Question

Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are $A(2,1),B(4,3)$ and $C(2,5)$.

Answer 1

Given: A triangle whose vertices are $A(2,1),B(4,3)$ and $C(2,5)$

Let D, E and F are the midpoints of the sides CB, CA and AB respectively of $\Delta ABC$, as shown in the figure.

Find vertices of D, E and F:

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

Vertices of D:

$=(\frac{4+2}{2},\frac{3+5}{2})$

$=(\frac{6}{2},\frac{8}{2})=(3,4)$

Vertices of E:

$=(\frac{2+2}{2},\frac{5+1}{2})$

$=(\frac{4}{2},\frac{6}{2})=(2,3)$

Vertices of F:

$=(\frac{2+4}{2},\frac{1+3}{2})$

$=(\frac{6}{2},\frac{4}{2})=(3,2)$

Area of triangle DEF:

We know that:

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

Area of triangle DEF$=$

$=\frac{1}{2}\left[3\right(3-2)+2(2-4)+3(4-3\left)\right]$

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

$=\frac{1}{2}\times 2=1$ sq. units.