Question

Find the area of the quadrilateral $ABCD$ whose vertices are $A(1,1),B(7,3),C(12,2)$ and $D(7,21)$

Answer 1

The coordinates of $A,B,C$ & $D$ are $A(1,1),B(7,3),C(12,2),D(7,21).$

Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

Area of triangle = $\frac{1}{2}|\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]|$

Area of triangle ABC = $\frac{1}{2}|\left[1\right(3-2)+7(2-1)+12(1-3\left)\right]|$

$=\frac{1}{2}|[1+7-24]|$

$=\frac{1}{2}\left(16\right)=8$ sq. units

Area of triangle ACD = $\frac{1}{2}|\left[1\right(2-21)+12(21-1)+7(1-2\left)\right]|$

$=\frac{1}{2}|[-19+240-7]|$

$=\frac{1}{2}\left(214\right)=107$ sq. units

Hence, area of quadrilateral ABCD = $107+8=115$ sq. units