Question:hard

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?

Show 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.
Updated On: Jul 20, 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
  • \( 8\sqrt{35} \) units
Show Solution

The Correct Option is A

Solution and Explanation

Here is a numeric-first cross-check of the same problem.

Each original pyramid is a square pyramid, base side 8, with every edge equal to 8. Its height is $h = 8/\sqrt2 \approx 5.657$ units.
Volume of one pyramid $\approx \frac{1}{3}(8^2)(5.657) \approx \frac{1}{3}(64)(5.657) \approx 120.68$
Combined volume of both pyramids $\approx 241.36$ cubic units.

The new hexagonal pyramid has base area $= \frac{3\sqrt3}{2}(8)^2 = 96\sqrt3 \approx 166.28$.
Its height $H$ satisfies $241.36 = \frac{1}{3}(166.28)H$, so $H \approx \frac{3(241.36)}{166.28} \approx 4.354$.

The circumradius of the hexagonal base equals its side, $R = 8$.
Slant height (apex to base vertex) $l = \sqrt{H^2+R^2} \approx \sqrt{4.354^2+8^2} \approx \sqrt{18.96+64} \approx \sqrt{82.96} \approx 9.108$

Checking this against option (a): $\frac{8}{3}\sqrt{35/3} = 2.667 \times 3.416 \approx 9.109$, which matches the numeric value obtained above almost exactly, confirming the exact form is correct.
\[\boxed{l=\frac{8}{3}\sqrt{\frac{35}{3}}\ \text{units}}\]
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