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Find SURFACE AREA of a pyramid (hexagonal-base)?
How to find the surface area of a pyramid which has a regular hexagonal base of edge 6 cm and a height of 8 cm?
The area of the base should be simple, so the problem is the six triangular sides. Each triangle has area $\displaystyle \tfrac{1}{2}bl$, where b = 6 cm is the length of a side of the base, and l is the height of the triangle. Consider the right triangle formed by a line from the center of the base to the center of one of the sides of the base, by the center axis of the pyramid, and the height l of the side in question. Since b=6 cm, the distance from center of base to the side of the base is $\displaystyle 6 \cdot \tfrac{\sqrt{3}}{2} = 3\sqrt{3}$. Thus by the pythagorean theorem, $\displaystyle l^2 = (8 cm)^2 + (3\sqrt{3} cm)^2 = 64 cm^2 + 27 cm^2 = 91 cm^2$, and $\displaystyle l = \sqrt{91} \approx 9.54 cm$. From here, you can find the area of one of the triangles. Then just multiply by 6 to get the entire lateral area, and add the area of the base for the entire surface area.
--Kevin C.
For the base (img1):Originally Posted by kapitanvicki
You can divide it up into 6 equilateral triangles by drawing diagonals from opposite points.
The area of each of the equilateral triangles is
$\displaystyle A = \frac{1}{2}\left(6 \cdot \sqrt{6^2 -3^2}\right)=3\sqrt{27}=9\sqrt{3}$
So the Total Base Area is
$\displaystyle 6 \cdot 9\sqrt{3} = 54\sqrt{3}$
Now for the slanting plane areas (img2):
From before, we learnt that the length of $\displaystyle b$ is $\displaystyle \sqrt{6^2-3^2}=3\sqrt{3}$
Also, $\displaystyle h = 8$, as given.
Therefore, $\displaystyle a$ can be found by Pythagoras:
$\displaystyle a = \sqrt{(3\sqrt{3})^2+8^2}=\sqrt{91}$
$\displaystyle a$ is the height of the triangular plane, so we can now work out its area:
$\displaystyle \frac{1}{2}\cdot 6 \cdot \sqrt{91}=3\sqrt{91}$
Multiplying by 6 gives us the area of all of them = $\displaystyle 18\sqrt{91}$
So the total surface area is $\displaystyle 54\sqrt{3}+18\sqrt{91}$
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