Question

If $D,E,F$ are respectively the midpoints of the sides $AB,BC,CA$ of $\Delta ABC$ and the area of $\Delta ABC$ is $24sq.cm$, then the area of $\Delta DEF$ is:

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

Answer 1

The correct option is D $6{cm}^2$

Given:

$\Delta ABC,D,E$ and $F$ are mid points of $AB,BC,CA$ respectively.
In $\Delta ABC$
$F$ is mid point of $AC$ and $D$ is mid point of $AB$.
Thus, by Mid point theorem, we get
$FD=\frac{1}{2}CB$,

$FD=CE$ and $FD\parallel CE$ ...(1)
Similarly,
$DE=FC$ and $DE\parallel FC$ ...(2)
$FE=DB$ and $FE\parallel DB$ ...(3)

From (1), (2) and (3)
$\square ADEF$, $\square DBEF$, $\square DECF$ are parallelograms.

The diagonal of a parallelogram divides the parallelogram into two congruent triangles.
Hence, $\Delta DEF\cong \Delta ADF$
$\Delta DEF\cong \Delta DBE$
$\Delta DEF\cong \Delta FEC$
Or, $\Delta DEF\cong \Delta ADF\cong \Delta ECF\cong \Delta ADF$
Thus, mid points divide the triangle into $4$ equal parts.
Now,

$A\left(\Delta DEF\right)=\frac{1}{4}A\left(\Delta ABC\right)$

$A\left(\Delta DEF\right)=\frac{1}{4}\left(24\right)$

$A\left(\Delta DEF\right)=6{cm}^2$

Hence, option D.