Find the area of quadrilateral ABCD whose vertices are A(3, −1), B(9, −5),


Find the area of quadrilateral ABCD whose vertices are A(3, −1), B(9, −5), C(14, 0) and D(9, 19).     


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

Let $A\left(x_{1}, y_{1}\right)=A(3,-1), B\left(x_{2}, y_{2}\right)=B(9,-5), C\left(x_{3}, y_{3}\right)=C(14,0)$ and $D\left(x_{4}, y_{4}\right)=D(9,19)$. Then

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


$=\frac{1}{2}[-15+9+56]=25$ sq. units

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


$=\frac{1}{2}[-57+280-9]=107$ sq. units

So, the area of the quadrilateral is 25 + 107 = 132 sq. units.

Leave a comment


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