$△ A B C$ is such that $A B = 3 c m, B C = 2 c m$ and $C A = 2.5 c m .$ $△ D E F$ is similar to $△ A B C .$ If $E F = 4 c m$ , then the perimeter of $△ D E F$ is ____.

7.5 cm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

15 cm

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

22.5 cm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

30 cm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
15 cm

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

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

$∴ Perimeter of Δ D E F$
$= (D E + E F + D F)$
$= 6 + 4 + 5 = 15 c m .$

Suggest Corrections

Similar questions

△

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 find the perimeter of

△

DEF.

$△$ 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 ___

$△ A B C$ is such that $A B = 3 c m, B C = 2 c m$ and $C A = 2.5 c m .$ $△ D E F$ is similar to $△ A B C .$ If $E F = 4 c m$ , then the perimeter of $△ D E F$ is ____.

$△ A B C$ 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 -

$△ A B C$ is such that $A B = 3 c m, B C = 2 c m$ and $C A = 2.5 c m . △ D E F$ is similar to $△ A B C .$ If $E F = 4 c m,$ then the perimeter of $△ D E F$ is

Join BYJU'S Learning Program

△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=6 cm, DF=2×AC=5 cm ∴Perimeter of ΔDEF =(DE+EF+DF) =6+4+5=15 cm.

$△ A B C$ is such that $A B = 3 c m, B C = 2 c m$ and $C A = 2.5 c m .$ $△ D E F$ is similar to $△ A B C .$ If $E F = 4 c m$ , then the perimeter of $△ D E F$ is ____.

The correct option is B
15 cm

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

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

$∴ Perimeter of Δ D E F$
$= (D E + E F + D F)$
$= 6 + 4 + 5 = 15 c m .$