In the isosceles triangle ABC with the base BC, the median Am is drawn. Find Medina AM

In the isosceles triangle ABC with the base BC, the median Am is drawn. Find Medina AM, if the perimeter of the triangle ABC is 32cm, and the perimeter of the triangular ABM is 24cm

1. The perimeter of the triangle ABM = AB + BM + AM = 24 cm.

2. The perimeter of the triangle ABC = AB + BC + AC = 32 cm.

3. AB = AC, since the sides of an isosceles triangle are equal.

4. BM = CM, since the median AM divides the base BC in half.

5. BM + CM = BC. ВС = 2ВМ.

6. In the second formula, replace BC with AB, AC with 2BM:

2АВ + 2ВМ = 32 cm.Divide this expression by 2:

AB + BM = 16 cm. Substitute in the first formula:

16 + AM = 24 cm.

AM = 24 – 16 = 8 cm.

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