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प्रश्न
The base BC of triangle ABC is divided at D so that BD = `1/2`DC.
Prove that area of ΔABD = `1/3` of the area of ΔABC.
उत्तर
In ΔABC, ∵ BD = `1/2"DC" ⇒ "BD"/"DC" = 1/2`
∴ Ar.( ΔABD ) : Ar.( ΔADC ) = 1:2
But Ar.( ΔABD ) + Ar.( ΔADC ) = Ar.( ΔABC )
Ar.( ΔABD ) + 2Ar.( ΔABD ) = Ar.( ΔABC )
3 Ar.( ΔABD ) = Ar.( ΔABC )
Ar.( ΔABD ) = `1/3`Ar.( ΔABC )
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संबंधित प्रश्न
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