Question

A pyramid with a square base is cut by a plane that is parallel to its base and is $2$ units from the base. The surface area of the smaller pyramid that is cut from the top is half the surface area of the original pyramid. What is the altitude of the original pyramid?

$2$

$2+\sqrt2$

$1+2\sqrt2$

$4$

$4+2\sqrt2$

Solution:

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.

2007 AMC 10B Problem 23

Solution:

Problem 23 in Other Years