In Figure $(i)$ and $(ii),$ $DE || BC.$ Find $EC$ in $(i)$ and $AD$ in $(ii).$
(N/A) $(i)$ Let $EC = x \text{ cm}.$
It is given that $DE || BC.$
By using the Basic Proportionality Theorem,we obtain:
$\frac{AD}{DB} = \frac{AE}{EC}$
$\frac{1.5}{3} = \frac{1}{x}$
$x = \frac{3 \times 1}{1.5}$
$x = 2$
$\therefore EC = 2 \text{ cm}.$
$(ii)$ Let $AD = x \text{ cm}.$
It is given that $DE || BC.$
By using the Basic Proportionality Theorem,we obtain:
$\frac{AD}{DB} = \frac{AE}{EC}$
$\frac{x}{7.2} = \frac{1.8}{5.4}$
$x = \frac{1.8 \times 7.2}{5.4}$
$x = 2.4$
$\therefore AD = 2.4 \text{ cm}.$
It is given that $DE || BC.$
By using the Basic Proportionality Theorem,we obtain:
$\frac{AD}{DB} = \frac{AE}{EC}$
$\frac{1.5}{3} = \frac{1}{x}$
$x = \frac{3 \times 1}{1.5}$
$x = 2$
$\therefore EC = 2 \text{ cm}.$
$(ii)$ Let $AD = x \text{ cm}.$
It is given that $DE || BC.$
By using the Basic Proportionality Theorem,we obtain:
$\frac{AD}{DB} = \frac{AE}{EC}$
$\frac{x}{7.2} = \frac{1.8}{5.4}$
$x = \frac{1.8 \times 7.2}{5.4}$
$x = 2.4$
$\therefore AD = 2.4 \text{ cm}.$
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