In the given figure, ∠ABC = 90° and BD ⊥ AC.

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, $B C=8.1 \mathrm{~cm}$

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