Question
Let $\triangle ABC \sim \triangle DEF$ and their areas be respectively, $64 cm^2$ and $121 cm^2$. If $EF =15 \cdot 4 cm$, find BC .
✓
Answer
Given: $\triangle\text{ABC}\sim\triangle\text{DEF}$ We know ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides,$\frac{\text{ar}\triangle\text{ABC}}{\text{ar}\triangle\text{DEF}}=\Big(\frac{\text{BC}}{\text{EF}}\Big)^2$
$\Rightarrow\frac{64}{121}=\Big(\frac{\text{BC}}{15.4}\Big)^2$
$\Rightarrow\Big(\frac{8}{11}\Big)^2=\Big(\frac{\text{BC}}{15.4}\Big)^2$
$\Rightarrow\frac{8}{11}=\frac{\text{BC}}{15.4}$
$\Rightarrow\text{BC}=\frac{8\times15.4}{11}=11.2\text{cm}$
Thus, $\text{BC}=11.2\text{cm}$
$\Rightarrow\frac{64}{121}=\Big(\frac{\text{BC}}{15.4}\Big)^2$
$\Rightarrow\Big(\frac{8}{11}\Big)^2=\Big(\frac{\text{BC}}{15.4}\Big)^2$
$\Rightarrow\frac{8}{11}=\frac{\text{BC}}{15.4}$
$\Rightarrow\text{BC}=\frac{8\times15.4}{11}=11.2\text{cm}$
Thus, $\text{BC}=11.2\text{cm}$
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