∆ABC ∼ ∆DEF, ar(∆ABC) = 9 cm2, ar(∆DEF) = 16 cm2.


$\triangle \mathrm{ABC} \sim \triangle \mathrm{DEF}, \operatorname{ar}(\triangle \mathrm{ABC})=9 \mathrm{~cm}^{2}, \operatorname{ar}(\triangle \mathrm{DEF})=16 \mathrm{~cm}^{2} .$ If $\mathrm{BC}=2.1 \mathrm{~cm}$, then the measure of $\mathrm{EF}$ is

(a) 2.8 cm
(b) 4.2 cm
(c) 2.5 cm
(d) 4.1 cm


Given: $\operatorname{Ar}(\triangle \mathrm{ABC})=9 \mathrm{~cm}^{2}, \operatorname{Ar}(\triangle \mathrm{DEF})=16 \mathrm{~cm}^{2}$ and $\mathrm{BC}=2.1 \mathrm{~cm}$

To find: measure of EF

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\frac{\operatorname{Ar}(\triangle \mathrm{ABC})}{\operatorname{Ar}(\Delta \mathrm{DEF})}=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}$

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


$\mathrm{EF}=2.8 \mathrm{~cm}$

Hence the correct answer is

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