Sample Question 23:

A pyramid with a square base is cut by a plane that is parallel to its base and \(2\) 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?

\(\textbf{(A) } 2 \qquad\textbf{(B) } 2+\sqrt{2} \qquad\textbf{(C) } 1+2\sqrt{2} \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 4+2\sqrt{2}\)

---

Answer Keys

Question 23: E

---

Solutions

Question 23
Step 1: Realize that since the two pyramids are similar, the ratio of the altitudes is the square root of the ratio of the surface areas.

Step 2: Let \(a\) be the altitude of the larger pyramid and \(a-2\) be the altitude of the smaller pyramid.

Step 3: Write down the equation based on the ratio of the altitudes and the ratio of the surface areas: \(\frac{a}{a-2}=\sqrt{\frac21}=\frac{\sqrt{2}}{1}\).

Step 4: Solve this equation by rearranging and isolating variable \(a\): \(a= a\sqrt{2} - 2\sqrt{2}\) hence \(a\sqrt{2}-a=2\sqrt{2}\).

Step 5: Simplify the equation to find \(a\): \(a=\frac{2\sqrt{2}}{\sqrt{2}-1}\).

Step 6: Continue simplifying the equation until you reach the final result: \(a= \frac{4+2\sqrt{2}}{2-1} = \textbf{(E)}~4+2\sqrt{2}\).

So, the altitude of the original pyramid is \(4+2\sqrt{2}\) units.