Question
Let $\triangle ABC \sim \triangle DEF$ and their areas be respetively, $64 cm^2$ and $121 cm^2$. If $EF =15.4 cm$, find $BC .$
✓
Answer
$\because \Delta ABC \sim \triangle DEF$
$\therefore \frac{\text { area of }(\Delta ABC)}{\text { area of }(\triangle DEF)}=\frac{(BC)^2}{(EF)^2} \text { (by conversion of }$
$\text { thales theorem) }[1 / 2]$
$\text { Given, area of }(\triangle ABC)=64 sq \ cm,$
$\text { area of }(\Delta DEF)=121 sq \cdot \ cm \text { and } EF=15.4 \ cm$
$\therefore \frac{64}{121}=\frac{(BC)^2}{(15.4)^2}$
$\Rightarrow \frac{BC}{15.4}=\sqrt{\frac{64}{121}}$
$\Rightarrow \frac{BC}{15.4}=\frac{8}{11} \Rightarrow BC=\frac{8 \times 15.4}{11}=11.2 \ cm$
Hence, $B C=11.2 \ cm$.
$\therefore \frac{\text { area of }(\Delta ABC)}{\text { area of }(\triangle DEF)}=\frac{(BC)^2}{(EF)^2} \text { (by conversion of }$
$\text { thales theorem) }[1 / 2]$
$\text { Given, area of }(\triangle ABC)=64 sq \ cm,$
$\text { area of }(\Delta DEF)=121 sq \cdot \ cm \text { and } EF=15.4 \ cm$
$\therefore \frac{64}{121}=\frac{(BC)^2}{(15.4)^2}$
$\Rightarrow \frac{BC}{15.4}=\sqrt{\frac{64}{121}}$
$\Rightarrow \frac{BC}{15.4}=\frac{8}{11} \Rightarrow BC=\frac{8 \times 15.4}{11}=11.2 \ cm$
Hence, $B C=11.2 \ cm$.
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