2007 AMC 10B Problem 23

Let $h$ be the altitude of the original pyramid; the smaller pyramid has altitude $h-2.$ The two pyramids are similar, so the ratio of their surface areas is the square of the ratio of their altitudes. The smaller surface area is half the original, so $\left(\dfrac{h-2}{h}\right)^2=\dfrac12,$ giving $\dfrac{h}{h-2}=\sqrt2.$ Then $h=\sqrt2(h-2),$ so $h(\sqrt2-1)=2\sqrt2$ and $h=\dfrac{2\sqrt2}{\sqrt2-1}=4+2\sqrt2.$ Thus, the correct answer is E.