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.

AcademicMathematicsNCERTClass 10

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

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

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