how do you find the volume and surface area of a truncated tetrahedron?
544661
The surface area:\nS = 0.5*(B1+B2)*Hs  +  A1 + A2\nWhere B1 & B2 are the perimeters of the 2 bases\n      Hs = Slant height\n      A1 & A2 are the areas of the 2 bases\n\nThe volume:\nV = (1/3)*h*(A1+A2+ sqrt(A1*A2))\nWhere h = the height of the truncated tetrahedron
218464
If by area you mean "Surface area" then\nSA(sphere) = 4(pi)(r)^2\nand\nV(sphere) = 4(pi)(r)^3/3\n\nSince SA = 196 cm^2, then\n\n196cm^n = 4(pi)r^2\nr^2 = 49/pi\nr = 7/sqrt(pi) or approx. 3.949 cm\n\nfor the volume,\nV = 4(pi)(r)^3/3\nV = 4(pi)(7/pi^0.5)^3/3\nV = (4pi/3)(343/pi^1.5)\nV = (1372/(3pi^.5)) or approx. 258.02 cm^3\n\n^_^ Hope this helps ^_^
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!
574480
The volume of the whole sphere is V = (4/3)πr^3 = (4/3)π(1000) = 4000π/3.  The unfilled part is a spherical cap with a volume V = (π/6)(3a²+h²)h, where a is the radius of the base of the cap and h is its height.  So subtracting 14 from 20, h = 6.  Using Pythagoras, a² = 10²-4² = 84.  That makes the volume of the cap π*(3*84+36)*6/6 = 288π.  Subtracting cap from sphere leaves 4000π/3 - 864π/3 = 3136π/3 (exactly) = 3284.01 cubic meters (approximately).
981062
The surface area of a sphere of radius 'r' is given by:\n\nsurface area = \n 4*pi*r^2\n \n\nNOTE: The value of p can never be known exactly, so surface areas of spheres cannot be calculated exactly. Common approximations for p are: 3.14, and 22/7. \n\nExample 1  if 'r' = 5 for a given sphere, and p = 3.14, then the surface area of the sphere is:\n\nsurface area = 4*pi*r^2\n\n         = 4 × 3.14 × 25\n\n         = 314
594683
To find area you multiply the length times the width. It's pretty easy once you know
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
191332
Find the dementions and multiply.
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)
820928
I am guessing that you want to know how big one of the sections are, so that's what I will try to determine.\n\nThe first step is to find the surface area of the hemisphere with a radius of 18 meters:\nThe surface area of a sphere, with r as the radius, is: A = (4)(π)(r²). Since this is a hemisphere, we can divide the expression equal to the surface area by 2, leaving us: A = (2)(π)(r²).\nA = 2(π)(18²) = 2(π)(324) = 648(π)\n\nSince we know that the surface area of the hemisphere is 648π m², we can divide this by 12 to find out the surface area of one of the sections:\n648π / 12 = 54π meters² ≈ 169.65 meters²
223583
If you are looking for the surface area or volume of a pyramid, check this out:
852456
Total surface area of a tin can = A = 2pr^2 + 2prh\nVolume of a tin can = V = pr^2h = 1000 cm^3\nTherefore, h = 1000 / pr^2\n\nSubstituting this into the area equation gives :\nA =  2pr^2 + 2pr * 1000 / pr^2\n= 2pr^2 + 2000 / r\n\nNow take the derivative of A with respect to r :\ndA / dr = 4pr - 2000 / r^2\n\nSet this equal to zero to find the minimum :\n4pr - 2000 / r^2 = 0\nThis can be rearranged to show that : 1000 / pr^2 = 2r, which equals h, as shown above.\nContinuing to find r, we get : r = (500 / p) ^ (1/3), or approx. 5.42 cm.\n\nNow h = 2r = 2 * (500 / p) ^ (1/3), or approx. 10.84 cm.\n\nThus, ratio is h / r = 2r / r = 2\n\nI'm a bit confused about the second question, but I think what you want is a square sheet of aluminium, such that all 3 pieces of a can will fit, with the minimum of wastage. This would be a square sheet with the dimensions : (perimeter of can) by (height + 2 * radius).\n\nThus, Perimeter = Height + 2 * Radius, because the sides of a square are equal.\nTherefore, 2*pi*r = h + 2r,\nso, h = 2r(pi - 1)\n\nBut, V = pi*r^2*h = 1000, so h = 1000 / (pi*r^2)\n\nEquating the 2 values for h gives :\n2r(pi - 1) = 1000 / (pi*r^2)\nfrom which we get : r  = {500 / [pi(pi - 1)]} ^ (1/3)\n\nSo now we know that h = 2r(pi - 1), so that after substituting for r, we get : \nh = 2 * (pi - 1) ^ (2/3) * (500 / pi) ^ (1/3)\nAnd we find that the ratio h / r = 2r(pi - 1) / r = 2(pi - 1)\n\nWell, that was a bit of speculation, but even if I've got it wrong, then perhaps you can adapt it to your needs.\n\nEdit\nCan't help feeling I've misunderstood the second question the first time around, but on re-reading it many times, I still don't understand it.
549797
enclose the ireegular shape with a rectangle and find the empty spaces. then subtract that with the area of the enclosed rectangle to get ur answer.
726531
A=1/2bh is for the area of a triangle.\nV=s^3 is for the volume of a cube.
16584
You can find the volume using a triple integral
724216
Find the volume of the cuboid(a 3D rectangle) and the whole cylinder. DIVIDE ONLY THE VOLUME OF THE CYLINDER BY 2. Then add it to the cuboid's volume.
801553
lol.\nfind formula for area of spiral.  subtract smaller from bigger.  sry no more help.