The perimeter of an isosceles triangle is 64 cm, its lateral side is 11 cm larger. Find the height lowered to its side.

Let x denote the length of the lateral side of this triangle.

Then the length of the other side of this triangle is also x cm.

It is known from the problem statement that the lateral side of this triangle is 11 cm larger than its base.

Then the length of the base should be equal to x – 11 cm, and since the perimeter of this triangle is 64 cm, we can draw up the following equation:

x + x + x – 11 = 64,

solving which, we get:

3x – 11 = 64;

3x = 11 + 64;

3x = 75;

x = 75/3 = 25 cm.

Therefore, the base length is 25 – 11 = 14 cm.

Using the Pythagorean theorem, we find the length of the height lowered to the base:

√ (25 ^ 2 – (14/2) ^ 2) = √ (25 ^ 2 – 7 ^ 2) = √ (625 – 49) = √576 = 24.

Find the area S of this triangle:

S = 14 * 24/2 = 14 * 12 = 168 cm ^ 2.

We find the length of the height lowered to the side:

2 * S / 25 = 2 * 168/25 = 336/25 = 13.44.

Answer: 13.44 cm.