The perimeter of an isosceles triangle is 36cm, and its lateral side is 13cm. Find the median of the triangle, drawn to the base.

Let an isosceles triangle, given by condition, be a triangle ABC, AB = BC, AC is the base, BM is the median.
Knowing the perimeter, we find the base of the ABC triangle:
AB + BC + AC = 36;
13 + 13 + AC = 36;
AC = 36 – 26 = 10 (cm).
From the properties of an isosceles triangle, it is known that its bisector, median and height are equal to each other, so we can conclude that ВM is not only the median of the ABC triangle, but also its height, so it is perpendicular to its base AC.
ВM divides triangle ABC into two equal right-angled triangles. Let’s consider one of them AВM.
In the triangle ABM, the hypotenuse is 13 cm. Let’s find the AM leg. Since ВM is the median, it divides the AC in half into two parts AM and MС:
AM = MS = AC / 2 = 10/2 = 5 (cm).
Thus, in the AВM triangle, we know the hypotenuse and one of the legs, we need to find the second ВM leg, which is the height and median of the ABC triangle:
AB ^ 2 = AM ^ 2 + BM ^ 2;
13 ^ 2 = 5 ^ 2 + BM ^ 2;
BM ^ 2 = 169 – 25;
BM ^ 2 = 144;
BM = √144 = 12 (cm)
Answer: ВM = 12 cm