Question

Question 5
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 the vertices of the triangle be A (4, -6), B (3, -2), and C (5, 2).
Let D be the mid-point of side BC of $\Delta ABC$.
Therefore, AD is the median in $\Delta ABC.$
Coordinates of point $D$
$=(\frac{3+5}{2},\frac{-2+2}{2})=(4,0)$
Area of a triangle
=$\frac{1}{2}\left\{x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right\}$
Area of $\Delta ABD$
$=\frac{1}{2}\left[\right(4)\left\{\right(-2)-(0\left)\right\}+3\left\{\right(0)-(-6\left)\right\}$
$+\left(4\right)\left\{\right(-6)-(-2\left)\right\}]$
$=\frac{1}{2}(-8+18-16)$
$=-3squareunits$
However, area cannot be negative. Therefore, area of $\Delta ABD$ is 3 square units.
Area of $\Delta ABD$
$=\frac{1}{2}\left[\right(4)\{0-(2\left)\right\}+4\left\{\right(2)-(-6\left)\right\}$
$+\left(5\right)\left\{\right(-6)-(0\left)\right\}]$
$=\frac{1}{2}(-8+32-30)$
$=-3squareunits$
However, area cannot be negative.
Therefore, areas of $\Delta$ ABD and $\Delta$ ACD are equal i.e. 3 square units.
Thus, median AD has divided $\Delta$ ABC in two triangles of equal areas.