# 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.

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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 $\mathrm{SC}=625 \mathrm{cm}$.

Updated on 10-Oct-2022 13:28:09