what is the volume of a triangular pyriamid?
340579
Find the area of the triangular base, then multiply that by (1/3) times height.\nThe formula is (1/3)*(area of base)*(height)\nSay the triangular base has base = 3 and height 4.\nArea of triangle = (1/2)(3)(4) = 6\nSay the height of the pyramid is 10.\nThe volume is  \n(1/3)(6)(10) = 20
1099275
Oh c'mon! Smiddy hasn't had geometry for 20 years! but this is basic. Pythagorean theorem states that on a triangle A-squared plus B-squared = C-squared where C is the hypotenuse. The hypotenuse of a triangle is the diagonal bit if it's two halves of a square and we know by definition that on a square A and B are equal. Therefore 2X squared = 8-squared or 2x^2 = 64\n\nx^2 = 32\nx = sq.rt 32\n\nthe area of the original square is x times x so you're back at 32 square centimeters
63817
tsniperx\n2 months ago\n\nVolume of one block cube = 10cm*10cm*10cm = 1000cm^3\nVolume of two block cube = 2000 cm^3\nVolume of one sphere = (4/3)*pi*r^3 = (4/3)*pi*(5cm)^3\nVolume of one sphere= 523.59877 cm^3\n\nno spheres= Volume of clay / Volume of sphere= 2000cm^3 / 523.59877cm^3\nNo spheres must be integer = 3 spheres
625137
Volume is L * W * H ... if the Volumen is 1000^3 and it's a cube, then L, W and H all = 1000cm\n\nDuh.
953022
There sure is!  \n\nFirst, recall this: given a tetrahedron, choose any of its faces to be its base.  Now find the area of the base (remember all the faces are triangles).  Next, find the perpendicular distance from the fourth vertex of the tetrahedron to the plane that contains the base.  This distance is the height of the tetrahedron.  Now the volume of the tetrahedron is 1/3 X area of its base X its height.  \n\nNow, using three-dimensional Cartesian co-ordinates, consider the tetrahedron ABCD where A=(0,0,0), B=(27,0,0), C=(0,16,0), and D=(0,0,3).  Face ABC is a right triangle in the x-y plane.  The two perpendicular legs of ABC have lengths 16 and 27, so the area of face ABC is 1/2 X 16 X 27 = 216.  The distance of D from the x-y plane is 3, so the volume of ABCD is 1/3 X 3 X 216 = 216.  \n\nLet E=(8,2,0).  E lies in the x-y plane in the interior of ABC.  Join E to A, B, C, and D with line segments.  Triangle ABC is subdivided into triangles ABE, BCE, and ACE, and tetrahedron ABCD is subdivided into tetrahedra ABED, BCED, and ACED, and the heights of these three sub-tetrahedra are also 3.\n\nThe distance from E to the x-axis is 2, so triangle ABE has \narea = 1/2 X 2 X 27 = 27 and tetrahedron ABED has \nvolume = 1/3 X 3 X 27 = 27.  \n\nThe distance from e to the y-axis is 8, so triangle BCE has \narea = 1/2 X 8 X 16 = 64 and tetrahedron BCED has \nvolume = 1/3 X 3 X 64 = 64.  \n\nArea of ABC = 216, area of ABE = 27, and area of BCE = 64, so triangle ACE has area = 216 - 27 - 64 = 125 and tetrahedron ACED has area 1/3 X 3 X 125 = 125.  \n\nThat does it!
598778
Here's the correct solution.\n\nA cylindrical package with height(length) h and radius r will provide the maximum volume under the restriction that the sum of the girth and length equals a given constant C.\n(In your case C = 108")\n\nGeneral case, for arbitrary C\nFor a cylinder, the girth = 2pi*r, so our restriction is:\n\nC = 2pi*r + h\n\nThe volume V = pi*r^2*h\n\nFirst solve for h in terms of C:\n\nh = C - 2pi*r  Substitute this into volume formula:\n\nGives  V = pi*(r^2)*(C-2pi*r)\n\nV = pi*C*(r^2) - 2(pi^2)r^3\n\ndetermine dV/dr = 2*pi*r*C - 6*(pi^2)r^2\n\nSet dV/dr = 0, implies 6*(pi^2)r^2 = 2*pi*r*C\n\nSo 2pi*r = 2C/3, and therefore h = C/3, since 2pi*r + h= C\n\nSo if C = 108" then for maximum volume h= 108/3 =36\nand 2pi*r = 2C/3 so r = C/3pi and r = 108/3pi = 11.46 \n\nMax Volume = pix(11.46^2)x36 = 14,853 cubic in.\n\nWe can test that this is a maximum volume by testing the neighborhood. for example 3C/10 < C/3 < 3C/8\n\nlet h=3C/8 = 40.5 then r = 67.5/2pi= 10.74\n\nand Volume = pi*(10.74^2)*40.5 = 14,684 cubic in. This is less than the Max volume of 14,853 \n\nIf h= 3x108/10 then Volume = pi(12.03^2)*32.4 = 14,730\n\nThis is again less than Max volume, and difference increases the more h varies from C/3 = 108/3 = 36
890633
1. total area of a box of sides a,b,c is 2(ab+bc+ca)\na=4,b=3,c=6 => surface area = 108 sq.ft\n..\n2.Lateral area = pi*radius*slant ht\npi=3.1415,radius=2m(4m/2), slant ht=15m\nans is 188.49sq.metres\n..\n3.   volume of pyramid = (1/3)* height * base area\n= (1/3)* (6) * (6*6)\n= 72 cu.metre\n..\n4. lateral area of cylinder = 2*pi* radius* height\n=2*3.1415*12*60\n=4523.76cu. mm\n..\n5. total surface area = outer surface + inner surface + top surface + bottom surface\n= 2*pi*8*height + 2*pi*5*height + 2 times ( pi*(8-5)^2)\n[the last term is the sum of top and bottom surfaces which are circular rings]\nheight is not given :(
423638
Volume Formulas \nNote: "ab" means "a" multiplied by "b". "a2" means "a squared", which is the same as "a" times "a". "b3" means "b cubed", which is the same as "b" times "b" times "b".\nBe careful!! Units count. Use the same units for all measurements. Examples\ncube = a 3  \n\nrectangular prism = a b c  \n\nirregular prism = b h  \n\ncylinder = b h = pi r 2 h  \n\npyramid = (1/3) b h  \n\ncone = (1/3) b h = 1/3 pi r 2 h  \n\nsphere = (4/3) pi r 3  \n\nellipsoid = (4/3) pi r1 r2 r3  \n\n  \n\n  Units \n\nVolume is measured in "cubic" units. The volume of a figure is the number of cubes required to fill it completely, like blocks in a box. \n\nVolume of a cube = side times side times side. Since each side of a square is the same, it can simply be the length of one side cubed. \n\nIf a square has one side of 4 inches, the volume would be 4 inches times 4 inches times 4 inches, or 64 cubic inches. (Cubic inches can also be written in3.) \n\nBe sure to use the same units for all measurements. You cannot multiply feet times inches times yards, it doesn't make a perfectly cubed measurement. \n\nThe volume of a rectangular prism is the length on the side times the width times the height. If the width is 4 inches, the length is 1 foot and the height is 3 feet, what is the volume?\n\nNOT CORRECT .... 4 times 1 times 3 = 12 \n\nCORRECT.... 4 inches is the same as 1/3 feet. Volume is 1/3 feet times 1 foot times 3 feet = 1 cubic foot (or 1 cu. ft., or 1 ft3).
1000601
The method to finnd the volume of sphereis computed by The mathematician, Archimedes.\nThe formula for find the volume =4/3 pi  r cube\nie, pi = 3.14\n     r is the radius of the sphere.\nFor eg, r=1.2 m\n v =4/3 pi*(1.2*1.2*1.2) \n    =4/3*3.14*1728\n     =7.2346 cu.m\n     1 cu.m = 1000 litres\nV   =7.2346*1000\n       =7234.6\n\nDon't u understand.
543695
volume = (1/3) b h = 1/3 pi r 2 h \nSurface area = pr2 + prs\ns= the height of the outside of the cone, while h = the height of the cone at its center
466950
Volume = Area x height\nArea = r x r x pi\npi is approximated by the fraction 22/7\nTherefore Volume of cylinder = 4 x4 x 22/7 x7 = 352 units^3
762991
Volume = pi * r^2 * h\n\nwhere\nr is the radius\nand\nh is the height\n\nFor example:\nr = 0.9m\nh = 1.2m\nvolume = 3.14159 * 0.9 * 0.9 * 1.2 = 3.0536 cubic metres or 3053.6 litres\n\n_
492961
Yes 152 cm is correct.\n\nTo solve the problem factor 2002 = 2x7x11x13\n\nThe closer the box is to a perfect cube the lower the ratio of the surface area to the volume.\n\nThis occurs when the sides are  11, 13 and 2x7=14\n\nIn the cube, each edge appears 4 times and so the sum of the edges will be 4 times (11+13+14) = 4x38 = 152 cm
75959
think of a pyramid as half of a cube, with the base sides equal to Sqrt(2) times the side of the cube.  a whole cube would have twice the volume and be 60cm^3, So, the dimensions would be 60^(1/3) or 3.91cm sides and 3.91*2^(1/2) or 5.54cm base lengths.
116020
Volume = pi*r^2*h\n\nWhere pi = 3.14.159...\nr = raduis of cylinder in feet\nh = height of cylinder in feet
430210
Volume of a shphere  = 4/3 * pi * R^3\n\nwhere pi = 3.142 (approx)\nand R is the radius
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