Question

Two solid pyramids are melted together. These pyramids had a number of edges equal to the length of each of their edges, both equal to 8 units. They are moulded to form a hexagonal pyramid with the length of each side of the base equal to 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
• E. \( 8\sqrt{35} \) units

Hint:

A pyramid has 2n edges for an n-sided base, so 8 edges means a square-base pyramid whose 8 edges are all 8 units; use its volume to find the height of the resulting hexagonal pyramid, then add the hexagon's circumradius using Pythagoras.

Answer 1

The Correct Option is A

Step 1: Identify the shape of each melted pyramid.
A pyramid built on an n-sided base has n base edges plus n lateral edges, i.e. 2n edges in all. Since the number of edges equals 8, we get \( 2n = 8 \), so \( n = 4 \): each solid is a pyramid on a square base, and every one of its 8 edges (4 base + 4 lateral) measures 8 units. This means all four triangular faces are equilateral triangles of side 8.

Step 2: Find the height and volume of one such pyramid.
Let base side \( a = 8 \). The apex sits above the centre of the square base. Distance from the centre to a base vertex \( = \frac{a}{\sqrt{2}} \).
Height \( h = \sqrt{a^2 - \left(\frac{a}{\sqrt2}\right)^2} = \sqrt{a^2 - \frac{a^2}{2}} = \frac{a}{\sqrt2} \)
With \( a = 8 \): \( h = \frac{8}{\sqrt2} = 4\sqrt2 \)
Volume of one pyramid \( = \frac{1}{3}a^2h = \frac{1}{3}(64)(4\sqrt2) = \frac{256\sqrt2}{3} \)
Two such pyramids melted together give total volume \( = 2 \times \frac{256\sqrt2}{3} = \frac{512\sqrt2}{3} \) cubic units.

Step 3: Set up the hexagonal pyramid.
Area of a regular hexagon of side 8 \( = \frac{3\sqrt3}{2}(8)^2 = 96\sqrt3 \)
Let \( H \) be the height of the new hexagonal pyramid. Since volume is conserved:
\( \frac{512\sqrt2}{3} = \frac{1}{3}(96\sqrt3)H \)
\( H = \frac{512\sqrt2}{96\sqrt3} = \frac{16\sqrt2}{3\sqrt3} = \frac{16\sqrt6}{9} \)

Step 4: Find the slant height (apex to a base vertex).
For a regular hexagon, the distance from the centre to a vertex (circumradius) equals the side length, so \( R = 8 \).
\( H^2 = \frac{256 \times 6}{81} = \frac{1536}{81} = \frac{512}{27} \)
Slant height \( l = \sqrt{H^2+R^2} = \sqrt{\frac{512}{27}+64} = \sqrt{\frac{512+1728}{27}} = \sqrt{\frac{2240}{27}}\)
\( = \sqrt{\frac{64 \times 35}{27}} = 8\sqrt{\frac{35}{27}} = \frac{8}{3}\sqrt{\frac{35}{3}}\)

So the slant height of the new pyramid is \( \frac{8}{3}\sqrt{\frac{35}{3}} \) units, matching option (a).