In an isosceles triangle ABC with base BC, the median am is drawn, find the median AD

In an isosceles triangle ABC with base BC, the median am is drawn, find the median AD if the perimeter of triangle ABC is 32 cm and the perimeter of triangle ABC is 24 cm

Since AM is the median of triangle ABC, it divides the base of BC in half, then BC = CM / 2.

Since ABC is an isosceles triangle, AB = BC.

The sum of the segments (AB + BM) is equal to half the perimeter of the triangle ABC.

(AB + BM) = Ravs / 2 = 32/2 = 16 cm.

The perimeter of the triangle ABM is:

Ravm = 24 = (AB + BM) + AM.

AM = 24 – (AB + BM) = 24 – 16 = 8 cm.

Answer: The length of the median AM is 8 cm.