ΔABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. ΔDEF ∼ ΔABC. If EF = 4 cm, then the perimeter of ΔDEF is :

1. 7.5 cm

2. 15 cm

3. 22.5 cm

4. 30 cm

Given,

ΔDEF ∼ ΔABC

Perimeter of ΔABC = AB + BC + CA = 3 + 2 + 2.5 = 7.5 cm

Since the two triangles are similar, we can write:

$\Rightarrow \dfrac{\text{Perimeter of ΔABC}}{\text{Perimeter of ΔDEF}} = \dfrac{BC}{EF} \\[1em] \Rightarrow \dfrac{7.5}{\text{Perimeter of ΔDEF}} = \dfrac{2}{4} \\[1em] \Rightarrow \text{Perimeter of ΔDEF} = \dfrac{7.5 \times 4}{2} \\[1em] \Rightarrow \text{Perimeter of ΔDEF} = \dfrac{30}{2} \\[1em] \Rightarrow \text{Perimeter of ΔDEF} = 15 \text{ cm.}$

Hence, option 2 is the correct option.