Question

Q11.AM is a median of a triangle ABC. Prove that AB + BC + CA > 2 AM
B4
C
012. Find the perimeter of the rectanglah​

Answer 1

Answer:

As we know that the sum of lengths of any two sides in a triangle should be greater than the length of third side.

Therefore,

In △ABM

AB+BM>AM.....(i)

In △AMC

AC+MC>AM.....(2)

Adding eq

n

(1)&(2), we have

(AB+BM)+(AC+MC)>AM+AM

⇒AB+(BM+MC)+AC>2AM

⇒AB+BC+AC>2AB

Hence AB+BC+AC>2AB

Step-by-step explanation:

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