# Volume and surface area question??

• Sep 4th 2008, 07:21 PM
pyrosilver
Volume and surface area question??
"The lateral edges of a regular hexagonal pyramid are all 20 cm long, and the base edges are all 16 cm long. To the nearest cc, whta is the volume of this pyramid? To the nearest square cm, what is the combined area of the base and six lateral faces?"

For the volume, I was very confused. I tried to take one third of the base and multiplied it by the height, but I don't know if that is only for regular pyramids. I got 3200 but I don't know if I'm right because I tried it again and I got a diff answer? I haven't done the surface area yet but I'm just so confused, could someone explain to me how this is done.
• Sep 4th 2008, 07:40 PM
11rdc11
What do you mean by lateral edges, is that the sides that slant up to form a point on the pyramid?
• Sep 4th 2008, 07:41 PM
pyrosilver
Quote:

Originally Posted by 11rdc11
What do you mean by lateral edges?

Slant height.
• Sep 4th 2008, 07:55 PM
11rdc11
Ok I'm not sure if this is right but you would have to find you center point in your base and then use pythrageon theorem to figure out your height since you know a lateral edge.
• Sep 4th 2008, 08:01 PM
11rdc11
Ok just wondering by lateral edge you mean the lateral edge is side A of the triangle or is the lateral side if I bisect the triangle at the midpoint of side B?

1
101
10001
a 1000001 a
100000001
10000000001

b
• Sep 4th 2008, 08:42 PM
11rdc11
So im going to say $20^2= 16^2 + h^2$

so

$h = 12$

B is equal to $384\sqrt{3}$

So now $\frac{1}{3}(b)(h)$

That should be your volume which I get it to be 2660cm^3
• Sep 4th 2008, 09:33 PM
Dr Zoidburg
slant height (or edge length) can be found:
$s^2=h^2+a^2$
where s=slant height, h=height and a=base length.
It's easy enought to rearrange the formula to get height = 12cm.
so now you've got:
base side = 16
slant height = 20
height = 12

Volume is
http://mathworld.wolfram.com/images/...dEquation2.gif
and Surface area
http://mathworld.wolfram.com/images/...dEquation3.gif