Now let us look at following fig.
You may observe that planar region formed by figure T is made up of two planar regions formed by figures P and Q. You can easily see that
Area of figure T = Area of figure P + Area of figure Q.
You may denote the area of figure A as ar(A), area of figure B as ar(B), area of figure T as ar(T), and so on. Now you can say that area of a figure is a number (in some unit) associated with the part of the plane enclosed by the figure with the following two properties:
(1) If A and B are two congruent figures, then ar(A) = ar(B); and
(2) if a planar region formed by a figure T is made up of two non-overlapping planar regions formed by figures P and Q, then
ar(T) = ar(P) + ar(Q).
ABCD is a parallelogram. E is a point on BA such that BE = 2 EA and F is a point on DC
such that DF = 2 FC. Prove that AE CF is a parallelogram whose area is one third of the
area of parallelogram ABCD.
In the below Fig, ABC and ABD are two triangles on the base AB. If line segment CD is
bisected by AB at O, show that ar (Δ ABC) = ar (Δ ABD)
D is the mid-point of side BC of ΔABC and E is the mid-point of BD. if O is the mid-point
of AE, prove that ar (ΔBOE) = `1/8` ar (Δ ABC).
If P is any point in the interior of a parallelogram ABCD, then prove that area of the
triangle APB is less than half the area of parallelogram.