Question

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

(i) If $CD=10cm$ and $BD=8cm$; find $AD$.

Answer 1

Step 1 : Drawing the figure

Given : $\angle B={90}^{\circ }$ and $BD\perp AC$, $CD=10cm,BD=8cm$.


Step 2 : Finding the value of $AD$

In $\Delta ABC$ and $\Delta ADB,$

$\angle A=\angle A$ [Common angle]

$\angle ABC=\angle ADB$ [ Both are ${90}^{\circ }$]

By $AA$ criterion, $\Delta ABC\sim \Delta ADB$

$\Rightarrow \frac{AB}{AD}=\frac{BC}{DB}=\frac{AC}{AB}$

[Corresponding sides of similar triangles are proportional]

Consider, $\frac{AB}{AD}=\frac{BC}{DB}$

$\Rightarrow \frac{AB}{BC}=\frac{AD}{DB}$ … (1)

Similarly, $\Delta ABC\sim \Delta BDC$

$\Rightarrow \frac{AB}{BD}=\frac{BC}{DC}$

$\Rightarrow \frac{AB}{BC}=\frac{BD}{DC}$ … (2)

On comparing (1) and (2), we get;

$\frac{AD}{DB}=\frac{BD}{DC}$

$\Rightarrow \frac{AD}{8}=\frac{8}{10}$

$\Rightarrow AD=\frac{8\times 8}{10}$

$\Rightarrow AD=\frac{64}{10}=6.4cm$

Hence, $AD$ is $6.4cm$.