Answer:
Surface Area = (24√3 + 28√2) ft²
Step-by-step explanation:
A hexagonal pyramid comprises:
- hexagonal base
- 6 congruent isosceles triangles
Area of a Regular Polygon
[tex]\textsf{A}=\sf \dfrac{n\:s\:a}{2}[/tex]
where:
- n = number of sides
- s = length of one side
- a = apothem (the line drawn from the center of any polygon to the midpoint of one of the sides)
From inspection of the diagram:
- n = 6 (hexagon)
- s = 4 ft
- a = 2√3 ft
Substitute the given values into the formula to find the area of the hexagonal base:
[tex]\implies \sf A=\dfrac{6 \cdot 4 \cdot 2\sqrt{3}}{2}[/tex]
[tex]\implies \sf A=24 \sqrt{3}\:\:ft^2[/tex]
Area of an isosceles triangle
[tex]\sf A=\dfrac{1}{2}\left(b\sqrt{a^2-\dfrac{b^2}{4}}\right)[/tex]
where:
- a = length of the equal sides
- b = length of the base
From inspection of the diagram:
Substitute the given values into the formula to find the area of one of the triangles:
[tex]\implies \sf A=\dfrac{1}{2}\left(4\sqrt{6^2-\dfrac{4^2}{4}}\right)[/tex]
[tex]\implies \sf A=\dfrac{1}{2}\left(4\sqrt{32}\right)[/tex]
[tex]\implies \sf A=8\sqrt{2}\:\:ft^2[/tex]
Surface Area of the Hexagonal Prism
[tex]\implies \sf SA=\textsf{Area of hexagonal base}+6 \times \textsf{Area of isosceles triangle}[/tex]
[tex]\implies \sf SA=24\sqrt{3}+6(8\sqrt{2})[/tex]
[tex]\implies \sf SA=24\sqrt{3}+48\sqrt{2}\:\:ft^2[/tex]
Learn more about areas of regular polygons here:
https://brainly.com/question/27949277
Learn more about surface areas of prisms here:
https://brainly.com/question/27708763
