the surface area of a hexigonal pyramid with a slant hight of about 17 and edge of 8

Question

I was given a diagram of a hexagonal pyramid that was marked with the slant hight not complet it was ten and picked up at a coresponding point with a four square root three. I rounded it to seventeen. I need to find the surface area and it is quite frankly blowing my mind.

Answer 1

Calculate the surface area of a Regular Pyramid by using Formula : SA = (Pl)/2 + B, where P = perimeter of the base, B = area of the base and l = slant height.

Here the pyramid is regular hexagonal pyramid, so the base is hexagonal and its side is 8 units and slant height is 17 units.

The perimeter of the hexagonal P = 6a = 6(8) = 48 units.

The area of the hexagonal B = (3√3 * a²)/2 = (3√3 * 8²)/2 = 166.28 square units.

Surface area of a Regular Pyramid : SA = (Pl)/2 + B.

SA = (48 * 17)/2 + 166.28

      = 408 + 166.28

      = 574.28 square units.