Question

In △ABC,∠ABC=90o,BD⊥AC△ABC,∠ABC=90o,BD⊥AC△ABC,∠ABC=90o,BD⊥AC. If $AB=75AB=75$

Answer 1

In $△ABC$ and $△BDA$, it is given that $\angle ABC={90}^0$, $BD\perp AC$, $AB=75$ cm, $AC=1.25$ m or $AC=125$ cm and $BC=1$ m or $BC=100$ cm. ( as we know that $1$ m$=100$ cm)

Therefore,

$A{B}^2=AC\times AD\Rightarrow {75}^2=125\times AD\Rightarrow AD=\frac{75\times 75}{125}=45$

Also, $△ABC$ and $△CDB$ are similar right angled triangle, thus we have,

$B{C}^2=AC\times DC\Rightarrow {100}^2=125\times DC\Rightarrow DC=\frac{100\times 100}{125}=80$

Finally, $△ABD$ and $△CBD$ are similar right angled triangle, therefore,

$B{D}^2=AD\times DC=45\times 80=3600\Rightarrow BD=\sqrt{3600}=60$

Hence $BD=60$ cm.