Question

In the given figure, MN∥BC and AM : MB = 1 : 2.
Find$\frac{\mathrm{area}\left(△\mathrm{AMN}\right)}{\mathrm{area}\left(△\mathrm{ABC}\right)}$

Answer 1

We have
AM : MB = 1 : 2
$\Rightarrow \frac{\mathrm{MB}}{\mathrm{AM}}=\frac{2}{1}$
Adding 1 to both sides, we get
$$\begin{gathered} \Rightarrow \frac{\mathrm{MB}}{\mathrm{AM}}+1=\frac{2}{1}+1 \\ \Rightarrow \frac{\mathrm{MB}+\mathrm{AM}}{\mathrm{AM}}=\frac{2+1}{1} \\ \Rightarrow \frac{\mathrm{AB}}{\mathrm{AM}}=\frac{3}{1} \end{gathered}$$
Now, In △AMN and △ABC
∠AMN = ∠ABC (Corresponding angles in MN∥BC)
∠ANM = ∠ACB (Corresponding angles in MN∥BC)
By AA similarity criterion, △AMN ∼ △ABC
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
$\therefore \frac{\mathrm{area}\left(△\mathrm{AMN}\right)}{\mathrm{area}\left(△\mathrm{ABC}\right)}={\left(\frac{\mathrm{AM}}{\mathrm{AB}}\right)}^2={\left(\frac{1}{3}\right)}^2=\frac{1}{9}$