D and E are points on the sides AB and AC respectively of a

Solution:

(i)

In $\triangle A B C$, it is given that $D E \| B C$.

Applying Thales' theorem, we have :

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

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

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

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

$\Rightarrow x=4 \mathrm{~cm}$

(ii)

In $\triangle \mathrm{ABC}$, it is given that $\mathrm{DE} \| \mathrm{BC}$.

Applying Thales' theorem, we have :

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

$\Rightarrow \frac{4}{\mathrm{x}-4}=\frac{8}{3 \mathrm{x}-19}$

$\Rightarrow 4(3 \mathrm{x}-19)=8(\mathrm{x}-4)$

$\Rightarrow 12 \mathrm{x}-76=8 \mathrm{x}-32$

$\Rightarrow 4 \mathrm{x}=44$

$\Rightarrow \mathrm{x}=11 \mathrm{~cm}$

(iii)

In $\triangle \mathrm{ABC}$, it is given that $\mathrm{DE} \| \mathrm{BC}$.

Applying Thales' theorem, we have:

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

$\Rightarrow \frac{7 \mathrm{x}-4}{3 \mathrm{x}+4}=\frac{5 \mathrm{x}-2}{3 \mathrm{x}}$

$\Rightarrow 3 \mathrm{x}(7 \mathrm{x}-4)=(5 \mathrm{x}-2)(3 \mathrm{x}+4)$

$\Rightarrow 21 \mathrm{x}^{2}-12 \mathrm{x}=15 \mathrm{x}^{2}+14 \mathrm{x}-8$

$\Rightarrow 6 \mathrm{x}^{2}-26 \mathrm{x}+8=0$

$\Rightarrow(\mathrm{x}-4)(6 \mathrm{x}-2)=0$

$\Rightarrow \mathrm{x}=4, \frac{1}{3}$

$\because \mathrm{x} \neq \frac{1}{3} \quad$ (as if $x=\frac{1}{3}$ then $A E$ will become negative)

$\therefore \mathrm{x}=4 \mathrm{~cm}$