Question

D and E are points on the sides AB and AC, respectively, of a $∆ABC$, such that $DE\parallel BC.$


(i) If AD = 3.6 cm, AB = 10 cm and AE = 4.5 cm, find EC and AC.
(ii) If AB = 13.3 cm, AC = 11.9 cm and EC = 5.1 cm, find AD.
(iii) If $\frac{AD}{DB}=\frac{4}{7}\mathrm{and}AC=6.6cm,$ find AE.
(iv) If $\frac{AD}{AB}=\frac{8}{15}\mathrm{and}EC=3.5cm,$ find AE.

Answer 1

(i)
$\mathrm{In}△\mathrm{ABC},\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{DE}\parallel \mathrm{BC}.$
Applying Thales' theorem, we get:
$\frac{AD}{DB}=\frac{AE}{EC}$

$\because$ AD = 3.6 cm , AB = 10 cm, AE = 4.5 cm​
$\therefore$ DB = 10 $-$ 3.6 = 6.4 cm
$$\begin{gathered} or,\frac{3.6}{6.4}=\frac{4.5}{EC} \\ or,EC=\frac{6.4\times 4.5}{3.6} \\ or,EC=8cm \\ \mathrm{Thus},AC=AE+EC \\ =4.5+8=12.5cm \end{gathered}$$


(ii)

$$\begin{gathered} \mathrm{In}△\mathrm{ABC},\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{DE}\parallel \mathrm{BC}. \\ \mathrm{Applying}\mathrm{Thales}\text{'}\mathrm{Theorem},\mathrm{we}\text{get}: \\ \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}} \\ \mathrm{Adding}1\mathrm{to}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ \frac{\mathrm{AD}}{\mathrm{DB}}+1=\frac{\mathrm{AE}}{\mathrm{EC}}+1 \\ \Rightarrow \frac{\mathrm{AB}}{\mathrm{DB}}=\frac{\mathrm{AC}}{\mathrm{EC}} \\ \Rightarrow \frac{13.3}{\mathrm{DB}}=\frac{11.9}{5.1} \\ \Rightarrow \mathrm{DB}=\frac{13.3\times 5.1}{11.9}=5.7\mathrm{cm} \end{gathered}$$
$\mathrm{Therefore},AD=AB-DB=13.5-5.7=7.6\mathrm{cm}$


(iii)
$$\begin{gathered} \mathrm{In}△\mathrm{ABC},\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{DE}\parallel \mathrm{BC}. \\ \mathrm{Applying}\mathrm{Thales}\text{'}\mathrm{theorem},\mathrm{we}\text{get}: \\ \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}} \\ \Rightarrow \frac{4}{7}=\frac{\mathrm{AE}}{\mathrm{EC}} \\ \mathrm{Adding}1\mathrm{to}\mathrm{both}\mathrm{the}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ \frac{11}{7}=\frac{\mathrm{AC}}{\mathrm{EC}} \\ \Rightarrow \mathrm{EC}=\frac{6.6\times 7}{11}=4.2\mathrm{cm} \\ \mathrm{Therefore}, \\ \mathrm{AE}=\mathrm{AC}-\mathrm{EC}=6.6-4.2=2.4\mathrm{cm} \end{gathered}$$


(iv)

$$\begin{gathered} \mathrm{In}△\mathrm{ABC},\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{DE}\parallel \mathrm{BC}. \\ \mathrm{Applying}\mathrm{Thales}\text{'}\mathrm{theorem},\mathrm{we}\text{get}: \\ \frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}} \\ \Rightarrow \frac{8}{15}=\frac{\mathrm{AE}}{\mathrm{AE}+\mathrm{EC}} \\ \Rightarrow \frac{8}{15}=\frac{\mathrm{AE}}{\mathrm{AE}+3.5} \\ \Rightarrow 8\mathrm{AE}+28=15\mathrm{AE} \\ \Rightarrow 7\mathrm{AE}=28 \\ \Rightarrow \mathrm{AE}=4\mathrm{cm} \end{gathered}$$