If a cube shape is divided into 2 equal parts, how would its surface area be divided/multiplied?
915365
initial surface area=6a^2\nnew surface area=2[2a^2+4(a^2/2)]\n=2*4a^2=8a^2\nit will be 4/3 times the original surface area
166652
For a specific kind of solid, yes.  In general, no.\n\nFor example, a cube with a side length of s will have a volume of s^3 and a surface area of 6s^2.  So, for any cube, the volume and surface area will be in the ratio of s/6.\n\nFor a right cylinder, the volume is (pi)r^2*h, while the surface area is given by 2(pi)r^2 + 2(pi)r*h, which through some algebraic manipulation can be reduced to 2(r+h)(pi*r).  So the ratio of volume to surface area for a cylinder will be r*h to 2(r+h)... \n\nFor other solids, you could observe similar relationships, but in general, the relationship depends on the shape of the solid, so you could not make a definitive statement about the surface area enclosing a certain volume unless you also knew what shape that volume was in.
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.
976222
Area A = area of two ends  +  Curved surfaces.\n  A  =  2π r²  +  2π r h\n\nVolume V = Area of end  x  Heigh.\n  V  = π r²  *  h\n  V  = π r² h\n  h  =  V / π r²\n Now substitute for h into the area formula.\n\n  A  =  2π r²  +  2π r h\n  A  =  2π r²  +  2π r (V / π r²)\n  A  =  2π r²  +  2V / r\nThe volume of the can is 1000cm^3, so substitue in.\n  A  =  2π r²  +  2πV / r\n  A  =  2π r²  +  2000 / r\n\nTake first derivide and let it equat to zero (zero for max. or min. pt.).\n  A  =  2π r²  +  2000 / r\n  δA/δr  =  4π r  -  2000 / r²\n  4π r  -  2000 / r² = 0\n  4π r  =  2000 / r²\n  4π r^3  =  2000\n  r^3  =  2000/ 4π\n  r  =  (2000/ 4π)^1/3    (that is, cubed root).\n  r  =  5∙419 260 701\n  r  ≈  5∙42 cm.\n\nLet's use the r value to calculate h.\n  V  = π r²  *  h\n  1000  = π (5∙419 260 701)²  *  h\n  h  =  1000 /π (5∙419 260 701)²\n  h  =  10∙838 5214\n  h  ≈  10 51 cm.\n\nFor ratio:\n h/r = 10∙838 5214 / 5∙419 260 701\n h/r = 2/1   That is, a two to one ratio.\n\nFor queation two, follow the same proceedure after you set up your formulas.\nLike wise for question three.\n\nWith question two it's not clear what value you are taken for the length of the side of the square edge.  Because you are still using r, I'm assuming you are taken the largest size of a square that fits into the circle or radius r.  Based on this assumsion, the calculations are are follows:\n\nTo identify the length x of one side of the square of radius r:\nDraw two diameters of the circle at 90º to each other.  Where these diameters make contact with the circumference of the circle will be the corner points of the square.  The area of this square now consists of four triangles.  These triangles will have two sides of length r with the enclosed angle of 90º.  So the length of x will be:\n\n x² = r² + r²\n x² = 2 r²\n x = √(2 r²)   That is, length of one side of the square.\n\nArea A = area of two square ends + parameter of square * heigh.\nA = (2r² * 2) + [(√2)r * 4 * h]\nA =  4r²  +  4√(2)r h\n\nVolume V = square area x heigh.\nV =  2 r² * h\nV =  2 r² h\nh = V / 2 r²\nNow substitute for h into the area formula.\n\nA =  4r²  +  4√(2)r h\nA =  4r²  +  4√(2)r V / 2 r²\nA =  4r²  +  2√(2) V/r\nA =  4r²  +  2000√(2) r^-1     (Volume is 1000cm^3).\n\nTake first derivide and let it equat to zero (zero for max. or min. pt.).\nA =  4r²  +  2000√(2) r^-1\nδA/δr = 8r  -  2000√(2) r^-2\n 8r  -  2000√(2) r^-2  =  0\n 8r  =  2000√(2) r^-2\n 8r  =  [2000√(2)] /r²\n 8r^3  =  2000√(2)\n r^3  =  [2000√(2)]/ 8\n r  =  ([2000√(2)]/ 8)^1/3             (that is, cubed root).\n r  =  (353∙55 33 906...)^1/3\n r  =  7∙071 0678 12\n r  ≈  7∙07 cm.\n\nLet's use the r value to calculate h.\nV = 2r² * h\n1000 = 2r² * h\n1000 = 2 (50) * h\n1000 = 100h\n h = 10 cm.\n\nFor ratio:\nh/r = 10 / 7∙071 0678 12\nh/r = 1∙845 270 149 / 1\n h : r  ≈ 1∙85 : 1\nThese ratios do not take into account the scrap metal from the circle when the square is cut out of it.
1050164
Divide the L-shape into two rectangles, and compute the area of each rectangle separately. Then add the two areas.
769834
100 and 1000, respectively. Surface area is proportional to the square of the radius, so if a cell's surface ares is 4πr², then the surface are of a cell with 10 times the radius is 4π(10r)² = 100(4πr²). Similar logic applies to volume (which is proportional to the cube of the radius).
287728
Surface Area = [Base Area] +\n1/2 × Perimeter × [Side Length]\n\nVolume = 1/3 × [Base Area] × Height\n\nThe Surface Area has two parts: the area of the sides (the Lateral Area) and the area of the base (the Base Area). \n\nThe Base Area depends on the shape, there are different formulas for triangle, square, etc. \n\nBut the Lateral Area is surprisingly simple. Just multiply the perimeter by the side length and divide by 2. This is because the sides are always triangles and the triangle formula is base times height divided by 2
515325
A cube has 6 square sides and the surface area of the original cube is 6 m^2 or 6x(1000mm)^2 = 6 million mm^2\n\nSince there are 1000 mm in a m, "crushing" the cube results in (1000)^3 = a billion little cubes and each little cube has a surface area of 6mm^2. So the surface area of all the little cubes added together is 6 billion mm^2\n\nThe surface area has increased by a factor of 6 billion over 6 million = 1000
45249
Let S be a surface in three-dimensional space ( R^3 ) and let f be a function defined on this surface. To describe the surface, we will choose a parameterization of the surface by two variables which we shall call u and v. (For instance, if f is a sphere, u and v might be the latitude and longitude.) In terms of this parameterization, the function f can be expressed as a function of u and v. Then the integral of f with respect to the surface area is usually notated as \n\nSurface Integral S[f(u,v)d^2 A]\n\nJust as with the ordinary integral, this quantity may be understood as a limit of sums.1 To form an approximating sum, we subdivide the surface S into a miniscule pieces, multiply the area of each piece by the value of f at a point located inside that piece, sum over all the miniscule pieces into which we have subdivided the surface. The limiting value of these sums as the size of the pieces shrinks to zero is the integral with respect to surface area. \nTo compute this quantity, one may use the following formula to convert it to a double integral: \n\n\nIf the surface is described as the graph of a function , then we may also use the following formula:
205480
Surface area of a sphere is = 4 pi r^2\nThis means that surface area of a sphere is equal to 4 times the biggest circle in the sphere Or 4 times the area of the circle swept by its radius or 4 times the area of the base of its hemisphere.For example if the radius of a sphere is 1 unit then its surface area is = 4 pi square units.When it is divided by 4 it is equal to the area of the base its hemisphere.ie. area of the base of its hemisphere is = pi square units.
87971
The area of a parallelogram is area = base x height. Now, drawing a diagonal of the parallelogram splits the shape into two congruent triangles. Therefore, each triangle will have an area equal to (1/2)base x height.\n\nThis is true for any triangle; that is, any triangle has a corresponding parallelogram associated with it, which can be cut in half.
1033784
A cube has 6 square faces so divide the surface area by six = 81 sq. cm\n\nThe area of a suare is s^2 so so you have to get the suqre root of 81 sq. cm = 9 cm\n\nVolume of a cube is s^3 so raise 9 to the third power  9^3 = 729 cu. cm ---->volume of the cube
270216
in a 2d object or representation, it would be the area. In a 3 dimensional object it would be the volume.
572270
triangle area = 1/2(height * width)\n\ncircle area = pi * radius^2
726531
A=1/2bh is for the area of a triangle.\nV=s^3 is for the volume of a cube.
985825
Figure out the surface area A1 of the uncut box. You will be making two cuts to divide the box into 3 pieces which adds 4 new surfaces. Calculate their area A2. From A1 and A2, figure out the total surface area P to be painted in cm^2. Figure out the number N of 1x1x1 cm cubes you can cut a 7x6x5 cm box into. The total area of the little cubes is 6*N, and the unpainted area is 6*N-P cm^2.
856302
Surface Area = (2 •  p   • r²) + (2 • p  • r • height)\n\nWhere (2 • p • r² ) is the surface area of the "ends" and\n(2 • p • r • height) is the lateral area (the area of the "side").\n\nVolume   =   p • r² • height   =   ¼ • p • d² • height\n\np = Pi