$△ A B C$ is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. $△ D E F$ is similar to $△ A B C$ . If EF = 4 cm, then the perimeter of $△ D E F$ 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

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

$D E = 2 \times A B = 6 c m, D F = 2 \times A C = 5 c m .$

$∴$ Perimeter of $△ D E F = (D E + E F + D F) = 15 c m$ .

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

$△ A B C$ is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. $△ D E F$ is similar to $△ A B C$ . If EF = 4 cm, then the perimeter of $△ D E F$ is -

The correct option is B
15 cm

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

$D E = 2 \times A B = 6 c m, D F = 2 \times A C = 5 c m .$

$∴$ Perimeter of $△ D E F = (D E + E F + D F) = 15 c m$ .