Find the area of quadrilateral ABCD whose vertices are A(−3, −1), B(−2, −4), C(4, −1) and D(3, 4).


Find the area of quadrilateral ABCD whose vertices are A(−3, −1), B(−2, −4), C(4, −1) and D(3, 4).


By joining A and C, we get two triangles ABC and ACD.

Let $A\left(x_{1}, y_{1}\right)=A(-3,-1), B\left(x_{2}, y_{2}\right)=B(-2,-4), C\left(x_{3}, y_{3}\right)=C(4,-1)$ and $D\left(x_{4}, y_{4}\right)=D(3,4)$. Then

Area of $\Delta A B C=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$


$=\frac{1}{2}[9-0+12]=\frac{21}{2}$ sq. units

Area of $\Delta A C D=\frac{1}{2}\left[x_{1}\left(y_{3}-y_{4}\right)+x_{3}\left(y_{4}-y_{1}\right)+x_{4}\left(y_{1}-y_{3}\right)\right]$


$=\frac{1}{2}[15+20+0]=\frac{35}{2}$ sq. units

So, the area of the quadrilateral $A B C D$ is $\frac{21}{2}+\frac{35}{2}=28$ sq. units sq. units.


Leave a comment