Discussion
For that same pyramid the slant height of a triangular face is about
Answer: C
The face slant height runs from the apex to the midpoint of a base edge, so its horizontal reach is half the base side, \( 20 \) mm, and not half the diagonal. It is the hypotenuse on that and the axis height \( h = 50 \) mm: \[ h_{s} \;=\; \sqrt{h^{2} + \left(\tfrac{a}{2}\right)^{2}} \;=\; \sqrt{50^{2} + 20^{2}} \;=\; \sqrt{2500 + 400} \;=\; \sqrt{2900} \;\approx\; 53.9\ \text{mm}. \] This is shorter than the lateral edge \( \sqrt{3300} \approx 57.4 \) mm because the corner of the base lies further out in plan than an edge midpoint, by a factor \[ \frac{d/2}{a/2} \;=\; \sqrt{2}. \] Using the whole base side, \( \sqrt{50^{2} + 40^{2}} = \sqrt{4100} \approx 64.0 \) mm, is the usual slip.
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