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the base BC of a triangle ABC is divided at D so that BD=1/3 BC .prove that ar(triangle ABD)=1/2 ar (triangle ADC)

  • MATHS -

    isn't the height of the triangle the same, so the only thing different about the two new triangles is the base?

    Area ABD=1/2 Base*height= 1/2 1/3BC*h
    area ADC= 1/2 Base*height= 1/2 2/3BC*h

    dividefirst area by second

    ratio of areas= 1/2

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