!in isosceles triangle ∆abc with base bc draw the median am . find am, if the perimeter of ∆abc is 32 cm and perimeter of ∆abm is 24 cm.

Length of line segment AM = 8

I'm not going to show you a drawing of triangle ABC with the line AM, but will describe the solution.

Since the triangle is an isosceles triangle, you know that the sides AB and AC have the same length. And since point M is the median of line segment, you know that the length of BM is half the length of BC. With that in mind, let's write down some equations.

The perimeter of triangle ABC is 32, so
32 = AB + AC + BC

And since it's an isosceles triangle, we can simplify the equation to
32 = 2AB + BC

Now we know the perimeter of triangle ABM is 24, so
24 = AB + BM + AM

And since BM is half the length of BC, let's substitute that value, so
24 = AB + BC/2 + AM

Now looking at the two equations, you'll notice that "AB + BC/2" is exactly half of the value of the equation for the perimeter of ABC. So let's take that equation and divide it by 2, giving:
32 = 2AB + BC
16 = AB + BC/2

And subtract that equation from the equation for the perimeter of ABM
24 = AB + BC/2 + AM
minus
16 = AB + BC/2
equals
8 = AM

So the length of line segment AM is 8.

3x+9+4x+x=8x+9that's the answer for your question.