Question

In given figure, $\angle CAB={90}^0$ and $AD\perp BC$. If AC=75 cm. AB=1 m and BD=1.25 m, find AD

Answer 1

We have,

AB=1 m=100 cm, AC=75 cm and BD=125 cm.

In $△BAC$ and $△BDA$, we have

$\angle BAC=\angle BDA$ [Each equal to ${90}^{\circ }$]

and, $\angle B=\angle B$

So, by AA-criterion of similarity, we have

$△BAC\sim △BDA$

$\Rightarrow$ $\frac{BA}{BD}=\frac{AC}{AD}$

$\Rightarrow$ $\frac{100}{125}=\frac{75}{AD}$

$\Rightarrow$ $AD=\frac{125\times 75}{100}cm=93.75cm$