Question

Two solid pyramids are melted together. These pyramids had the number of edges equal to the length of each of their edges, equal to 8 units. They are moulded to form a hexagonal pyramid with the length of each side of its base being 8 units. What is the slant height of the new pyramid?

• A. \(\frac{8}{3}\sqrt{\frac{35}{3}}\) units
• B. \(8\sqrt{\frac{35}{3}}\) units
• C. \(2\sqrt{\frac{35}{3}}\) units
• D. \(3\sqrt{35}\) units

Hint:

A pyramid with 8 edges, all equal in length, is a regular square pyramid; find its volume first, then use the total volume to find the new hexagonal pyramid's height.

Answer 1

The Correct Option is A

Step 1: Identify the shape of each original pyramid.
A pyramid with an n-gon base has 2n edges. "Number of edges = length of each edge = 8" means each pyramid has 8 edges (so n = 4, a square base) and every edge, base and lateral, is 8 units.
Step 2: Find the height of one square pyramid.
Base is a square of side 8, so half its diagonal is \(\frac{8\sqrt2}{2}=4\sqrt2\). Since the lateral edge is also 8, the height is \(h_1=\sqrt{8^2-(4\sqrt2)^2}=\sqrt{64-32}=\sqrt{32}=4\sqrt2\).
Step 3: Find the volume of one pyramid, then of both together.
\(V_1=\frac{1}{3}\times8^2\times4\sqrt2=\frac{256\sqrt2}{3}\). Two pyramids melted together: \(V=2V_1=\frac{512\sqrt2}{3}\).
Step 4: Find the height of the new hexagonal pyramid.
Base area of a regular hexagon of side 8: \(\frac{3\sqrt3}{2}\times8^2=96\sqrt3\).
\(\frac{1}{3}\times96\sqrt3\times H=\frac{512\sqrt2}{3}\Rightarrow H=\frac{512\sqrt2}{96\sqrt3}=\frac{16\sqrt6}{9}\).
Step 5: Find the slant height (apex to base vertex).
For a regular hexagon, the circumradius equals the side, so R = 8. The slant edge is \(L=\sqrt{H^2+R^2}=\sqrt{\left(\frac{16\sqrt6}{9}\right)^2+8^2}=\sqrt{\frac{1536}{81}+64}=\sqrt{\frac{6720}{81}}=\frac{8\sqrt{105}}{9}=\frac{8}{3}\sqrt{\frac{35}{3}}.\)\[\boxed{Slant\ height=\frac{8}{3}\sqrt{\frac{35}{3}}\ units}\]