Answer :

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)

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A villager RamayyAP- Mathematics