$△ A B C \sim D E F$ . 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 $E F$ .

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Solution

We have,

$\frac{A r e a (△ A B C)}{A r e a (△ D E F)} = \frac{{B C}^{2}}{{E F}^{2}}$

$\Rightarrow$ $\frac{9}{16} = \frac{{(2.1)}^{2}}{{E F}^{2}}$

$\Rightarrow$ $\frac{3}{4} = \frac{2.1}{E F}$

$\Rightarrow E F = \frac{4 \times 2.1}{3} c m = 2.8 c m$

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△ A B C \sim △ D E F

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

In given figure $△$ ABC and $△$ DEF are similar, BC = 3cm, EF = 4cm, and area of triangle ABC = 54 $c m^{2}$ find the area of $△$ DEF .

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△ABC∼DEF. If area of △ABC=9 sq.cm., area of △DEF=16 sq.cm and BC=2.1 cm, find the length of EF.

We have, Area(△ABC)Area(△DEF)=BC2EF2 ⇒ 916=(2.1)2EF2 ⇒ 34=2.1EF ⇒EF=4×2.13cm=2.8cm

$△ A B C \sim D E F$ . 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 $E F$ .

We have,

$\frac{A r e a (△ A B C)}{A r e a (△ D E F)} = \frac{{B C}^{2}}{{E F}^{2}}$

$\Rightarrow$ $\frac{9}{16} = \frac{{(2.1)}^{2}}{{E F}^{2}}$

$\Rightarrow$ $\frac{3}{4} = \frac{2.1}{E F}$

$\Rightarrow E F = \frac{4 \times 2.1}{3} c m = 2.8 c m$