In ΔABC, P is a point on AB such that PB : AP = 3 : 4 and PQ is parallel to AC. If AR and QS are perpendicular to PC and QS = 9 cm, what is the length (in cm) of AR?

The correct answer is 21.

Step 1: Understand the given geometric relationships
We are given triangle ABC where P is a point on AB such that the ratio PB : AP = 3 : 4.

From this ratio, we can write the proportions of segment AP and segment PB relative to the full side AB:

$\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{4}{4+3}=\frac{4}{7}$

and

$\frac{\mathrm{PB}}{\mathrm{AB}}=\frac{3}{7}$

Step 2: Apply the Basic Proportionality Theorem
Since PQ is parallel to AC (PQ ∥ AC), by the Basic Proportionality Theorem (Thales' Theorem), PQ divides sides AB and BC proportionally. Therefore:

$\frac{\mathrm{BQ}}{\mathrm{QC}}=\frac{\mathrm{PB}}{\mathrm{AP}}=\frac{3}{4}$

This implies that:

$\frac{\mathrm{QC}}{\mathrm{BC}}=\frac{4}{3+4}=\frac{4}{7}$

Step 3: Determine the areas of ΔPAC and ΔPQC
Let Area(ΔABC) represent the total area of triangle ABC.

Triangles PAC and ABC share the same vertex C and have their bases AP and AB on the line AB. Therefore, their areas are in proportion to their base lengths:

$\text{Area}\left(\mathrm{\Delta PAC}\right)=\frac{\mathrm{AP}}{\mathrm{AB}}\times \text{Area}\left(\mathrm{\Delta ABC}\right)=\frac{4}{7}\times \text{Area}\left(\mathrm{\Delta ABC}\right)$

Similarly, for triangle PBC:

$\text{Area}\left(\mathrm{\Delta PBC}\right)=\frac{\mathrm{PB}}{\mathrm{AB}}\times \text{Area}\left(\mathrm{\Delta ABC}\right)=\frac{3}{7}\times \text{Area}\left(\mathrm{\Delta ABC}\right)$

Now consider triangle PQC within triangle PBC. Since they share vertex P and their bases QC and BC lie on line BC:

$\text{Area}\left(\mathrm{\Delta PQC}\right)=\frac{\mathrm{QC}}{\mathrm{BC}}\times \text{Area}\left(\mathrm{\Delta PBC}\right)=\frac{4}{7}\times (\frac{3}{7}\times \text{Area}(\mathrm{\Delta ABC}\left)\right)=\frac{12}{49}\times \text{Area}\left(\mathrm{\Delta ABC}\right)$

Step 4: Relate the perpendiculars AR and QS to the areas
We are given that AR ⊥ PC and QS ⊥ PC. This means AR is the perpendicular altitude from vertex A to base PC in ΔPAC, and QS is the perpendicular altitude from vertex Q to base PC in ΔPQC.

Expressing the areas using base PC:

$\text{Area}\left(\mathrm{\Delta PAC}\right)=\frac{1}{2}\times \mathrm{PC}\times \mathrm{AR}$

and

$\text{Area}\left(\mathrm{\Delta PQC}\right)=\frac{1}{2}\times \mathrm{PC}\times \mathrm{QS}$

Dividing the area of ΔPAC by the area of ΔPQC:

$\frac{\text{Area}\left(\mathrm{\Delta PAC}\right)}{\text{Area}\left(\mathrm{\Delta PQC}\right)}=\frac{\frac{1}{2}\times \mathrm{PC}\times \mathrm{AR}}{\frac{1}{2}\times \mathrm{PC}\times \mathrm{QS}}=\frac{\mathrm{AR}}{\mathrm{QS}}$

Step 5: Solve for length AR
Substitute the area expressions into the ratio:

$\frac{\mathrm{AR}}{\mathrm{QS}}=\frac{\frac{4}{7}\times \text{Area}\left(\mathrm{\Delta ABC}\right)}{\frac{12}{49}\times \text{Area}\left(\mathrm{\Delta ABC}\right)}=\frac{4}{7}\times \frac{49}{12}=\frac{7}{3}$

Given that QS = 9 cm:

$\frac{\mathrm{AR}}{9}=\frac{7}{3}$

$\mathrm{AR}=9\times \frac{7}{3}=21\text{cm}$

Thus, the length of AR is 21 cm.