Question

In a $△ABC$, G is centroid Then area of $△AGC$ is_______of area of $△ABC$

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{2}{3}$

Answer 1

The correct option is B $\frac{1}{3}$

$\Rightarrow$ We know that, the median of a triangle divide it into two triangles of equal area.

$\Rightarrow$ In $△ABC$, $AD$ is the median

$\therefore$ $ar\left(△ABD\right)=ar\left(△ACD\right)$ ...(1)

$\Rightarrow$ In $△GBC$, $GD$ is the median.

$\therefore$ $ar\left(△GBD\right)=ar\left(△GCD\right)$ ...(2)

Subtracting (2) from (1), we get

$\Rightarrow$ $ar\left(△ABD\right)-ar\left(△GBD\right)=ar\left(△ACD\right)-ar\left(△GCD\right)$

$\therefore$ $ar\left(△AGB\right)=ar\left(△AGC\right)$ ...(3)

$\Rightarrow$ Similarly, $ar\left(\Delta △AGB\right)=ar\left(△BGC\right)$ ...(4)

From (3) and (4), we get

$\Rightarrow$ $ar\left(△AGB\right)=ar\left(△AGC\right)=ar\left(△BGC\right)$ ...(5)

$\Rightarrow$ Now, $ar\left(△AGB\right)+ar\left(△AGC\right)+ar\left(△BGC\right)=ar\left(△ABC\right)$

$\Rightarrow$ $ar\left(△AGC\right)+ar\left(△AGC\right)+ar\left(△AGC\right)=ar\left(△ABC\right)$ [ By using equation ( 5 ) ]

$\Rightarrow$ $3ar\left(△AGC\right)=ar\left(\Delta ABC\right)$

$\therefore$ $ar\left(△AGC\right)=\frac{1}{3}ar\left(△ABC\right)$