$△$ ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. $△$ DEF is similar to $△$ ABC. If EF = 4 cm, then the perimeter of $△$ DEF in cm is equal to ___

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Solution

$\frac{A B}{D E}$ = $\frac{A C}{D F}$ = $\frac{B C}{E F}$ = $\frac{2}{4}$ = $\frac{1}{2}$

DE = 2 AB = 6 cm, DF = 2 AC = 5 cm.

$∴$ Perimeter of $△$ DEF = (DE + EF + DF) = 15 cm.

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△ ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm.△ DEF is similar to △ABC. If EF = 4 cm, then the perimeter of △DEF in cm is equal to ___

ABDE =ACDF = BCEF = 24 = 12 DE = 2 AB = 6 cm, DF = 2 AC = 5 cm. ∴ Perimeter of △DEF = (DE + EF + DF) = 15 cm.

$△$ ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. $△$ DEF is similar to $△$ ABC. If EF = 4 cm, then the perimeter of $△$ DEF in cm is equal to ___

$\frac{A B}{D E}$ = $\frac{A C}{D F}$ = $\frac{B C}{E F}$ = $\frac{2}{4}$ = $\frac{1}{2}$

DE = 2 AB = 6 cm, DF = 2 AC = 5 cm.

$∴$ Perimeter of $△$ DEF = (DE + EF + DF) = 15 cm.