Question

in the given figure point D is the midpoint of side BC and point G is the centroid of triangle ABC

Answer 1

Answer:

The value of $\frac{A(\triangle AGB)}{A(\triangle ADB)}$ is $\frac{2}{3}$.

Step-by-step explanation:

Given information: D is midpoint of BC and G is centroid.

Let the area of triangle ABC be x.

Since D is midpoint of BC and AD is median, therefore AD divide the area of ABC in two equal parts.

$A(\triangle ADB)=\frac{x}{2}$

Centroid divides the median in 2:1. Since AD is median, therefore AG:GD=2:1.

$\frac{A(\triangle AGB)}{A(\triangle ADB)}=\frac{\frac{1}{2}\times BG\times AG}{\frac{1}{2}\times BG\times AD}=\frac{AG}{AD}=\frac{2}{3}$

Therefore the value of $\frac{A(\triangle AGB)}{A(\triangle ADB)}$ is $\frac{2}{3}$.