmetry. But one could also prove the congruence. Not sure if it was required.)

But once we're here, we can just look at the diagram and see that \(Area_{\triangle BQC} = Area_{\triangle BQS} + Area_{\triangle QSC} = A_1 = 2A_1 = 3A_1\). You now have all the ingredients to finish it off.

[Jefferson]

I did it "my own way", but I think it does somewhat coincide with their approach.

The intuition behind my approach is that I kept using \( Area=\frac12 a b \sin C\) throughout parts ii) and iii). For ii) I chose to show that \( \frac{Area_{\triangle CAS}}{Area_{\triangle BQS}}=4 \) and \( \frac{Area_{\triangle BSA}}{Area_{\triangle QSC}}=1 \). These were all done using the proportional sides from part i):
\[ \frac{CS}{BS}=\frac{AS}{QS}=\frac{CA}{BQ}=2. \]
Let \(A_1 = Area_{\triangle BQS}\) and \(A_2 = Area_{\triangle BSA}\). Observe that the question claims the area of the rectangle is \(12 A_1\). This means that somehow \(A_2\) needs to have left the puzzle.

This leads to the question: Can we express \(A_2\) in terms of \(A_1\)?
\begin{align*}
\frac{A_2}{A_1} &= \frac{Area_{\triangle BSA}}{Area_{\triangle BQS}}\\
&= \frac{\frac12 \cdot BS \cdot SA \cdot \sin\angle BSA}{\frac12 \cdot BS \cdot QS \cdot \sin\angle BSQ}\\
&= \frac{AS}{QS} \cdot \frac{\sin \angle BSA}{\sin \angle BSQ}\\
&= 2\cdot 1\\
&= 2\\
\therefore A_2 &= 2A_1.
\end{align*}
In my method, the tricky bit was to see that these sines also cancel out. This is a consequence of the second quadrant property: \( \boxed{\sin (180^\circ - \theta) = \sin\theta }\). Here, we know that \( 180^\circ - \angle BSQ = \angle BSA \) since \(ASQ\) is a straight line, so indeed we have \( \sin \angle BSQ = \sin \angle BSA \).

Now that we've found that \(A_2 = 2A_1\), we can consider all the areas we have so far. Using parts ii) and iii), the area of trapezium \(BQAC\) is \( A_1 + 4A_1 + 2A_1 + 2A_1 = 9A_1\). So if we want the area of the rectangle to be \(12A_1\), fingers crossed that \(Area_{\triangle APB} = 3A_1\).

This is the only time in the entire question where I just used \( \frac12 bh \). I only did it because I wanted to avoid the angles.
\begin{align*}
Area_{\triangle APB} &= \frac12 \cdot PB \cdot PA\\
&= \frac12 \cdot BQ\cdot QC\\
&= Area_{\triangle BQC}
\end{align*}
(Note: In fact, \(\triangle APB\) and \(\triangle BQC\) are congruent. The loose explanation is because they're identical triangles from symmetry. But one could also prove the congruence. Not sure if it was required.)

But once we're here, we can just look at the diagram and see that \(Area_{\triangle BQC} = Area_{\triangle BQS} + Area_{\triangle QSC} = A_1 = 2A_1 = 3A_1\). You now have all the ingredients to finish it off.

Hi RuiAce,
That was very clear and thoroughly explained.
I still haven't figured out what the sample solutions were trying to get at with those ratios of 1:3 altitude into 1:3 area, so this helps a lot.
Thank you so much!