Question

If D, E, F are respectively the mid-points of the sides BC, CA, and AB of $△$ABC, and the area of $△$ ABC is 24 $c{m}^2$, then find the area of $△$ DEF.

• A. 24 $c{m}^2$
• B. 12 $c{m}^2$
• C. 8 $c{m}^2$
• D. 6 $c{m}^2$

Answer 1

The correct option is D

6 $c{m}^2$

Given: D, E, and F are the mid-points of sides BC, AC, and AB respectively of $\Delta$ABC.

We know that DE = $\frac{1}{2}$ (AB) (using mid-point theorem)

Similarly, FE = $\frac{1}{2}$ (BC)
and FD = $\frac{1}{2}$ (AC)

$\Delta$ABC will be similar to $\Delta$DEF (by SSS similarity criterion)

$\therefore$$\frac{\text{Area}\left(\Delta DEF\right)}{\text{Area}\left(\Delta ABC\right)}={\left(\frac{DE}{AB}\right)}^2$

$\frac{\text{Area}\Delta DEF}{\text{Area}\Delta ABC}={\left(\frac{DE}{2DE}\right)}^2$

Clearly, $ar\left(△DEF\right)$ = $\frac{1}{4}\times$ $ar\left(△ABC\right)$

= $(\frac{1}{4}\times 24)$ = 6 $c{m}^2$

$\therefore$ $Ar\left(△\text{D}EF\right)=6c{m}^2$