Question

In the given figure, $AE$ and $BD$ are two medians of a $△ABC$ meeting at $F.$ The ratio of the area of $△ABF$ to the area of the quadrilateral $FDCE$ is

• A. 1:1
• B. 2:1
• C. 1:2
• D. 2:3

Answer 1

The correct option is A

1:1

area $△ABD=\frac{1}{2}\left(\text{area}△ABC\right)$

area $△AEC=\frac{1}{2}\left(\text{area}△ABC\right)$
$\therefore \text{area}△ABD=\text{area}△AEC$

$\Rightarrow \text{area}△ABD-\text{area}△AFD=\text{area}△AEC-\text{area}△AFD$

$\Rightarrow \text{area}△ABF=\text{area}\left(\text{quadrilateral}FDCE\right)$

So, the ratio of the area of $△ABF$ to the area of the quadrilateral $FDCE$ is $1:1$