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:

The area enclosed by the curve $|x + y| + |x - y| = 11$ is

$P(2, 1), Q(4, -1), R(3, 2)$ 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:

The sides of a triangle are $4 \, cm, 5 \, cm$ and $6 \, cm$. The area of the triangle is equal to

The area (in square units) of the triangle formed by the points with polar coordinates $(1, 0)$,$(2, \frac{\pi}{3})$,and $(3, \frac{2\pi}{3})$ is:

$OPQR$ is a square and point $Q$ is $(\alpha, \alpha)$. If $M$ and $N$ are the midpoints of sides $PQ$ and $QR$ respectively,then what is the ratio of the area of the square to the area of triangle $OMN$?