Question

In an equilateral ∆ABC, D is the midpoint of AB and E is the midpoint of AC. Then, ar(∆ABC) : ar(∆ADE) = ?


(a) 2 : 1
(b) 4 : 1
(c) 1 : 2
(d) 1 : 4

Answer 1

(b) 4:1
In ∆ABC, D is the midpoint of AB and E is the midpoint of AC.
Therefore, by midpoint theorem, DE$\parallel$BC.
Also, by Basic Proportionality Theorem,
$\frac{AD}{DB}=\frac{AE}{EC}$

$$\begin{gathered} Also, \\ AB=AC=BC\left(\because ∆ABC\mathrm{is}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}\right) \\ So,\frac{AD}{DB}=\frac{AE}{EC}=1 \\ \mathrm{In}△ABCand△ADE,\mathrm{we}\mathrm{have}: \\ \angle A=\angle A \\ \frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{2} \\ \therefore △ABC~△ADE(SAS\mathrm{criterion}) \\ \therefore ar\left(△ABC\right):ar\left(△ADE\right)={\left(AB\right)}^2:{\left(AD\right)}^2 \\ \Rightarrow ar\left(△ABC\right):ar\left(△ADE\right)=2^2:1^2 \\ \Rightarrow ar\left(△ABC\right):ar\left(△ADE\right)=4:1 \end{gathered}$$