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

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Solution

Given: D, E, and F are the mid-points of sides BC, AC, and AB respectively of $Δ$ 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)

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

$∴$ $\frac{Area (Δ D E F)}{Area (Δ A B C)} = {(\frac{D E}{A B})}^{2}$

$\frac{Area Δ D E F}{Area Δ A B C} = {(\frac{D E}{2 D E})}^{2}$

Clearly, $a r (△ D E F)$ = $\frac{1}{4} \times$ $a r (△ A B C)$

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

$∴$ $A r (△ D E F) = 6 c m^{2}$

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