In a square $$ABCD$$ of side 6 units $$P$$, $$Q$$ are mid points of $$BC$$, $$CD$$ respectively. The line segments $$BQ$$, $$DP$$ intersect in $$R$$ then find the area of the quadrilateral $$ABRD$$.

Take $$A(0, 0)$$, $$B(6, 0)$$, $$C(6, 6)$$, $$D(0, 6)$$, so $$P(6, 3)$$ and $$Q(3, 6)$$. Line $$BQ$$ is $$(6 - 3t, 6t)$$ and line $$DP$$ is $$(6s, 6 - 3s)$$; solving gives $$s = \frac{2}{3}$$ and $$R(4, 4)$$. By the shoelace formula, the area of $$ABRD$$ with vertices $$(0, 0)$$, $$(6, 0)$$, $$(4, 4)$$, $$(0, 6)$$ is $$\frac{1}{2}(24 + 24) = 24$$ sq. units.