Question:
In the given figure, $\angle \mathrm{ABC}=90^{\circ}$ and $\mathrm{BD} \perp \mathrm{AC}$. If $\mathrm{AB}=5.7 \mathrm{~cm}, \mathrm{BD}=3.8 \mathrm{~cm}$ and $\mathrm{CD}=5.4 \mathrm{~cm}$, find $\mathrm{BC}$.
Solution:
It is given that $B D \perp A C, A B=5.7 \mathrm{~cm}, D B=3.8 \mathrm{~cm}, C D=5.4 \mathrm{~cm}$ and $\angle A B C=90^{\circ}$
We have to find $B C$.
Since $\triangle A B C \sim \triangle B D C$
$\Rightarrow \frac{A B}{B D}=\frac{B C}{C D}$
So
$\Rightarrow \frac{5.7 \mathrm{~cm}}{3.8 \mathrm{~cm}}=\frac{B C}{5.4 \mathrm{~cm}}$
$\Rightarrow B C=\frac{5.7 \mathrm{~cm} \times 5.4 \mathrm{~cm}}{3.8 \mathrm{~cm}}$
$=8.1 \mathrm{~cm}$
Hence, $B C=8.1 \mathrm{~cm}$