# In figure below, $\mathrm{ABC}$ is a triangle right angled at $\mathrm{B}$ and $\mathrm{BD} \perp \mathrm{AC}$. If $\mathrm{AD}=4 \mathrm{~cm}$, and $C D=5 \mathrm{~cm}$, find $B D$ and $A B$."

Given:

$\mathrm{ABC}$ is a triangle right angled at $\mathrm{B}$ and $\mathrm{BD} \perp \mathrm{AC}$.

$\mathrm{AD}=4 \mathrm{~cm}$, and $C D=5 \mathrm{~cm}$

To do:

We have to find $B D$ and $A B$.

Solution:

In $\Delta A D B$ and $\Delta C D B$,

$\angle A D B =\angle C D B=90^o$

$\angle B A D =\angle D B C=90^o-\angle C$

Therefore, by AA similarity,

$\Delta D B A \sim \Delta D C B$

This implies,

$\frac{D B}{D A} =\frac{D C}{D B}$

$D B^{2} =D A \times D C$

$D B^{2} =4 \times 5$

$D B =2 \sqrt{5} \mathrm{~cm}$

In $\triangle BDC$,

$B C^{2} =B D^{2}+C D^{2}$

$B C^{2}=(2 \sqrt{5})^{2}+(5)^{2}$

$=20+25$

$=45$

$B C=\sqrt{45}$

$=3 \sqrt{5}$

$\Delta D B A \sim \Delta D C B$

This implies,

$\frac{D B}{D C}=\frac{B A}{B C}$

$\frac{2 \sqrt{5}}{5}=\frac{B A}{3 \sqrt{5}}$

$B A=\frac{2 \sqrt{5} \times 3 \sqrt{5}}$

Hence $BD=2 \sqrt{5} \mathrm{~cm}$ and $BA=6 \mathrm{~cm}$.

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