Question

In the given figure, AE and BD are two medians of $△$ABC meeting at F. The ratio of the area of $△$ABF and 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) = $\frac{1}{2}$ (area $△$ABC),

Area ($△$AEC) = $\frac{1}{2}$ (area $△$ABC),
$\therefore$ area ( $△$ABD) = area ($△$AEC)

$\Rightarrow$ area ( $△$ABD) - area ($△$AFD)

= area ($△$AEC) - area ($△$AFD)

$\Rightarrow$ area ($△$ABF) = area (quad. FDCE).