Let's walk through the problem step by step.

Problem Setup:

• The pyramid has a square base $ABCD$.
• The sides of the square are each 8 cm.
• The diagonals $AC$ and $BD$ intersect at point $M$, the center of the base.
• The height of the pyramid $PM$ (distance from the apex $P$ to the center of the base $M$) is 5 cm.
• We need to calculate the length of the edge $PB$.

Step-by-Step Solution:

1. Find the length of diagonal $AC$: Since the base is a square with side length $8 \, \text{cm}$, the diagonal $AC$ is: $AC = \sqrt{8^2 + 8^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2} \, \text{cm}$ Similarly, $BD = 8\sqrt{2} \, \text{cm}$.

2. Find the distance from $A$ to $M$: $M$ is the midpoint of the diagonal $AC$. Thus, the distance from $A$ to $M$ is: $AM = \frac{AC}{2} = \frac{8\sqrt{2}}{2} = 4\sqrt{2} \, \text{cm}.$

3. Form a right triangle $\triangle PBM$: In this triangle:

   • $PM = 5 \, \text{cm}$ (height of the pyramid),
   • $BM = 4\sqrt{2} \, \text{cm}$ (half of the diagonal),
   • $PB$ is the hypotenuse (the edge we need to find).

4. Use Pythagoras' Theorem: In $\triangle PBM$: $PB^2 = PM^2 + BM^2$ Substituting the known values: $PB^2 = 5^2 + (4\sqrt{2})^2 = 25 + 32 = 57.$ $PB = \sqrt{57} \, \text{cm}.$

Final Answer:

$PB = \sqrt{57} \, \text{cm} \approx 7.55 \, \text{cm}.$

Would you like me to explain any step in further detail or clarify any concept?

Related Questions:

1. How do you find the area of the square base?
2. How would you calculate the total surface area of the pyramid?
3. What is the volume of the pyramid?
4. How would the result change if the height $PM$ were doubled?
5. Can we apply trigonometry to find the angles inside $\triangle PBM$?

Tip: Whenever dealing with pyramids, right-angled triangles often appear when drawing perpendicular heights from the apex to the base!