It is given that $ \triangle \mathrm{ABC} \sim \Delta \mathrm{EDF} $ such that $ \mathrm{AB}=5 \mathrm{~cm} $, $ \mathrm{AC}=7 \mathrm{~cm}, \mathrm{DF}=15 \mathrm{~cm} $ and $ \mathrm{DE}=12 \mathrm{~cm} $. Find the lengths of the remaining sides of the triangles.
Given:
\( \triangle \mathrm{ABC} \sim \Delta \mathrm{EDF} \) such that \( \mathrm{AB}=5 \mathrm{~cm} \), \( \mathrm{AC}=7 \mathrm{~cm}, \mathrm{DF}=15 \mathrm{~cm} \) and \( \mathrm{DE}=12 \mathrm{~cm} \).
To do:
We have to find the lengths of the remaining sides of the triangles.
Solution:
\( \triangle \mathrm{ABC} \sim \Delta \mathrm{EDF} \)
This implies,
$\frac{A B}{E D}=\frac{A C}{E F}=\frac{B C}{D F}$.........(i)
$A B=5 \mathrm{~cm}, A C=7 \mathrm{~cm}, D F=15 \mathrm{~cm}$ and $D E=12 \mathrm{~cm}$
Substituting the above values in (i), we get,
$\frac{5}{12}=\frac{7}{E F}=\frac{B C}{15}$
Therefore,
$\frac{5}{12}=\frac{7}{E F}$
$E F=\frac{7 \times 12}{5}$
$=16.8 \mathrm{~cm}$
$\frac{5}{12} =\frac{B C}{15}$
$B C=\frac{5 \times 15}{12}$
$=625 \mathrm{~cm}$
Hence, the lengths of the remaining sides of the triangles are $\mathrm{EF}=16.8 \mathrm{~cm}$ and $\mathrm{SC}=625 \mathrm{cm}$.
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