Question

Question 13
In the given figure, if $\angle ACB=\angle CDA,AC=8cm$ and $AD=3cm$, then find BD.

Answer 1

Given, $AC=8cm,AD=3cm$ and $\angle ACB=\angle CDA$
From the figure,
$\angle CDA={90}^{\circ }$
$\therefore \angle ACB={90}^{\circ }$

In right angled $\Delta ADC$,
$A{C}^2=A{D}^2+C{D}^2$ (by using pythagoras theorem)
$\Rightarrow (8{)}^2=(3{)}^2+(CD{)}^2$
$\Rightarrow 64-9=C{D}^2$
$\Rightarrow CD=\sqrt{55}cm$

In $\Delta CDB$ and $\Delta ADC$,
$\angle BDC=\angle ADC\left[each{90}^{\circ }\right]$
$\angle DBC=\angle DCA[eachequalto{90}^{\circ }-\angle A]$
$\therefore \Delta CDB\sim \Delta ADC$

Then, $\frac{CD}{BD}=\frac{AD}{CD}$

$\Rightarrow C{D}^2=AD\times BD$

$\therefore BD=\frac{C{D}^2}{AD}=\frac{(\sqrt{55}{)}^2}{3}=\frac{55}{3}cm$