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correspoesment28. If the median of aABC intersect atshow that ar(AGB) ar(AGC) - ar(BGC -1/3 ar(ABC) |
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Answer» given ,AM , BN & CL are mediansto prove -ar. ΔAGB = ar.ΔAGC , ar.ΔAGB = ar.ΔBGC & ΔAGB= 1/3ar.ΔABCproof ,inΔAGB &ΔAGC AG is the median∴ar. ΔAGB = ar.ΔAGCsimilarly , BG is the median∴ar.ΔAGB = ar.ΔBGCso we can say thatar. ΔAGB = ar.ΔAGC =ΔBGCnow ,ΔAGB +ΔAGC +ΔBGC = ar.ΔABC1/3ΔAGB +1/3ΔAGC + 1/3ΔBGC ( they are equal in area)=ar.ΔABCso we can say that , ΔAGB = ar.ΔAGC =ΔBGC = ar 1/3ΔABC ( PROVED ) |
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