A villager Ramayya has a plot of land in the shape of a quadrilateral. The grampanchayat of the village decided to take over some portion of his plot from one of the corners to construct a school. Ramayya agrees to the above proposal with the condition that he should be given equal amount of land in exchange of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented. (Draw a rough sketch of plot).

The shape of plot is quadrilateral but actual shape is not mentioned so we can take any quadrilateral

Here let us consider shape of plot to be square as shown

Consider O as midpoint of AB and join DO as shown

Thus AO = OB …(i)

Area(ΔAOD) is the area given by ramayya to construct school

Now extend DO and CB so that they meet at point R as shown

Area(ΔBOR) is given to Ramayya so that now his plot is ΔDRC

We have to prove that Area(ΔAOD) = Area(ΔBOR)

∠DAO = 90° and ∠OBR = 90° …(ABCD is a square)

∠DOA = ∠BOR …(opposite air of angles)

By AA criteria

ΔDOA ~ ΔROB …(ii)

Area(ΔDOA) = 1/2 × DA × OA …(iii)

Area(ΔROB) = 1/2 × BR × OB

But from (i) OA = OB

⇒ Area(ΔROB) = 1/2 × BR × OA …(iv)

Now looking at (iii) and (iv) if we prove DA = BR then it would imply Area(ΔAOD) = Area(ΔBOR)

Using (ii)

=

But from (i) OA = OB

⇒ = 1

⇒ DA = BR

⇒ Area(ΔAOD) = Area(ΔBOR)