If ∆ABC ∼ ∆DEF such that AB = 9.1 cm and DE = 6.5 cm.

Given: ΔABC is similar to ΔDEF such that AB= 9.1cm, DE = 6.5cm. Perimeter of ΔDEF is 25cm.

To find: Perimeter of ΔABC.

We know that the ratio of corresponding sides of similar triangles is equal to the ratio of their perimeters.

Hence,

$\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{DE}}=\frac{\mathrm{PI}}{\mathrm{P} 2}$

$\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{P}(\Delta \mathrm{ABC})}{\mathrm{P}(\triangle \mathrm{DEF})}$

$\frac{9.1}{6.5}=\frac{\mathrm{P}(\Delta \mathrm{ABC})}{25}$

$\mathrm{P}(\Delta \mathrm{ABC})=\frac{9.1 \times 25}{6.5}$

$\mathrm{P}(\triangle \mathrm{ABC})=35 \mathrm{~cm}$

Hence the correct answer is $(d)$