Since M is the midpoint of the square base ABCD, we can draw lines from M to each of the vertices A, B, C, and D. This divides the square base into four smaller triangles, each with M as the midpoint of the side and a vertex.
Now, we create a right triangle VMC with legs of 4 cm and 5 cm (half the side length of the square base).
Using the Pythagorean theorem, we can find the length of VM:
VM^2 = 4^2 + 5^2
VM^2 = 16 + 25
VM^2 = 41
VM = √41
VM is approximately 6.4 cm.
Use the pyramid and the given information to find the length of VM.
Given:
M is the midpoint of the pyramid's square base.
The square base ABCD has side lengths of 5 cm.
Lengths VA, VB, VC, and VD are each 8 cm.
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