Question
It is given that $\triangle\text{ABC}\sim\triangle\text{EDF}$ such that AB = 5cm, AC = 7cm, DF= 15cm and DE = 12cm. Find the lengths of the remaining sides of the triangles.
✓
Answer
$\triangle\text{ABC}\sim\triangle\text{EDF}$ [Given]$\therefore\frac{\text{AB}}{\text{ED}}=\frac{\text{AC}}{\text{EF}}=\frac{\text{BC}}{\text{DF}}$
$\Rightarrow\frac{5}{12}=\frac{7}{\text{y}}=\frac{\text{x}}{15}$
$\Rightarrow\frac{5}{12}=\frac{7}{\text{y}}$
$\Rightarrow\text{y}=\frac{7\times12}{5}$
$\Rightarrow\text{y}=\frac{84}{5}$
$\Rightarrow\text{y}=16.8\text{cm}$
$\Rightarrow\text{x}=\frac{5\times15}{12}$
$\Rightarrow\text{x}=\frac{25}{4}$
$\Rightarrow\text{x}=6.25\text{cm}$
Hence, the length of BC = 6.25cm and EF = 16.8cm.
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