Question
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)
Answers
bobpursley
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
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
Greeshma
Not informative concept vise.Worst answer everrrrr.Shame of the person who gave this solution.