Question
$\triangle\text{ABC}\sim\triangle\text{DEF}$ and their areas are respectively $64cm^2 and 121cm^2$. If $EF = 15.4cm$, find $BC.$
✓
Answer
It is given that $\triangle\text{ABC}\sim\triangle\text{DEF}$
Therefore, ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{BC}^2}{\text{EF}^2}$
Let BC be x cm.
$\Rightarrow\frac{64}{121}=\frac{\text{x}^2}{(15.4)^2}$
$\Rightarrow\text{x}^2=\frac{64\times15.4\times15.4}{121}$
$\Rightarrow\text{x}=\sqrt{\frac{64\times15.4\times15.4}{121}}$
$=\frac{8\times15.4}{11}$
$=11.2$
Hence, BC = 11.2cm
Therefore, ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{BC}^2}{\text{EF}^2}$
Let BC be x cm.
$\Rightarrow\frac{64}{121}=\frac{\text{x}^2}{(15.4)^2}$
$\Rightarrow\text{x}^2=\frac{64\times15.4\times15.4}{121}$
$\Rightarrow\text{x}=\sqrt{\frac{64\times15.4\times15.4}{121}}$
$=\frac{8\times15.4}{11}$
$=11.2$
Hence, BC = 11.2cm
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