Question

One of the equilateral triangular faces of a regular square pyramid is glued to a triangular face of a congruent regular square pyramid. If the side length of the regular square pyramid is 8, find the surface area of the new solid.

Answer 1

128 + 96√3 or 294.28 unit^3 to the nearest hundredth.

Explanation:

The surface area of one of the pyramids = area of the square base + area of the 4 equilateral triangles

= 8^2 + 4 * 1/2 * 8 * 4√3

= 64 + 64√3.

So the area of the new solid = 2(64 + 64√3) - area of 2 triangles

= 2(64 + 64√3) - 32√3

= 128 + 128√3 - 32√3

= 128 + 96√3.