Question

Find the area of $\Delta ABC$ with vertices $A(0,-1),B(2,1)$ and $C(0,3)$. Also, find the area of the triangle formed by joining the midpoints of its sides. Show that the ratio of the areas of these two triangles is $4:1$.

Answer 1

Vertices of $\Delta ABC$ are $A(0,-1),B(2,1)$ and $C(0,3)$

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

Area of triangle ABC:

$=\frac{1}{2}\left[0\right(1-3)+2(3+1)+0(-1-1\left)\right]$

$=\frac{1}{2}[0+2\times 4+0]=\frac{1}{2}\times 8=4$ sq. units.

From figure: Points D, E and F are midpoints of sides BC, CA and AB respectively.

Now the coordinates of D, E and F :

Coordinates of D

$=(\frac{2+0}{2},\frac{1+3}{2})$

$=(1,2)$

Coordinates of E

$=(\frac{0+0}{2},\frac{3-1}{2})$

$=(0,1)$

Coordinates of F

$=(\frac{0+2}{2},\frac{-1+1}{2})$

$=(1,0)$

Area of triangle DEF using the same formula:

$=\frac{1}{2}\left[1\right(1-0)+0(0-2)+1(2-1\left)\right]$

$=1$ sq. units

Therefore,

The ratio of the area of triangles ABC and DEF $=\frac{4}{1}=4:1$.