Question

In the given figure ABC is a right-angled triangle with $\angle BAC={90}^{\circ }$ [4 MARKS]

i) Prove that $\Delta ADB\sim \Delta CDA$
ii) IF BD = 18 cm, CD = 8 cm, find AD
iii) Find the ratio of the area of $\Delta ADB$ is to area of $\Delta CDA$

Answer 1

Applying theorems: 2 Marks
Calculation: 2 Marks

i) In $\Delta ADB$ and $\Delta ABC$,
$\angle ADB=\angle CAB={90}^{\circ }$ [Given]
and $\angle ABD=\angle ABC$ [Common]
$\therefore \Delta ADB\sim \Delta ABC$ ..... (1)
In $\Delta ABC$ and $\Delta CDA$,
$\angle CDA=\angle CAB={90}^{\circ }$
and $\angle ACB=\angle ACD$ [Common]
$\therefore \Delta ABC\sim \Delta CDA$ ..... (2)
From equations (1) and (2),
$\Delta ADB\sim \Delta CDA$ Hence proved

ii) Since two triangles ADB and CDA are similar (just proved), we get:
$\frac{CD}{AD}=\frac{AD}{BD}$
$\Rightarrow A{D}^2=8\times 18=144$
$\therefore AD=12cm$

iii) $\frac{ar\left(\Delta ADB\right)}{ar\left(\Delta CDA\right)}=\frac{A{D}^2}{C{D}^2}$
$=\frac{{12}^2}{8^2}=\frac{144}{64}=\frac{9}{4}=9:4$