Two similar triangles ABC and DEF such that area of Δ ABC = 64 sq. unit and area of Δ DEF = 121 sq. unit.

Statement (1) : $\dfrac{\text{Perimeter of Δ DEF}}{\text{Perimeter of Δ ABC}} = \dfrac{8}{11}$.

Statement (2) : The ratio of perimeters of two similar triangles is equal to the ratio of their areas.

1. Both the statement are true.

2. Both the statement are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Given, △ ABC ∼ △ DEF.

We know that,

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$\therefore \dfrac{\text{Area of Δ DEF}}{\text{Area of Δ ABC}} = \dfrac{DE^2}{AB^2}\\[1em] \Rightarrow\dfrac{121}{64} = \dfrac{DE^2} {AB^2}\\[1em] \Rightarrow \dfrac{AB^2}{DE^2} = \dfrac{64}{121} \\[1em] \Rightarrow \dfrac{AB}{DE} = \sqrt{\dfrac{64}{121}}\\[1em] \Rightarrow \dfrac{AB}{DE} = \dfrac{8}{11}$

Since, corresponding sides of similar triangle are proportional.

$\Rightarrow \dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}$

We know that,

For any two or more equal ratios, each ratio is equal to the ratio between sum of their antecedents and sum of their consequents.

$\Rightarrow \dfrac{AB}{DE} = \dfrac{AB + BC + AC}{DE + EF + DF} \\[1em] \Rightarrow \dfrac{\text{Perimeter of Δ ABC}}{\text{Perimeter of Δ DEF}} = \dfrac{AB}{DE} \\[1em] \Rightarrow \dfrac{\text{Perimeter of Δ DEF}}{\text{Perimeter of Δ ABC}} = \dfrac{DE}{AB} = \dfrac{11}{8}\\[1em]$

So, statement 1 is false.

The ratio of perimeters of two similar triangles is equal to the ratio of their corresponding sides.

So, statement 2 is false.

Hence, option 2 is the correct option.