Find area of the triangle formed by joining the midpoints of the sides of the triangle

The vertices of the triangle are A(2, 1), B(4, 3) and C(2, 5).

Coordinates of midpoint of $A B=P\left(x_{1}, y_{1}\right)=\left(\frac{2+4}{2}, \frac{1+3}{2}\right)=(3,2)$

Coordinates of midpoint of $B C=Q\left(x_{2}, y_{2}\right)=\left(\frac{4+2}{2}, \frac{3+5}{2}\right)=(3,4)$

Coordinates of midpoint of $A C=R\left(x_{3}, y_{3}\right)=\left(\frac{2+2}{2}, \frac{1+5}{2}\right)=(2,3)$

Now

Area of $\Delta P Q R=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

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

$=\frac{1}{2}[3+3-4]=1$ sq. unit

Hence, the area of the required triangle is 1 sq. unit.