In figure below, $∠\ ABC\ =\ 90^o$ and $BD\ ⊥\ AC$. If $AC\ =\ 5.7\ cm$, $BD\ =\ 3.8\ cm$ and $CD\ =\ 5.4\ cm$, find $BC$.
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Given:

In the given figure $∠\ ABC\ =\ 90^o$ and $BD\ ⊥\ AC$.

$AC\ =\ 5.7\ cm$, $BD\ =\ 3.8\ cm$ and $CD\ =\ 5.4\ cm$.

To do:

We have to find $BC$.

Solution:

In $\vartriangle ABC$ and $\vartriangle BDC$,

$\angle ABC=\angle BDC=90^o$

$\angle C=\angle C$   (common)

Therefore,

$\vartriangle ABC∼\vartriangle BDC$  (By AA similarity)

$\frac{AB}{BD} = \frac{BC}{DC}$   (Corresponding parts of similar triangles are proportional)

$\frac{5.7}{3.8} = \frac{BC}{5.4}$

$BC = \frac{5.7\times 5.4}{3.8}$

$BC = \frac{16.2}{2} = 8.1\ cm$

The measure of $BC$ is $8.1\ cm$.

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Updated on: 10-Oct-2022

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