Let $\triangle \mathrm{ABC} \sim \Delta \mathrm{DEF}$ and their areas be $64 \mathrm{~cm}^{2}$ and $121 \mathrm{~cm}^{2}$ respectively.
Question.
Let $\triangle \mathrm{ABC} \sim \Delta \mathrm{DEF}$ and their areas be $64 \mathrm{~cm}^{2}$ and $121 \mathrm{~cm}^{2}$ respectively. If $\mathrm{EF}=15.4 \mathrm{~cm}$, find BC.
Let $\triangle \mathrm{ABC} \sim \Delta \mathrm{DEF}$ and their areas be $64 \mathrm{~cm}^{2}$ and $121 \mathrm{~cm}^{2}$ respectively. If $\mathrm{EF}=15.4 \mathrm{~cm}$, find BC.
Solution:
$\triangle \mathrm{ABC} \sim \Delta \mathrm{DEF}($ Given $)$
$\Rightarrow \frac{\operatorname{ar}(A B C)}{\operatorname{ar}(D E F)}=\frac{B C^{2}}{E F^{2}}$ (By theorem 6.7)
$\Rightarrow \frac{64}{121}=\frac{B C^{2}}{E F^{2}} \quad \Rightarrow\left\{\frac{B C}{E F}\right\}^{2}=\left\{\frac{8}{11}\right\}^{2}$
$\Rightarrow \frac{\mathrm{BC}}{\mathrm{EF}}=\frac{8}{11} \quad \Rightarrow \mathrm{BC}=\frac{8}{11} \times \mathrm{EF}$
$\Rightarrow \mathrm{BC}=\frac{8}{11} \times 15.4 \mathrm{~cm}=11.2 \mathrm{~cm}$
$\triangle \mathrm{ABC} \sim \Delta \mathrm{DEF}($ Given $)$
$\Rightarrow \frac{\operatorname{ar}(A B C)}{\operatorname{ar}(D E F)}=\frac{B C^{2}}{E F^{2}}$ (By theorem 6.7)
$\Rightarrow \frac{64}{121}=\frac{B C^{2}}{E F^{2}} \quad \Rightarrow\left\{\frac{B C}{E F}\right\}^{2}=\left\{\frac{8}{11}\right\}^{2}$
$\Rightarrow \frac{\mathrm{BC}}{\mathrm{EF}}=\frac{8}{11} \quad \Rightarrow \mathrm{BC}=\frac{8}{11} \times \mathrm{EF}$
$\Rightarrow \mathrm{BC}=\frac{8}{11} \times 15.4 \mathrm{~cm}=11.2 \mathrm{~cm}$