Find the area of a triangle given that midpoints of its sides are $(2, 7)$, $(1, 1)$, and $(10, 8)$.

$P(2, 1), Q(4, -1), R(3, 2)$ are the vertices of a triangle. If lines parallel to the opposite sides are drawn through $P$ and $R$ to intersect at $S$,then the area of $PQRS$ is:

If $(1, 1)$,$(3, 4)$,$(5, -2)$,and $(4, -7)$ are the vertices of a quadrilateral,find its area.

The midpoints of the three sides of a triangle are $(1, 2)$,$(-1, 1)$,and $(0, 3)$. The area of this triangle is (in sq. units):

The vertices of the triangle $ABC$ are $(2, 1)$,$(4, 3)$,and $(2, 5)$. If $D$,$E$,and $F$ are the mid-points of the sides,then the area of the triangle $DEF$ is:

$A$ triangular corner is cut from a rectangular piece of paper and the resulting pentagon has sides $5, 6, 8, 9, 12$ in some order. The ratio of the area of the pentagon to the area of the rectangle is