Question

You have studied in Class IX, that a median of a triangle divides it into two triangles of equal areas. Verify this result for $\Delta ABC$ whose vertices are $A(4,-6),B(3,-2)$ and $C(5,2)$.

Answer 1

Let $AD$ be the median of $△ABC$.

Hence, $D$ is the midpoint of $BC$.

We know that the coordinates of the midpoint of the line segment joining $(x_1,y_1)and(x_2,y_2)$ are:

$P(x,y)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$\therefore$ Coordinates of $D=(\frac{3+5}{2},\frac{-2+2}{2})=(4,0)$

Median $AD$ divides the $△ABC$ into two triangles.

$\therefore$ Area of $△ABD=\left|\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]\right|$

Here, $(x_1,y_1)=(4,-6),(x_2,y_2)=(3,-2)and(x_3,y_3)=(4,0)$

Hence, area of $△ABD=\frac{1}{2}\left[4\right(-2-0)+3(0+6)+4(-6+2\left)\right]$

$=\frac{1}{2}[-8+18-16]=\left|\frac{-6}{2}\right|=3$

Also, area of $△ADC=\left|\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]\right|$

Here, $(x_1,y_1)=(4,-6),(x_2,y_2)=(4,0)and(x_3,y_3)=(5,-2)$

Hence, area of $△ADC=\frac{1}{2}\left[4\right(0+2)+4(-2+6)+5(-6-0\left)\right]$

$=\frac{1}{2}[8+16-30]=\left|\frac{-6}{2}\right|=3$

$\therefore$ Area of $△ABD=$ Area of $△ADC=3$

Hence, it is proved that a median of a triangle divides it into two triangles of equal areas.