Question.
$\mathrm{D}, \mathrm{E}$ and $\mathrm{F}$ are respectively the mid-points of sides $\mathrm{AB}, \mathrm{BC}$ and $\mathrm{CA}$ of $\triangle \mathrm{ABC}$. Find the ratio of the areas of $\Delta \mathrm{DEF}$ and $\triangle \mathrm{ABC}$.
$\mathrm{D}, \mathrm{E}$ and $\mathrm{F}$ are respectively the mid-points of sides $\mathrm{AB}, \mathrm{BC}$ and $\mathrm{CA}$ of $\triangle \mathrm{ABC}$. Find the ratio of the areas of $\Delta \mathrm{DEF}$ and $\triangle \mathrm{ABC}$.
Solution:
$\mathrm{DF}=\frac{1}{2} \mathrm{BC}, \mathrm{DE}=\frac{1}{2} \mathrm{AC}, \mathrm{EF}=\frac{1}{2} \mathrm{AB}$
[By midpoint theorem]
So, $\frac{D F}{B C}=\frac{D E}{A C}=\frac{E F}{A B}=\frac{1}{2}$
$\therefore \quad \Delta \mathrm{DEF} \sim \Delta \mathrm{CAB}$
So, $\frac{\operatorname{ar} \Delta \mathrm{DEF}}{\operatorname{ar} \Delta \mathrm{ABC}}=\left(\frac{\mathrm{DE}}{\mathrm{AC}}\right)^{2}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$ or $1: 4$
$\mathrm{DF}=\frac{1}{2} \mathrm{BC}, \mathrm{DE}=\frac{1}{2} \mathrm{AC}, \mathrm{EF}=\frac{1}{2} \mathrm{AB}$
[By midpoint theorem]
So, $\frac{D F}{B C}=\frac{D E}{A C}=\frac{E F}{A B}=\frac{1}{2}$
$\therefore \quad \Delta \mathrm{DEF} \sim \Delta \mathrm{CAB}$
So, $\frac{\operatorname{ar} \Delta \mathrm{DEF}}{\operatorname{ar} \Delta \mathrm{ABC}}=\left(\frac{\mathrm{DE}}{\mathrm{AC}}\right)^{2}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$ or $1: 4$