∆ABC ∼ ∆DEF. If BC = 3 cm, EF = 4 cm and ar(∆ABC) = 54 cm2, then ar(∆DEF) =

$\triangle \mathrm{ABC} \sim \triangle \mathrm{DEF}$. If $\mathrm{BC}=3 \mathrm{~cm}, \mathrm{EF}=4 \mathrm{~cm}$ and $\operatorname{ar}(\triangle \mathrm{ABC})=54 \mathrm{~cm}^{2}$, then $\operatorname{ar}(\triangle \mathrm{DEF})=$

(a) $108 \mathrm{~cm}^{2}$

(b) $96 \mathrm{~cm}^{2}$

(c) $48 \mathrm{~cm}^{2}$

(d) $100 \mathrm{~cm}^{2}$


Given: In Δ ABC and Δ DEF

$\triangle \mathrm{ABC} \sim \triangle \mathrm{DEF}$

$\operatorname{Ar}(\triangle \mathrm{ABC})=54 \mathrm{~cm}^{2}$

$B C=3 \mathrm{~cm}$ and $E F=4 \mathrm{~cm}$

To find: Ar(Δ DEF)

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

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

$\frac{54}{\operatorname{Ar}(\triangle \mathrm{DEF})}=\frac{3^{2}}{4^{2}}$

$\frac{54}{\operatorname{Ar}(\Delta \mathrm{DEF})}=\frac{9}{16}$

$\operatorname{Ar}(\Delta \mathrm{DEF})=\frac{16 \times 54}{9}$

$\operatorname{Ar}(\triangle \mathrm{DEF})=96 \mathrm{~cm}^{2}$

Hence the correct answer is (b)


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