Question

State true or false:

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

Area $(△AGB)=\frac{1}{3}\times$ Area $(△ABC)$

• 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, $AG:GD$ $=$ $2:1$
$AG$ $=\frac{2}{3}$ $AD$ (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$$=\frac{2}{3}\times$ Area of $△ABD$ (2)
Since $G$ is centroid of triangle and $AD$ passes through $G$, i.e. $AD$ is a median.
using property, median of a triangles divides triangle into two triangles of equal area.
Therefore, area of $△ABD=$ Area of $△ACD=\frac{1}{2}\times$ Area of $△ABC$
Area of $△ABD=\frac{1}{2}\times$ Area of $△ABC$ (3)
From 2,
Area of $△AGB$$=\frac{2}{3}\times$ Area of $△ABD$
Area of $△AGB=\frac{2}{3}\times \frac{1}{2}\times$ Area of $△ABC$ (From 3)
Area of $△AGB=\frac{1}{3}\times$ Area of $△ABC$