Volume of a slightly irregular Tetrahedron?
825601
its surface area q=q1+q2+q3+q4; area of any triangle =sqrt(p*(p-a)*(p-b)*(p-c)), a, b & c being edges of triangle & p=(a+b+c)/2. Here q1 is area of LSS facet, q2=q3 is area of two LLS facets each, q4 is area of LLL facet.\nIts volume is not as easy: first let’s find the height of facet LSS h1=2*q1/L=srqt(S^2-(L/2)^2), and that of LLL facet h4=2*q4/L=L*sqrt(3)/2, now the area of section Lh1h4=q5=(as above), and the height of this section:\nh5=2*q5/h4=L*q5/q4 that is the height your Tetrahedron.\nV=(1/3)*q4*L*q5/q4=L*q5/3 – oh poor kids!
518004
You would need to disect the polygon into two shapes.  A triangle and a rectangle.  Figure the volume of those individual pieces and then add them together.
889476
you can take a graduated cylinder. fill it with an mount of water to be above object. measure water-write,. now, place object in cylinder. now, record new volume. now, subtract old- new. thi will give you the volume of the irregular object itself. hope this helps!
826376
An irregular solid is like a stone, where you can not measure the sides. To get the volume of an irregular solid you can use water displacement.
656895
It had been conjectured that two equal partial spheres sharing a boundary of a flat disk separate two volumes of air using a total surface area that is less than any other boundary. This equal-volume case was proved by Hass in 1995, who reduced the problem to a set of 200260 integrals which they carried out on a PC. Frank Morgan, Michael Hutchings, Manuel Ritoré, and Antonio Ros finally proved the conjecture for arbitrary double bubbles in early 2000. \n\nIn this case of two unequal partial spheres, Morgan et al. showed that the separating boundary which minimizes total surface area is a portion of a sphere which meets the outer spherical surfaces at dihedral angles of . Furthermore, the curvature of the partition is simply the difference of the curvatures of the two bubbles.\n\nProbably not any huge influence on society, however.
920579
use synthetic  divsion method \ni think this link would help\nhttp://www.purplemath.com/modules/synthdiv.htm
196800
Surface area. Seriously. 8D
637876
v=4/3 pi r^3 (four thirds x pi x (r cubed))
214724
take a look at Pascal's Triangle:\n\n1...1\n1...2...1\n1...3...3...1\n1...4...6...4...1\n1...5..10..10...5...1\n1...6..15..20..15...6...1\n1...7..21..35..35..21...7...1\n1...8..28..56..70..56..28...8...1\n\nYou see how your tetrahedal numbers are in the fourth column?  Just follow it down to whatever you mean by "5th layer."  These numbers can be computed directly as nC3 on a TI-83.
626725
First you need the area of a single face, to do this you need to multiply one side of the face (square) by another adjacent (next to) to the first, this will give you the area if a single face; then you just need to multiply that times six wich is the number of faces in a cube. Depending on the mesuring system that you are using, i.e. metric, standard, etc., you will end up with a number followed by cm.2 or ft.2, that "2" means that you are talking about area and not distance nor volume or any other; it avoids confussion.
183956
Half the diameter, or circumference divided by 2.pi
787617
(m)\n\n14 sides: tetradecogon
1047892
OK, let p be pi, OK?\n\nThe volume of the sphere is 4*p*r^3/3...\n\nNOW, the first derivative of that, with respect to the radius, is:\n\n4*p*r^2 dr\n\nSO, you can approximate the answer by:\n\n(4*p*3^2)*0.01\n\nThat'd give you 0.36*pi cubic inches, which would be b
934849
291.13 L is the correct volume.\n\nCheck your algebra when you are solving for V1.
716964
A shape with minimum surface area for a given volume!...a round shaped ball-like mass would have minmum area of 4 pi r square for a volume 4/3 pi r cube !
130046
Half of the diameter.\n\nor\n\nMeasure around it.  Divide that number by 3.14.  Then divide by 2.
1053343
A tetrahedron is a solid with four faces. Tetra- means four, -hedron means solid geometric object.\n\nYou can also think of it as a triangular pyramid.