Find SURFACE AREA of a pyramid (hexagonal-base)?

**kapitanvicki** · Dec 4th 2007, 11:43 PM

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?

**TwistedOne151** · Dec 5th 2007, 12:26 AM

The area of the base should be simple, so the problem is the six triangular sides. Each triangle has area $\tfrac{1}{2}bl$ , 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 $6 \cdot \tfrac{\sqrt{3}}{2} = 3\sqrt{3}$ . Thus by the pythagorean theorem, $l^2 = (8 cm)^2 + (3\sqrt{3} cm)^2 = 64 cm^2 + 27 cm^2 = 91 cm^2$ , and $l = \sqrt{91} \approx 9.54 cm$ . 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.

**DivideBy0** · Dec 5th 2007, 12:31 AM

Originally Posted by kapitanvicki

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?

For the base (img1):
You can divide it up into 6 equilateral triangles by drawing diagonals from opposite points.

The area of each of the equilateral triangles is

$A = \frac{1}{2}\left(6 \cdot \sqrt{6^2 -3^2}\right)=3\sqrt{27}=9\sqrt{3}$

So the Total Base Area is

$6 \cdot 9\sqrt{3} = 54\sqrt{3}$

Now for the slanting plane areas (img2):

From before, we learnt that the length of $b$ is $\sqrt{6^2-3^2}=3\sqrt{3}$

Also, $h = 8$ , as given.

Therefore, $a$ can be found by Pythagoras:

$a = \sqrt{(3\sqrt{3})^2+8^2}=\sqrt{91}$

$a$ is the height of the triangular plane, so we can now work out its area:
$\frac{1}{2}\cdot 6 \cdot \sqrt{91}=3\sqrt{91}$

Multiplying by 6 gives us the area of all of them = $18\sqrt{91}$

So the total surface area is $54\sqrt{3}+18\sqrt{91}$