$△$ 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 is -

7.5 cm

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15 cm

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22.5 cm

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30 cm

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Solution

The correct option is B
15 cm

Since, $Δ A B C \sim Δ D E F$

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

DE = $2 \times A B$ = 6 cm, DF = $2 \times A C$ = 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 is -

The correct option is B 15 cm Since, ΔABC∼ΔDEF So, 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 is -

The correct option is B
15 cm

Since, $Δ A B C \sim Δ D E F$

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

DE = $2 \times A B$ = 6 cm, DF = $2 \times A C$ = 5 cm.

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