Question

С Mis the midpoint of seg AB and seg CM is a median of A ABC ACA AMC) A(ABMC) BD А B M Fig. 1.8 State the reason.​

Answer 1

Answer:

Let H be the height of the given triangle ABC.

It is given that M is the mid point of the segment AB and segment CM is the median of triangle ABC, therefore

AM=BM=\frac{AB}{2}

2

AB

Now, Area of triangle= \frac{1}{2}{\times}base{\times}height

2

1

×base×height

therefore, \frac{ar({\triangle AMC})}{ar({\triangle BMC})}

ar(△BMC)

ar(△AMC)

=\frac{\frac{1}{2}{\times}AM{\times}h}{\frac{1}{2}{\times}BM{\times}h }

2

1

×BM×h

2

1

×AM×h

=\frac{AM}{BM}

BM

AM

=\frac{\frac{AB}{2}}{\frac{AB}{2}}

2

AB

2

AB

=11