Talk:Pyramid (geometry)/Archive 1

Archive 1

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I'm new to wikipedia so I wanted to check if adding a section about the formulas for finding the surface area of pyramids would be all right. Because I think it would be fine. --Learning is good 04:32, 26 November 2006 (UTC)

• Added a surface area formula MathsIsFun 05:04, 26 November 2006 (UTC)

Thanks --Learning is good 05:12, 26 November 2006 (UTC)

"Since the area of any shape is multiplied by the square of the shape's scaling factor, the area of a cross section at height y is ..." Is this confusing to anyone else? Is there a work missing like "Since the area of any shape is <the base> multiplied by the square of the shape's scaling factor...>. I know <the base> is not the right word, but I don't understand the sentence as it is. --Rusty Maths —Preceding unsigned comment added by 84.203.137.22 (talk) 11:00, 11 January 2008 (UTC)

MOSMATH

Please: WP:MOSMATH exists. Don't write things like n+1. The correct notation is n + 1 (the n is italicized; the 1 is not; proper spacing is used). Michael Hardy (talk) 13:23, 28 March 2009 (UTC)

Johnsons

This language: "Besides the triangular pyramid, only the square and pentagonal pyramids can be composed of regular convex polygons, in which case they are Johnson solids." is not correct. A Johnson Solid is any convex polyhedrod... so a pyramid, which is a conical solid with a polygon base of n size, will always be convex. If the statement is true, and for some reason I'm mis-reading the information about Johnson solids, then the above-quoted statement needs to be referenced/cited. Alphachimera (talk) 17:00, 5 August 2009 (UTC)

(Add new comments at the bottom, using the "new section" button.)

A Johnson solid is a convex nonuniform solid with regular faces. You can't make a pyramid with a regular heptagonal base and regular (equilateral) triangle sides, because the distance from the corners to the center of the heptagon is longer than the side. —Tamfang (talk) 05:17, 6 August 2009 (UTC)

Six pyramids forming a cube

I deleted the paragraph stating that the factor of 1/3 can be derived by the fact that a cube can be divided into 6 equal pyramids. The converse is not true; the only pyramid that can form a cube in this way has a side-to-base dihedral angle of precisely 45° and a base that is exactly square. Other pyramids cannot form a cube in this way. Even if you allow distortion of a cube into cuboidal shapes, the constituent pyramids will not be equal, and therefore you cannot derive the 1/3 factor from it in any trivial way. In other words, this intuition is completely crap. Why not just stick with the real mathematical derivation of the formula in the first place?—Tetracube (talk) 17:24, 21 September 2009 (UTC)

Because for a lot of people who are less mathematical, it's easier to understand. Sometimes what Ian Stewart and Terry Pratchett called lies-to-children (see Science of Discworld) may be incomplete or entirely false, and yet still useful. However, the main volume paragraph has been rewritten to be a little clearer since I added that (now deleted) paragraph, after someone had come to me and said they found the current page utterly incomprehensible. Barefootliam (talk) 05:17, 10 February 2010 (UTC)

Simple proof of piramide value

Divide piramide into 10 parts and each part of height c=1 and so h=1*10=10, a=10, $B=a^{2}=100$. So aproximate piramide value is:

$V=c(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}+a_{5}^{2}+a_{6}^{2}+a_{7}^{2}+a_{8}^{2}+a_{10}^{2})=1(1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}+7^{2}+8^{2}+9^{2}+10^{2})=385.$

And this pretty close to ${ha^{2} \over 3}={10\cdot 10^{2} \over 3}=333.3333$.

"pretty close" is not a proof. —Tamfang (talk) 05:45, 31 December 2009 (UTC)

Divide into more peaces and you will get infinitly close to 333.(3). Like for example:

$V=c(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}+a_{5}^{2}+a_{6}^{2}+a_{7}^{2}+a_{8}^{2}+a_{10}^{2}+...+a_{99}^{2}+a_{100}^{2})=$

$=0.1(0.1^{2}+0.2^{2}+0.3^{2}+0.4^{2}+0.5^{2}+0.6^{2}+0.7^{2}+0.8^{2}+0.9^{2}+0.1^{2}+1.1^{2}+1.2^{2}+1.3^{2}+1.4^{2}+...+9.8^{2}+9.9^{2}+10^{2})=$

$=333.520$.

And if $h=100$ and $a=10$, then it is also correct:

$V=c(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}+a_{5}^{2}+a_{6}^{2}+a_{7}^{2}+a_{8}^{2}+a_{10}^{2}+...+a_{99}^{2}+a_{100}^{2})=$

$=1(0.1^{2}+0.2^{2}+0.3^{2}+0.4^{2}+0.5^{2}+0.6^{2}+0.7^{2}+0.8^{2}+0.9^{2}+0.1^{2}+1.1^{2}+1.2^{2}+1.3^{2}+1.4^{2}+...+9.8^{2}+9.9^{2}+10^{2})=$

$=3335.20$.

So like $V={Bh \over 3}={a^{2}h \over 3}={10^{2}\cdot 100 \over 3}={10000 \over 3}=3333.(3)$.

It's like integral sums of peaces under parabola $a=x$, $\int _{0}^{10}x^{2}dx={x^{3} \over 3}|_{0}^{10}={10^{3} \over 3}-{0^{3} \over 3}=333.3333$ - it's answer when $a=h$. All peaces under parabola have same width $c=1$, but height $x^{2}$ of each peace like $1^{2}$, then $2^{2}$, then $3^{2}$,...,$9^{2}$, $10^{2}$. So this $y=x^{2}$ at any point on x axis, can be interpretated as $B=a^{2}$ and this each peace of pyramid height is $c=1$ and each peace value $V_{1}=ca_{1}^{2}$, $V_{2}=ca_{2}^{2}$, $V_{3}=ca_{3}^{2}$,..., $V_{9}=ca_{9}^{2}$, $V_{10}=ca_{10}^{2}$. And $V=V_{1}+V_{2}+V_{3}+...+V_{9}+V_{10}=385$.

And if $a=100$, $h=100$, $c=1$, then of course it's also correct:

$V=c(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}+a_{5}^{2}+a_{6}^{2}+a_{7}^{2}+a_{8}^{2}+a_{10}^{2}+...+a_{99}^{2}+a_{100}^{2})=$

$=1(1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}+7^{2}+8^{2}+9^{2}+10^{2}+11^{2}+12^{2}+13^{2}+14^{2}+...+98^{2}+99^{2}+100^{2})=333520$. Very close to $V={ha^{2} \over 3}={100\cdot 100^{2} \over 3}=333333.(3)$.

And if $a=100,$ $h=10$, $c=0.1$, dividing pyramid into 100 parts (c=h/100=0.1) and it is correct:

$V=c(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}+a_{5}^{2}+a_{6}^{2}+a_{7}^{2}+a_{8}^{2}+a_{10}^{2}+...+a_{99}^{2}+a_{100}^{2})=$

$=0.1(1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}+7^{2}+8^{2}+9^{2}+10^{2}+11^{2}+12^{2}+13^{2}+14^{2}+...+98^{2}+99^{2}+100^{2})=33352$. Very close to $V={ha^{2} \over 3}={10\cdot 100^{2} \over 3}=33333.(3)$.

No matter how awesomely close you get, it's not a proof without some rigorous reason for confidence in the limit. If you understand integration for area, why not simply integrate for volume? Nor is the value for a single case all that interesting. —Tamfang (talk) 05:25, 16 February 2010 (UTC)

proof of surface area of piramid

Pyramid base is square. Pyramide base $B=a^{2},$ a=1, pyramid height is H=1. Now need to find each of 4 triangles area, but then need to know h of triangle, because a=1 of triangle base. So $h={\sqrt {H^{2}+({a \over 2})^{2}}}={\sqrt {1+{1 \over 4}}}={\sqrt {5 \over 4}}={\sqrt {1.25}}=1.118033989$.

So total surface area of pyramid:

$S=B+4\cdot {ha \over 2}=1+4\cdot {1.118033989\cdot 1 \over 2}=1+2\cdot 1.118033989=1+2.236067978=3.236067978$.

This is still only for one particular pyramid. We already have a general formula given in the article ($A=B+{\frac {PL}{2}}$ where B is the base area, P is the base perimeter and L is the slant height). If you would like to generalise your formula, then it could be added as an alternative, but don't substitute values. Dbfirs 11:43, 11 October 2010 (UTC)