The problem asks for the length of the non-base edges of a right pyramid with a square base of side 10 cm and a height of 12 cm.

1. Problem Description

The problem asks for the length of the non-base edges of a right pyramid with a square base of side 10 cm and a height of 12 cm.

2. Solution Steps

The non-base edges are the edges connecting the vertex of the pyramid to the vertices of the square base. Let $s$ be the side length of the square base, which is 10 cm. Let $h$ be the height of the pyramid, which is 12 cm.

Let $e$ be the length of each non-base edge.

Consider a right triangle formed by the height of the pyramid, half of the diagonal of the square base, and a non-base edge. Let $d$ be the length of the diagonal of the square base.

The formula for the diagonal of a square with side length $s$ is:

$d = s\sqrt{2}$

In our case, $s = 10$ cm, so $d = 10\sqrt{2}$ cm.

Half of the diagonal is $\frac{d}{2} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$ cm.

Using the Pythagorean theorem, we can find the length $e$ of the non-base edge:

$e^2 = h^2 + (\frac{d}{2})^2$

$e^2 = h^2 + (\frac{s\sqrt{2}}{2})^2$

$e^2 = 12^2 + (5\sqrt{2})^2$

$e^2 = 144 + 25 \times 2$

$e^2 = 144 + 50$

$e^2 = 194$

$e = \sqrt{194}$

3. Final Answer

$\sqrt{194}$