Question

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