Consider a regular square pyramid
Let "a=" the base edge, "h=" the altitude, and "s=" the slant height.
The volume "V" of the frustum of a regular square pyramid is
Then
"a=\\sqrt{\\dfrac{3V}{h}}=\\sqrt{\\dfrac{3(196)}{12}}=7 (m)"
By the Pythagorean Theorem from the right triangle
"s=\\sqrt{(\\dfrac{a}{2})^2+h^2}=\\sqrt{(\\dfrac{7}{2})^2+(12)^2}=12.5(m)"
12.5 m
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