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

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(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:
$\frac{AD}{DB} = \frac{AE}{EC}$
$\frac{x}{7.2} = \frac{1.8}{5.4}$
$x = \frac{1.8 \times 7.2}{5.4}$
$x = 2.4$
$\therefore AD = 2.4 \text{ cm}.$

Explore More

Similar Questions

If the areas of two similar triangles are equal,prove that they are congruent.

If a line intersects sides $AB$ and $AC$ of a $\Delta ABC$ at $D$ and $E$ respectively and is parallel to $BC$,prove that $\frac{AD}{AB} = \frac{AE}{AC}$ (see Figure).

In the figure,$\Delta ABC$ and $\Delta AMP$ are two right-angled triangles,right-angled at $B$ and $M$ respectively. Prove that:
$(i)$ $\Delta ABC \sim \Delta AMP$
$(ii)$ $\frac{CA}{PA} = \frac{BC}{MP}$

In the figure,$\angle ACB = 90^{\circ}$ and $CD \perp AB$. Prove that $\frac{BC^{2}}{AC^{2}} = \frac{BD}{AD}$.

In the figure,if $AD \perp BC$,prove that $AB^2 + CD^2 = BD^2 + AC^2$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo