1] seg. CD is the median of ∆ABC .point D is the midpoint of seg. AB. A(∆ADC)/A(∆BDC =……. *
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Given : seg. CD is the median of ∆ABC . point D is the midpoint of seg. AB.
To find : A(∆ADC)/A(∆BDC =…….
Solution:
Direct Solution :
Median Divided triangle in 2 Equal area triangle
=> A(∆ADC) = A(∆BDC)
=> A(∆ADC)/A(∆BDC) = 1
Detailed :
Draw CM ⊥ AB , AD , BD as D is point of AB
Area of Triangle = (1/2) * base * Height
=> A(∆ADC) = (1/2) * AD * CM
A(∆BDC) = (1/2) * BD * CM
AD = BD as D is the midpoint of seg. AB
=> A(∆BDC) = (1/2) * AD * CM
A(∆ADC) = A(∆BDC) = (1/2) * AD * CM
Hence A(∆ADC)/A(∆BDC) = 1
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