Question:hard

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?

Show 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.
Updated On: Jul 21, 2026
  • \(\frac{8}{3}\sqrt{\frac{35}{3}}\) units
  • \(8\sqrt{\frac{35}{3}}\) units
  • \(2\sqrt{\frac{35}{3}}\) units
  • \(3\sqrt{35}\) units
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Use the half-octahedron trick for the original pyramid.
A square pyramid with all edges equal to e is exactly half of a regular octahedron of edge e. Volume of a regular octahedron of edge e is \(\frac{\sqrt2}{3}e^3\), so one such pyramid has volume \(\frac{\sqrt2}{6}e^3\).
With e = 8: \(V_1=\frac{\sqrt2}{6}\times512=\frac{512\sqrt2}{6}=\frac{256\sqrt2}{3}\) (matches the direct calculation).
Step 2: Combine both pyramids.
\(V=2V_1=\frac{512\sqrt2}{3}\).
Step 3: Set up the hexagonal pyramid's volume equation.
Base area of regular hexagon, side 8: \(A=\frac{3\sqrt3}{2}(8)^2=96\sqrt3\).
\(V=\frac{1}{3}AH\Rightarrow H=\frac{3V}{A}=\frac{3\times\frac{512\sqrt2}{3}}{96\sqrt3}=\frac{512\sqrt2}{96\sqrt3}=\frac{16\sqrt2}{3\sqrt3}=\frac{16\sqrt6}{9}.\)
Step 4: Find the slant edge using the circumradius.
Circumradius of a regular hexagon of side 8 is R = 8 (a special hexagon property). Slant height:\[L=\sqrt{H^2+R^2}=\sqrt{\frac{1536}{81}+64}=\sqrt{\frac{6720}{81}}=\frac{8}{3}\sqrt{\frac{35}{3}}\ units.\]\[\boxed{Slant\ height=\frac{8}{3}\sqrt{\frac{35}{3}}\ units}\]
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