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 $A(4,-6),B(3,-2)\mathrm{and}C(5,2).$

Answer 1

Step 1. Draw triangle $\Delta ABC$ with the coordinate of the vertices.

Give that $\Delta ABC$ whose vertices are $A(4,-6),B(3,-2)\mathrm{and}C(5,2).$

Triangle $\Delta ABC$ is drawn below,

$AD$ is the median of the triangle.

Since the median divides the opposite side into two equal parts.

Therefore, $D$ is the mid point of the side $BC$.

Co-ordinate of point $D$$=\left(\frac{3+5}{2},\frac{-2+2}{2}\right)=(4,0)$

Step 2. Calculate the area of the triangle $∆ABD$.

If the coordinates of vertices of the triangle are $(x_1,y_1),(x_2,y_2)\mathrm{and}(x_3,y_3)$. Then,

$\mathrm{Area}=\frac{1}{2}\left[x_1(y_2-y_3)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$

The area of the triangle $∆ABD$ is computed as,

$$\begin{gathered} \therefore ar(∆ABD)=\left|\frac{1}{2}\left[4(-2-0)+3(0+6)+4(-6+2)\right]\right| \\ \Rightarrow ar(∆ABD)=\left|\frac{1}{2}\left[-8+18-16\right]\right| \\ \Rightarrow ar(∆ABD)=\left|\frac{1}{2}\times (-6)\right| \\ \Rightarrow ar(∆ABD)=3 \end{gathered}$$

Step 3. Calculate the area of the $∆ACD$.

The area of the triangle $∆ACD$ is computed as,

$$\begin{gathered} \therefore ar(∆ABD)=\left|\frac{1}{2}\left[4(2-0)+5(0+6)+4(-6-2)\right]\right| \\ \Rightarrow ar(∆ABD)=\left|\frac{1}{2}\left[8+30-32\right]\right| \\ \Rightarrow ar(∆ABD)=\left|\frac{1}{2}\times \left(6\right)\right| \\ \Rightarrow ar(∆ABD)=3 \end{gathered}$$

$\because ar(∆ABD)=ar(∆ACD)$

Therefore, A median of a triangle divides it into two triangles of equal areas.