Q. 104.4( 26 Votes )

A villager Ramayy

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