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

In $\Delta ABC$,$A-D-B$,$A-E-C$ and $\overline{DE} \parallel \overline{BC}$. If $AB:AC = 3:4$,which of the following holds true?

The diagonals of a quadrilateral $ABCD$ intersect each other at the point $O$ such that $\frac{AO}{BO} = \frac{CO}{DO}$. Show that $ABCD$ is a trapezium.

Difficult
View Solution

In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at $D$. If $AB = 8$,$AC = 10$ and $BC = 9$,find $BD$ and $DC$.

In $\Delta XYZ$,$m\angle Y = 90^{\circ}$ and $\overline{YM}$ is an altitude to the hypotenuse $\overline{XZ}$. If $XM = \sqrt{12}$ and $ZM = \sqrt{3}$,find $YM$.

In $\Delta PQR$, the bisector of $\angle P$ intersects $\overline{QR}$ at $S$. If $PQ = 5$, $QS = 4$ and $QR = 9.6$, find $PR$.

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