Question

Find the area of the quadrilateral whose vertices are
(i).A(2,1),B(2,3),C(-2,2),D(-1,0)
(ii) A(-3, 1),B(-2, -2), C(4,1), D(2,3)

Answer 1

(I) Let there be two triangles $△ABD$ and $△BCD$.

Thus, Area(ABCD)=Area(ABD)+Area(BCD)

Using determinant form to find area, we get

Area(ABCD)=$\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ 2& 2& -1\\ 1& 3& 0\end{array}\right|+\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ 2& -2& -1\\ 3& 2& 0\end{array}\right|$

Area(ABCD)=12

(II) Let there be two triangles $△ABC$ and $△ACD$.

Thus, Area(ABCD)=Area(ABC)+Area(ACD)

Using determinant form to find area, we get

Area(ABCD)=$\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ -3& -2& 4\\ 1& -2& 1\end{array}\right|+\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ -3& 4& 2\\ 1& 1& 3\end{array}\right|$

Area(ABCD)=17.5 or $\frac{35}{2}$