Understanding Similar Triangles and Area Ratios

Two triangles are said to be similar if their corresponding angles are equal and their corresponding sides are in proportion. A key property of similar triangles relates the ratio of their areas to the ratio of their corresponding sides.

Property of Area Ratio for Similar Triangles

If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides. Mathematically, if $\Delta ABC \sim \Delta PQR$, then:

$$\frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta PQR)} = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{AC}{PR}\right)^2$$

Analyzing the Given Information

We are given that $\Delta ABC$ and $\Delta PQR$ are similar triangles. This is written as $\Delta ABC \sim \Delta PQR$.

This similarity statement tells us the correspondence between vertices:

• A corresponds to P
• B corresponds to Q
• C corresponds to R

This also means the corresponding sides are AB and PQ, BC and QR, and AC and PR.

We are also given the ratio of the lengths of a pair of corresponding sides:

$$\frac{BC}{QR} = \frac{1}{3}$$

Calculating the Ratio of Areas

Using the property of similar triangles, the ratio of the area of $\Delta ABC$ to the area of $\Delta PQR$ is the square of the ratio of their corresponding sides. We can use the given ratio $\frac{BC}{QR}$:

$$\frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta PQR)} = \left(\frac{BC}{QR}\right)^2$$

Substitute the given value $\frac{BC}{QR} = \frac{1}{3}$ into the formula:

$$\frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta PQR)} = \left(\frac{1}{3}\right)^2 = \frac{1^2}{3^2} = \frac{1}{9}$$

Finding the Required Ratio

The question asks for the ratio $\frac{\text{ar}(\Delta PQR)}{\text{ar}(\Delta BCA)}$.

First, note that $\Delta BCA$ refers to the same triangle as $\Delta ABC$. The order of vertices can be rearranged for the same triangle, but for similarity, the order matters to show correspondence. Here, $\Delta BCA$ just names the triangle, it doesn't imply a different correspondence for area calculation unless it's in the context of similarity with another triangle.

So, we need to find $\frac{\text{ar}(\Delta PQR)}{\text{ar}(\Delta ABC)}$.

We found that $\frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta PQR)} = \frac{1}{9}$.

To find the reciprocal ratio, we just flip the fraction:

$$\frac{\text{ar}(\Delta PQR)}{\text{ar}(\Delta ABC)} = \frac{1}{\frac{1}{9}} = 1 \times 9 = 9$$

Therefore, $\frac{\text{ar}(\Delta PQR)}{\text{ar}(\Delta BCA)} = 9$.

Final Answer Calculation

Given: $\Delta ABC \sim \Delta PQR$ and $\frac{BC}{QR} = \frac{1}{3}$.

Ratio of areas: $\frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta PQR)} = \left(\frac{BC}{QR}\right)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$.

Required ratio: $\frac{\text{ar}(\Delta PQR)}{\text{ar}(\Delta BCA)} = \frac{\text{ar}(\Delta PQR)}{\text{ar}(\Delta ABC)}$.

Since $\frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta PQR)} = \frac{1}{9}$, the reciprocal is $\frac{\text{ar}(\Delta PQR)}{\text{ar}(\Delta ABC)} = 9$.

Revision Table: Similar Triangles Area Ratio

Additional Information: Scale Factor and Area

The ratio of corresponding sides of similar figures is called the scale factor. In this problem, the scale factor from $\Delta ABC$ to $\Delta PQR$ using sides BC and QR is $k = \frac{BC}{QR} = \frac{1}{3}$.

The ratio of areas is the square of this scale factor.

$$\frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta PQR)} = k^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$$

If you consider the scale factor from $\Delta PQR$ to $\Delta ABC$, it would be the reciprocal, $\frac{QR}{BC} = 3$. Then the ratio of areas $\frac{\text{ar}(\Delta PQR)}{\text{ar}(\Delta ABC)}$ would be $(3)^2 = 9$. This confirms our result.

The ratio of perimeters of similar triangles is equal to the scale factor (the ratio of corresponding sides).

$$\frac{\text{Perimeter}(\Delta ABC)}{\text{Perimeter}(\Delta PQR)} = \frac{AB+BC+AC}{PQ+QR+PR} = \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = k$$