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.

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

Step-by-step 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.

You have 2/3 and 3/4
You need to find common denominator
You need to multiply both bottoms to get the same number on the bottom
Shorter story you need to multiply the bottom numbers by a number to get the same number on the bottom
When you get that make sure you multiply both tops by the same number
2/3 3/4

3x4= 12. 4x3=12.
Your denominator is 12
Now multiply the numerator by the same number you multiplied the denominator by.
2x4=8. 3x3=9
Answers
8/12 and 9/12