The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to ar(△ABC). 1/3 ar(△ABC). 1/4ar(△ABC). 1/2ar(△ABC).
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Given−InABCp,q,raremidpointsofAC,AB&BCtakeninorder.Tofindout−Thekind&areasofthequadrilateralspqCrorFqrBorAprq.Solution−p&qaremidpointsofAC&ADrespectively.∴pq∥Cr......(i)SimilarlypC∥qr.......(ii)∥∴from(i)&(ii)pqCrisaparallelogram.prisitsdiagonal.⟹arΔpCr=arΔpqr.InthesamewayarΔqBr=arΔpqr&arΔApp=arΔpqrSoallthefourΔshaveequalareas.∴areaofeachΔ=41arΔABCNowareaofeachparallelogram=sumoftheareaofthetwoΔsformedbyitsdiagonals=4
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