Question

In Fig., (i) and (ii), $DE||BC$. Find $EC$ in (i) and $AD$ in (ii).

Answer 1

(i) Given : $DE\parallel BC$ in $△$ ABC,

Using Basic proportionality theorem,

$\therefore \frac{AD}{DB}=\frac{AE}{EC}$

$\Rightarrow \frac{1.5}{3}=\frac{1}{EC}$

$\Rightarrow EC=\frac{3}{1.5}$

$EC=3\times \frac{10}{15}=2$ cm

$EC=2$ cm.

(ii) In $△ABC,DE\parallel BC$ (Given)

Using Basic proportionality theorem,

$\therefore \frac{AD}{DB}=\frac{AE}{EC}$

$\Rightarrow \frac{AD}{7.2}=\frac{1.8}{5.4}$

$\Rightarrow AD=1.8\times \frac{7.2}{5.4}=\frac{18}{10}\times \frac{72}{10}\times \frac{10}{54}=\frac{24}{10}$

$\Rightarrow AD=2.4$cm

So, $AD=2.4$ cm