Question

In triangle ABC, DE is parallel to BC; where D and E are the points on AB and AC respectively. Also, if AD = 12 cm, BD = 24 cm and BC = 8 cm, the length of DE is

• A. $\frac{2}{3}\text{cm}$
• B. $\frac{3}{4}\text{cm}$
• C. $2\frac{2}{3}\text{cm}$
• D. $2\frac{3}{4}\text{cm}$

Answer 1

The correct option is C

$2\frac{2}{3}\text{cm}$

In $\Delta ABC$, $DE\parallel BC$

$\therefore \angle ADE=\angle ABC$ (Corresponding angles)

and $\angle AED=\angle ACB$

Now, in $\Delta ADE$ and $\Delta ABC$

$\angle ADE=\angle ABC$ (Proved)

$\angle AED=\angle ACB$ (Proved)

$\angle A=\angle A$ (Common)

$\therefore \Delta ADE\sim \Delta ABC$ (AA Postulate)

$\therefore \frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}$

$\Rightarrow \frac{AD}{AD+DB}=\frac{DE}{BC}\Rightarrow \frac{2}{12+24}=\frac{DE}{8}$

($\therefore$ AD = 12 cm, DB = 24 cm and BC = 8 cm)

$\Rightarrow \frac{12}{36}=\frac{DE}{8}$

$\therefore DE=\frac{12\times 8}{36}=\frac{8}{3}=2\frac{2}{3}\text{cm}$