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

Question:

Find 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).

Solution:

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.

 

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now