Question

In the given figure, DE $\parallel$ BC.

(i) If $AD=3\cdot 6cm,AB=9cm$ and $AE=2\cdot 4cm,$ find $EC$

(ii) If $\frac{AD}{DB}=\frac{3}{5}$ and $AC=5\cdot 6cm,$ find $AE$

(iii) If $AD=xcm,DB=(x-2)cm,AE=(x+2)cm$ and $EC=(x-1)cm,$ find the value of $x.$

Answer 1

In $\Delta ABC$ & $\Delta ADE$

$\angle A=\angle A$ (common)

$\angle ABC=\angle ADE$ (corresponding angles)

$\angle ACB=\angle AED$ (corresponding angles)

$\therefore$ $\Delta ABC$ and $\Delta ADE$ are similar to each each by $AAA$ criterion.

Hence the corresponding sides will be proportional to each other.

i.e. $\frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}$

(i) $AD=3.6cm$ $AB=9cm$ $AE=2.4cm$

$\therefore$ $\frac{AD}{AB}=\frac{AE}{AC}\Rightarrow AC=\frac{AE\times BA}{AD}=\frac{2.4\times 9}{3.6}=6cm$

$\therefore$ $AC=6cm$

$\Rightarrow AE+EC=6cm$

$\Rightarrow EC=6-AE=6-2.4=3.6cm$

(ii) $\frac{AD}{BD}=\frac{3}{5}\Rightarrow \frac{BD}{AD}=\frac{5}{3}$ (taking reciprocal)

$\Rightarrow \frac{BD}{AD}+1=\frac{5}{3}+1$

$\Rightarrow \frac{BD+AD}{AD}=\frac{8}{3}$

$\Rightarrow \frac{AB}{BD}=\frac{8}{3}$

$\Rightarrow \frac{AD}{AB}=\frac{3}{8}$ (taking reciprocal)

Also given $AC=5.6cm$, we have to find $AE$

Now,

$\frac{AD}{AB}=\frac{AE}{AC}$

$\Rightarrow \frac{3}{8}=\frac{AE}{5.6}$

$\Rightarrow AE=\frac{3}{8}\times 5.6=2.1cm$

(iii) $AD=x$ cm

$DB=(x-2)$ cm

$AE=(x+2)$ cm

$EC=(x-1)$ cm

We have $\frac{AD}{AB}=\frac{AE}{AC}$

$\Rightarrow \frac{AD}{AD+DB}=\frac{AE}{AE+EC}$

$\Rightarrow \frac{x}{x+x-2}=\frac{x+2}{x+2+x-1}$

$\Rightarrow \frac{x}{2x-2}=\frac{x+2}{2x+1}$

$\Rightarrow x(2x+1)=(2x-2)(x+2)$

$\Rightarrow {2x}^2+x={2x}^2+4x-2x-4$

$\Rightarrow x=2x-4$

$\Rightarrow x=4$