finding the volume of a pyramid?
409815
uh... 1/3 times base times height.
148864
If your trench silo looks like the ones that I am familiar with, it's rectangular when you look down on it and  has sloping sides and sloping ends. The way to compute its volume is to break it down into pieces that are regular prisms or pyramids. I would break it down like this:\n\nStarting with the flat rectangular bottom, extend the edges of the rectangle to the boundry of the silo. That will divide your silo into 9 pieces.\n\nThe center section with a flat bottom is a rectanguar box.\n\nThe 4 corner pieces that slope in two dirctions. They are pyramids.\n\nThe 4 pieces sloping sides, two that lie along the long side of the rectangular box and two at the ends,  are triagular prisms.\n\nOf course, some of those silos get mounded up in the middle, so that adds another dimension to solving the problem.\n\nYours for better silage,\nGuru
493708
the formula for the area of any regular shape is \n(pa)/2, so there is no multipling the base by 2 involved.\np=perimeter\na=apothem
125346
Use Heron's formula to find the area of the base: 6 times the square root of 5.  Then multiply that by the height.\n\nThe volume is 30 times the square root of 5.
825337
NO! you have to replace V2 with given volumes:\nP = (V1P1)/V2\nP = (4 x 1)/3 \nP = 4/3 \n\nNow do the same thing for V2= 2. V2= 1 , V2= 5 and V2= 25
46232
The volume of the pyramid is 1/3 the volume of the prism.\n\nTo prove this, look at a horizontal cross section of pyramid. Let A = the area of the base.\n\n\nIf the the height of the pyramid is h, look at a cross section of the pyramid at a height t above the base. Any one dimentional measurement x in the base, will have size x(h-t)/n in the cross section. Any two dimensional measurement, (area) will be multiplied by ((h-t)/h)^2, i.e. A((h-t)/h)^2.\n\nNote\nNow integrate for t = 0 to h, A((h-t)/h)^2 dt\n\nYou get -A/3(h - t )^3/h^2.\nEvaluating at t = 0 and t = h you get 1/3Ah.\n\nThe volume of the prism is Ah.
563210
The volume of a pyramid is equal to one-third the base area multiplied by the height.\n\n\nImagine a step-pyramid made up of small square prisms with dimensions\na-by-a-by-b, arranged in square layers. If the layers are k-by-k for\nk = 1, 2, 3, ..., n, then the total volume of the step-pyramid is:\n\n   V = a^2*b*[1^2 + 2^2 + 3^2 + ... + n^2],\n     = a^2*b*n*(n + 1)*(2*n + 1)/6 (which can be proved by induction),\n     = a^2*b*n^3*(1/3 + 1/[2*n] + 1/[6*n^2]).\n\nNow, the height of the step-pyramid is h = b*n, and the area of the \nbase is:\n\n   B = a^2*n^2, \n\nso the volume is:\n\n   V = h*B*(1/3 + 1/[2*n] + 1/n^2).\n\nNow let n -> infinity, so the steps get smaller and smaller, and the\nvolume of the step-pyramid approaches the volume of the inscribed \npyramid. Then the right-hand side approaches h*B/3, as the volume V \napproaches the volume of the pyramid.\n\nThis same proof works for a pyramid with any polygonal base. For a \ncone, take a pyramid with a regular m-gon for a base, and let \nm -> infinity.\n\nThe base approaches a circle, and the pyramid approaches a cone, and \nthe formula V = h*B/3 still holds in the limit.\n\nThere is another proof, which cuts a rectangular solid into six pieces \nof equal volume, each of which is a triangular pyramid (or irregular\ntetrahedron). This proves that the volume of each is 1/6th the volume \nof the starting rectangular solid, and each has base of area 1/2 the \narea of one of the faces of the solid, and height equal to the \ndimension of the solid perpendicular to that face
738861
If you are talking about a 3-D rectangle,\n\nThe volume of a rectangle is height times width times depth.  \n\nYou can also use base times height if its easier.\n\nhope this helps!
690095
pyramid: V = (Area of base X height)/3\n\ncone: V = (πr2h)/3
421382
a pyramid. 3 sides and the bottom is 4 sides. but only 3 corners.
1068729
i think n=6..\nyes this can be done..if you know the formula..\nof volume and surface arae of a hypersphere and also the form of gamma fucntion..\nhope you can do it yourself..i saw these two links and based on the formula given for volume and surface ares and the gamma fucntion expression its evident that n=6.\nHope you can figure it out yourself.\nvolume= pi^3/6 (radius =1 here)
604203
to find the surface area of a pyramid take 1/2 of the perimeter of the base times the slant height and add on the area of the base.  First lets find the area of the base with 3 sides of 17.32 each. You need more information for this, unless you have been taught some simple trig functions. To get the area of the base we need to know the height of this base(if it were standing up).If I use the tangent function I find that the height of the triangle is approximately 15 inches. So the area of the base is 1/2 times 17.32 times 15. I did the tangent of 60 degrees = the height of the triangle divided by 1/2 the side of the triangle.  I hope this helps solve the rest of the problem.
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.
877943
volume of cyllinder = area of circle times height\n\n= Pi x radius x radius  x height\n\nremember its just the same as finding the volume of any prism.\nfind the area of the shape of the end then multiply by its length!!
580358
You are interested in the volume of the batter up to the half way mark.\n\nFirst determine the area the pan bottom = A = 30x20= 600 sq cm\n\nImagine the batter filled up to the halfway mark. You need to determine the suface area at this level. You do this by taking the average of the top and bottom dimensions. So one is (34.3 + 30)/2 = 32.15 cm, and the other is (22.86+20)/2 = 21.43 cm. The area at this level = B = 688.97 sq cm\n\nNow, we use the formula for the Volume of the pan up to the halfway level(a truncated pyramid formula): \n\nV = 1/3(A+B+sqrt(AxB))xh  Now h in this case = 5cm/2 = 2.5 cm.\n\nThe square root of AxB = 642.95 so our volume is \n\nV = 1/3(600+688.97+642.95)x2.5 = 1609.99 or about 1610 cubic cm or 1.61 litres (6.8 cups)
824
Volume of cylinder = area of base * height.\nSince the base is a circle, that part is Pi *r^2\nSo we have Pi * r^2 * height --> \n3.14 * 12^2 * 21 = ?\n\nVolume of square prism = area of base * height\nSince the base is a square, that part is side^2\nSo we have 12 * 12 * 21 = ? \n\nSubtract the results for your final answer.
726531
A=1/2bh is for the area of a triangle.\nV=s^3 is for the volume of a cube.