$I f △ A B C \sim △ D E F s u c h t h a t a r e a o f △ A B C i s 9 c m^{2} a n d t h e a r e a o f △ D E F i s 25 c m^{2} a n d B C = 2.1 c m . F i n d t h e l e n g t h o f E F .$

3.0cm

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

2.8cm

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

3.5cm

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

3.6cm

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

Open in App

Solution

The correct option is C
3.5cm

$W e h a v e \frac{A r e a o f (△ A B C)}{A r e a (△ D E F)} = \frac{B C^{2}}{E F^{2}} \frac{9}{25} = \frac{(2.1)^{2}}{E F^{2}} \frac{3}{5} = \frac{2.1}{E F} \Rightarrow E F = \frac{(5 \times 2.1)}{3} c m = 3.5 c m$

Suggest Corrections

Similar questions

△ A B C \sim D E F

. If area of

△ A B C = 9 s q . c m

., area of

△ D E F = 16 s q . c m

and

B C = 2.1 c m

, find the length of

E F

If $Δ A B C \sim Δ D E F$ such that area of $Δ A B C$ is $9 c m^{2}$ and the area of $Δ D E F$ is $16 c m^{2}$ and BC = 2.1 cm, then the length of EF is

Q. Triangles ABC and DEF are similar.

(i) If area (∆ABC) = 16 cm², area (∆DEF) = 25 cm² and BC = 2.3 cm, find EF.
(ii) If area (∆ABC) = 9 cm², area (∆DEF) = 64 cm² and DE = 5.1 cm , find AB.
(iii) If AC = 19 cm and DF = 8 cm, find the ratio of the area of two triangles.
(iv) If area (∆ABC) = 36 cm², area (∆DEF) = 64 cm² and DE = 6.2 cm, find AB.
(v) If AB = 1.2 cm and DE = 1.4 cm, find the ratio of the areas of ∆ABC and ∆DEF.

△ A B C \sim △ D E F

. IF

B C = 4 c m

E F = 5 c m

and area

(△ A B C) = 32 {c m}^{2}

, determine the area of

△ D E F

If $△$ ABC is similar to $△$ DEF such that BC = 3 cm, EF = 4 cm and area of $△$ ABC = 54 ${c m}^{2}$ . Determine the area of $△$ DEF.

Join BYJU'S Learning Program

If △ABC ∼ △DEF such that area of △ ABC is 9 cm2 and the area of △DEF is 25cm2 and BC=2.1cm. Find the length of EF.

The correct option is C 3.5cm We haveArea of(△ABC)Area(△DEF)=BC2EF2925=(2.1)2EF235=2.1EF⇒EF=(5×2.1)3cm=3.5 cm

$I f △ A B C \sim △ D E F s u c h t h a t a r e a o f △ A B C i s 9 c m^{2} a n d t h e a r e a o f △ D E F i s 25 c m^{2} a n d B C = 2.1 c m . F i n d t h e l e n g t h o f E F .$

The correct option is C
3.5cm

$W e h a v e \frac{A r e a o f (△ A B C)}{A r e a (△ D E F)} = \frac{B C^{2}}{E F^{2}} \frac{9}{25} = \frac{(2.1)^{2}}{E F^{2}} \frac{3}{5} = \frac{2.1}{E F} \Rightarrow E F = \frac{(5 \times 2.1)}{3} c m = 3.5 c m$