What is the formula for figuring out the volume of a trapeziodal prism?
101549
V=1/2(b1+b2)*h*w  where b1 and b2 are the two parallel sides of the trapezoid and h is the height and w is the width.
84844
Height = H = 10"\nRadius = r = 4"\n\nVOLUME (not area, volume is 3-dimensional) = (π/3)(r^2*H) = (160/3)*π\n\nSURFACE AREA = (πr)(sqrt(r^2+H^2)) = 4(sqrt(116))*pi
430786
The formula for the capacity or volume, V, of a cylinder is:\n\nV = Π(d^2)h/4\nwhere:\nΠ = 3.1416 (value up to 4 decimal places)\nd = diameter of the cylinder\nh = height of the cylinder\n\nSo, merely substitute the values that you have and you will get the capacity or volume of the cylinder. Just make sure to use similar unit of measures both for the diameter and the height.
389859
The volume of a sphere can be found by the formula: \nvolume = 4/3 π r3 \n\nhttp://grapevine.abe.msstate.edu/~fto/tools/vol/sphere.html\n\nTry this website to help you out:\nhttp://www.gomath.com/algebra/sphere.php
524937
look here\nhttp://education.yahoo.com/reference/weights_and_measures/index?p=volume
1068661
Perimeter of the triangle multiplied by the height of the prism.
675011
The angle of incidence inside the prism is 25. (the ray passes straight into the prism as it is incident normally).\n \n\nmu = sin a/ sin g( g is the angle in glass and a is the angle in air)\nSin a = mu sin g \n\nsin a = 1.662 sin 25\n\nr = 45 \n for 1.698 , r =45.8 = 46.
885732
well...lets call your known volume V.  so...what kind of shape are we dealing with...eh, lets claim we are dealing with a sphere...we know the volume of the sphere is determined by v=(4/3)*pi*r^3 so we could solve for r (the radius).\n\nso....\n\nV=(4/3)*pi*r^3\nr=((3V)/(4pi))^(1/3)\n\nwe can now substitute that r into our surface area of a sphere equation....which is SA=4*pi*r^2, so we would have\n\nSA=4*pi*((3V)/(4pi))^(2/3)\n\na similar series of actions would occur for any other shape\n\nhope this helps,\n\nMatttlocke
559607
A rectangular prism is simply made up of 6 rectangles.  One pair of rectangles would be 3cm x 3cm (the ends).  The other 4 rectangles would be 3cm x 20cm.  So the total surface area would be the sum.\n\n3cm x 3cm = 9 sq. cm\n3cm x 20cm = 60 sq. cm\n\nSo 9 + 9 + 60 + 60 + 60 + 60 = 258 sq. cm.\n\nIn this case two of the dimensions were the same, but if the sides were h,w,l the total surface area would be:\n= hw + hw + hl + hl + lw + lw\n= 2hw + 2hl + 2lw\n= 2 (hw + hl + lw)\n\nIn your example:\n= 2 (3x3 + 3x20 + 3x20)\n= 2 (129)\n= 258 sq. cm.   (Don't forget the units are sq. cm)
1124064
Because you have a fixed height, h, of 30m, and the Volume  of a triangular prism is given by: V = h*A where A is the area of the triangular base, your problem reduces to maximizing A, given the perimeter of A = 210m\nAn equilateral will enclose the maximum area(see link below) So the length of each side will be 210/3 = 70 m. The area of an equilateral triangle =  S^2 x sqrt(3)/4 where S = side length. So, in your case the area = 70^2*sqrt(3)/4 = 2,121.18 m^2 and the maximum volume for your prism is 30m*2121.18 m^2 = 63,652.86 m^3\n\nFor a polygon, the general rule is: the 'closer' the shape of a polygon is to a circle, the greater the area of the interior of the polygon. This implies that a equilateral triangle, a square(see below), a regular pentagon; and in general a regular n-gon will enclose the most area. \n\nConsider a square with side length X, then the area of the square = X^2 and the perimeter is 4X. Let's keep the perimeter fixed at 4X and see what happens to the area if we let one pair of opposing sides of the rectangle be X-y and the other two be X+y (Note the perimeter will still be 4X). However, the area is now: (X+y)*(X-y) = X^2 - y^2. We see that if y>0 then area decreases, while the perimeter stays the same. We see that a square encloses the maximum area for a fixed perimeter. \n\nHowever, if you allow the height, h, to vary as well then you have 3D maximization problem; but the solution is not difficult in this case either. In general, right-angle prisms with regular polygon bases will have a maximum volume for a fixed perimeter or surface area\n\nhttp://mathforum.org/library/drmath/view/53789.html\n\nhttp://mathworld.wolfram.com/Prism.html\n\nhttp://mathworld.wolfram.com/EquilateralTriangle.html
691753
maybe a hexagonal prism http://en.wikipedia.org/wiki/Hexagonal_prism http://en.wikipedia.org/wiki/Prism_%28geometry%29
63381
Conjecture A\nIf B is the area of the base pf a right rectangular prism and H is the height of the solid, then the formula for the volume is V= _?_\nConjecture B \nIf B is the area of the base of a right prism (or cylinder) and H is the height of the solid, then the formula for the volume is V= _? _\nConjecture C\nThe volume of an oblique prism (or cylinder) is the same as the column of the right prism (or cylinder) that has the same _? _ and the same _? _\nPrism-Cylinder Volume Conjecture\nThe volume of a prism or a cylinder is the _? _ multiplied by the _? _\nPyramid-Cone Volume Conjecture\nIf B is the area of the base of a pyramid or a cone and H is the height of the solid, then the formula for the volume is V= _? _ \nSphere Volume Conjecture\nThe volume of a sphere with radius r is given by the formula _?_\nSphere Surface Area Conjecture\nThe surface area, S, of a sphere with radius r is given by the formula _? _\n\nSolution 47916\n\n#1\nV = BH\n#2\nV = BH\n#3\nsame base area and same height\n#4 \nthe base area multiplies the height\n#5\nV = (1/3)BH\n#6\nV = (4/3)(pi)(r^3)\n#7\nS = 4(pi)(r^2)
16584
You can find the volume using a triple integral
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).
918429
use the formulas your given\nto find the volume of the cylinder\nthen pretend the semi hemi on  the top is a sphere find the volume and multiply by 1/2\n\nadd two totals to find volume
731102
consider the front face to be an A by A square that has depth length L.  The Volume of this rectangle would then be A*A*L.  since we know A + L =108. Then L = 108 - A.  Putting this into our volume formula we get 108A^2 - A^3.  In order to maximize A take the derivative and set it to zero.  So we have 216A - 3A^2 = 0.  This leads to A = 72 cm.  Then L would be 108 - 72 = 36cm.  The maximum volume would then be 72 * 72 * 36 = 186,624 cm^3
1119363
Easy... fill the cone with some liquid.... then pour this liquid into a measuring cylinder and measure the amount of liquid... you'll get the volume of the cone.\n\nMathematically it is 1/3 the volume of Cylinder\nthat is 1/3(pie*r*r*h)