Given that triangles $\triangle ABC$ and $\triangle PQR$ are similar, and $AD$ and $PM$ are the perpendiculars drawn from $A$ and $P$ to $BC$ and $QR$ respectively, the problem asks for the ratio of the area of $\triangle ABC$ to the area of $\triangle PQR$.

Similarity and Area Ratio

Since $\triangle ABC$ and $\triangle PQR$ are similar, the ratio of their corresponding sides (let's say $k$) will be the same. The ratio of their areas will be the square of the ratio of their corresponding sides.

Area Ratio Calculation

Given:

• $AD = 3$ cm (height in $\triangle ABC$)
• $PM = 4$ cm (height in $\triangle PQR$)

The ratio of the heights is: $\text{Ratio of heights} = \frac{AD}{PM} = \frac{3}{4}$

Since the triangles are similar, this is also the ratio of their corresponding sides, so: $\text{Ratio of areas} = \left(\frac{\text{corresponding sides}}{\text{corresponding sides}}\right)^2 = \left(\frac{3}{4}\right)^2 = \frac{9}{16}$

Thus, the ratio of the area of $\triangle ABC$ to the area of $\triangle PQR$ is $\frac{9}{16}$.

Would you like more details or have any questions?

Here are some related questions:

1. How can you verify if two triangles are similar?
2. What is the relationship between the sides of similar triangles?
3. How does the ratio of heights affect the area of similar triangles?
4. Can you find the ratio of the perimeters of $\triangle ABC$ and $\triangle PQR$?
5. What would happen to the area ratio if $AD$ was equal to $PM$?

Tip: When working with similar triangles, remember that corresponding angles are equal, and the ratio of corresponding sides is constant. This ratio squared gives you the ratio of their areas.