Question

In ΔABC, D, E and F are midpoints of BC, AC and AB. If area of parallelogram BDEF is $20c{m}^2$, then the area of ΔABC is equal to:

• A. $10c{m}^2$
• B. $20c{m}^2$
• C. $40c{m}^2$
• D. $80c{m}^2$

Answer 1

The correct option is C $40c{m}^2$

As we know, on joining the mid-points of the three sides of any triangle will give four congruent triangles.
Here, in the given figure D, E and F are midpoints of BC, AC and AB.
So, ΔAFE, ΔBDF, ΔFDE, ΔDCE are congruent to each other.

Area of parallelogram BDEF $=20c{m}^2$ [given]

$\Rightarrow$ Area of ΔBDF + Area of ΔFDE$=20c{m}^2$
Since, Area of ΔBDF = Area of ΔFDE
$\Rightarrow 2\times$Area of ΔBDF $=20c{m}^2$
$\Rightarrow$Area of ΔBDF $=10c{m}^2$

Since, ΔAFE, ΔBDF, ΔFDE, ΔDCE are congruent to each other
$\Rightarrow$Area of ΔBDF = Area of ΔFDE = Area of ΔDCE = Area of ΔAFE
$\Rightarrow$Area of ΔABC $=4\times$Area of ΔBDF
$=4\times 10c{m}^2$$=40c{m}^2$
$\therefore$ Area of ΔABC $=40c{m}^2$