We are given a right square pyramid with a base side length of 10 cm and a height of 12 cm. We need to find the length of each non-base edge in centimeters. A non-base edge is also called a lateral edge.

1. Problem Description

We are given a right square pyramid with a base side length of 10 cm and a height of 12 cm. We need to find the length of each non-base edge in centimeters. A non-base edge is also called a lateral edge.

2. Solution Steps

Let the square base be $ABCD$ and the apex of the pyramid be $V$.

The base side length is 10 cm, so $AB = BC = CD = DA = 10$ cm.

The height of the pyramid is 12 cm. Let $O$ be the center of the square base. Then $VO = 12$ cm, and $VO$ is perpendicular to the base.

We need to find the length of the non-base edges, which are $VA, VB, VC,$ and $VD$. Since it's a right pyramid, all these lengths are equal. Let's find the length of $VA$.

The distance from the center $O$ to vertex $A$ is half the length of the diagonal of the square.

The diagonal of the square base is $d = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}$ cm.

Therefore, $OA = \frac{1}{2}d = \frac{1}{2}(10\sqrt{2}) = 5\sqrt{2}$ cm.

Now we have a right triangle $\triangle VOA$ with $VO = 12$ cm and $OA = 5\sqrt{2}$ cm. We want to find $VA$.

By the Pythagorean theorem, $VA^2 = VO^2 + OA^2$.

$VA^2 = (12)^2 + (5\sqrt{2})^2 = 144 + 25(2) = 144 + 50 = 194$.

$VA = \sqrt{194}$ cm.

3. Final Answer

$\sqrt{194}$