$\Delta ABC$ and $\Delta PQR$ are equilateral triangles. If $\frac{AB}{PQ} = \frac{3}{2}$,prove that $4 \times \text{Area of } \Delta ABC = 9 \times \text{Area of } \Delta PQR$.

(N/A) Given that $\Delta ABC$ and $\Delta PQR$ are equilateral triangles,they are similar by the $AAA$ similarity criterion $(\Delta ABC \sim \Delta PQR)$.
According to the theorem of the ratio of areas of two similar triangles,the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Therefore,$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \left( \frac{AB}{PQ} \right)^2$.
Given $\frac{AB}{PQ} = \frac{3}{2}$,we substitute this into the equation:
$\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta PQR)} = \left( \frac{3}{2} \right)^2 = \frac{9}{4}$.
By cross-multiplying,we get $4 \times \text{Area}(\Delta ABC) = 9 \times \text{Area}(\Delta PQR)$.
Hence,the statement is proved.