Question

Question 9
In given figure, ABC is a triangle right angled at B and $BD\perp AC$. If AD =4 cm and CD = 5cm, then find BD and AB.

Answer 1

Given, $\Delta ABC$ in which $\angle B={90}^{\circ }$ and $BD\perp AC$.

Also, AD = 4cm and CD = 5cm
In $\Delta ADB$ and $\Delta CDB$,
$\angle ADB=\angle CDB$ [each equal to ${90}^{\circ }$]
$\angle BAD=\angle DBC$ [each equal to ${90}^{\circ }-\angle C$ ]
$\therefore$ $\Delta DBA\sim \Delta DCB$ [by AAA similarity criterion]
Then, $\frac{DB}{DA}=\frac{DC}{DB}$
$\Rightarrow D{B}^2=DA\times DC$
$\Rightarrow D{B}^2=4\times 5$
$\Rightarrow DB=2\sqrt{5}cm$

In right angled $\Delta BDC$,
$B{C}^2=B{D}^2+C{D}^2$ [ by Pyathogoras theorem]
$\Rightarrow B{C}^2=(2\sqrt{5}{)}^2+(5{)}^2$
$=20+25=45$
$\Rightarrow BC=\sqrt{45}=3\sqrt{5}$
Again, $\Delta DBA\sim \Delta DCB$,
$\therefore$ $\frac{DB}{DC}=\frac{BA}{BC}$
$\Rightarrow \frac{2\sqrt{5}}{5}=\frac{BA}{3\sqrt{5}}$
$\therefore$ $BA=\frac{2\sqrt{5}\times 3\sqrt{5}}{5}=6cm$
Hence, $BD=2\sqrt{5}cm$ and $AB=6cm$