Question

In the following figure, $G$ is centroid of the triangle $ABC$, then

Area $\left(△AGB\right)=\frac{2}{3}\times$Area $(△ADB)$

• A. True
• B. False

Answer 1

The correct option is A True

Given, In the figure $G$ is centroid and a line from $B$ meet $AD$ at $G$. Centroid is a point where median intersects.
Using property when the medians of triangles intersect each other, it divides into ratio of $2:1$.
So, $AG:GD$ $=$ $2:1$
$\Rightarrow AG$ $=\frac{2}{3}$ $AD$ (1)
Drawing a altitude $BF$ from $B$ to $AD$ at $F$.
So,Area of $△ABD$$=\frac{1}{2}\times \text{base}\times \text{altitude}=\frac{1}{2}\times AD\times BF$
Area of $△AGB$$=\frac{1}{2}\times \text{base}\times \text{altitude}=\frac{1}{2}\times AG\times BF$
$=\frac{1}{2}\times \frac{2}{3}AD\times BF$ ( $AG$ $=\frac{2}{3}$ $AD$, From 1)
Taking the ratio of both the areas ,
$\frac{\text{Area of triangle}AGB}{\text{Area of triangle}ADB}=\frac{\frac{1}{2}\times \frac{2}{3}AD\times BF}{\frac{1}{2}\times AD\times BF}=\frac{2}{3}$
Area of $△AGB$ $=\frac{2}{3}\times$ Area of $△ABD$