No. Tripling the dimensions (the side of the square base and the height) would increase the surface area of the pyramid by a factor of [tex]3^2[/tex] or [tex]9[/tex]
Let
[tex]h=\text{the height of the pyramid}\\s=\text{length of one side of the square base}\\t=\text{total surface area of the pyramid}[/tex]
The total surface area of the square-based pyramid is given by
[tex]t=b(b+\sqrt{4h^2+b^2})[/tex]
Let the tripled dimensions be given by [tex]B=3b[/tex] and [tex]H=3h[/tex], then the new surface area will be
[tex]T=B(B+\sqrt{4H^2+B^2})\\=3b(3b+\sqrt{4(3h)^2+(3b)^2})\\=3b(3b+\sqrt{9(4h^2+b^2)})\\=3b(3b+3 \sqrt{4h^2+b^2})\\=9b(b+\sqrt{4h^2+b^2})\\=9t[/tex]
The new surface area will be greater than the original by a factor of [tex]9[/tex]
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