Question

In the given figure, $\angle CAB={90}^o$ and $AD\perp BC$. Show that $\Delta BDA\sim \Delta BAC$. If AC = 75 cm, AB = 1 m and BC = 1.25 m, find AD.

Answer 1

$In△CABand△ADB\angle CAB=\angle ADB={90}^o\angle CBA=\angle ABD\left(Common\right)By,AAsimilarity,△CAB\sim △ADBSo,\frac{CA}{AD}=\frac{CB}{AB}\frac{75}{AD}=\frac{125}{100}AD=\frac{75\times 100}{125}=60cmAD=60cm$