What are the Lateral Area and Serface Area formulas?
457596
Surface area:\nCylinder(lateral):2 times pie times radius times height\ncylinder(total): 2 times pie times radiuse times height + 2 times pie radius squared\nTriangular prism(total): find the area of every side and add all together to get surface area(hint:area of a triangle=1/2bh or bh/2)\nCone(lateral):pie times radius times length\nCone(total):pie times radius times length + pie squared\npyramid(total):find area of ever side and add all together \narea of a triangle=1/2bh bottom of the pyramid is length times width
909179
the area of each end=pi*r^2\nthe area of circular part=2*pi*r*h\nsurface area of metal required\nS=2*pi*r^2+2*pi*r*h.....(1)\n\nvolume V= pi*r^2*h \ntherefore,h=V/pi*r^2\n                 =1000/pi*r^2\nsubstitute this value back into(1)\nS=2*pi*r^2+2000/r\nwe now differentiate S to find max/\n                      min value of S\ndS/dr=4*pi*r -2000/r^2=0 \nr^3=2000/4*pi at max/min point\n>r=cube root{500/pi}=5.4192607cm\nd^2S/dS^2= 4*pi+4000/r^3\nhence,d^2S/dS^2 is +ve for all r,\ntherefore,S is a minimum\n\nnow,h=1000/pi*r^2\n          =10.83852141cm\nratio h/r=10.83852141/5.4192607\n            =2\n\nyou'll have to be more explicit about the \nrest of the question because i am not \nquite sure what you want\n\nthis ratio makes sense,since h=2r this\nwould make the vertical plane of\nsymmetry a square which is the most\nefficient rectangular shape-couple this\nwith the top and bottom being circular,\nthe above answer is correct\n\ni hope that this helps
549577
The formula for surface area of a sphere is \nA = 4*pi(r^2)\n64 pi = 4*pi(r^2)\nwhich means r^2 = 16 or r = 4\nVolume of a sphere is \nV = (4/3)*pi (r^3)\nV = (4/3)*pi (4^3)\nV = (4/3)*pi (64)\nV = (256/3)*pi
520472
Let s = length of the cube\nsurface area of a cube = 6s^2\nThe diagonal is found by using the pythagorean theorem.\nLet c = length of the diagonal of a side, then\ns^2 + s^2 = c^2\nLet d = length of the diagonal, then\nd^2 = s^2 + c^2\nReplace c^2 with s^2 + s^2, then\nd^2 = s^2 +(s^2+s^2) \nd^2 = 3s^2\nSurface area(A) is twice the square of the diagonal means\nA = 2d^2\nNow A = 6s^2 and d^2 = 3s^2\nThus,\n6s^2 = 2(3s^2)\n6s^2 = 6s^2\n\n2) V = s^3\nWe want to show that\nV = (1/9)sqrt(3)*d^3\nd^2 = 3s^2, then d = sqrt(3s^2) = s*sqrt(3)\nd^3 = d^2*d = 3s^2(s*sqrt(3)\nd^3 = 3s^3*sqrt(3)\n\ns^3 = (1/9)sqrt(3)*3s^3*sqrt(3)\ns^3 = (1/9)(3)(3)*s^3\ns^3 = s^3
332600
to find areas of parallelograms use the formula: \n\nA = base×height.
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
192399
In the front?  You have to consider the whole cube to find volume.  Volume = base area * height  You could find the surface area of one side, but you'd need to know the side length.
440894
go to this website and try to understand how to get the area of the shade region \n\nhttp://www.algebralab.org/lessons/lesson.aspx?file=Geometry_AreaPerCricle.xml\n\ngood luck
669147
number of sides, congurency?, size of angles, distinguishing characteristics
186995
There is no specific word as far as i know, but we used to use the term "Total Surface Area.'\n\nThere are different formulas for calculating the total surface area depending upon the shape of the three-dimensional figure.  If you could elaborate your question, one can give you an exact answer; or is this enough?
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
219209
Surface Area of a Cylinder = 2 pi r 2 + 2 pi rh. (h is the height of the cylinder, r is the radius of the top).\n\nSurface Area = 2(pi r 2) + (2 pi r)* h \n\nSurface Area = Areas of top and bottom +Area of the side
728936
1. obtain formulas for surface area and volume of cylinder\n2. let volume equal 1000 and solve this volume equation for h\n3. substitute the expression from previous step into surface area    equation to get equation with only r's\n4. differentiate surface area formula with respect to r set this equal to 0 and solve for r i got   r=10/3rd root(pi)\n5. substitute value for r into volume equation solve for h  i got h=10/(pi)^(2/3)\n\nincidentally, i used 2nd derivative test to conclude r is minimum\n\n\n\ni'll get back to u with more
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!
586130
surface area = 3bh, where b is length of side, and h is the height dropped normally from center to any of the sides\n\nvolume = 6 times volumeof each triamgle, formed by all 3 diagonals
865884
A = lb = 98 in² (l = length and b = width, both in inches)\nSo b = 98/l\nWorking area W = (l - 8)(b - 4)\n= lb - 4l - 8b + 32\n= 98 - 4l - 8b + 32\n= 130 - 4l - 8b\n= 130 - 4l - 784/l\n\ndW/dl = - 4 + 784/l²\n= 0 for stationary points\nSo 4 = 784/l²\nie 4l² = 784\nThus l² = 196\nl = √196 (ignoring negative result)\n= 14 inches\nd²W/dl² = - 2 * 784/l³ < 0 when l = 14 so area is minimised when l = 14\n\nWhen l = 14 b = 7 (lb = 98)\nSo the dimensions to minimise paper used is 14 inches X 7 inches
548292
Assuming that an area may be dissected into triangles, then the area of each triangle is given by \n\nArea = (s(s-a)(s-b)(s-c))^(1/2)\n\nwhere s = (1/2)(a + b + c)\n\nand a, b, and c are the lengths of sides of each triangle -\n\nThe sought for area is (finally) the sum of all the areas of the triangles.\n\nGiven sufficient patience, any arbitrary area will be found.\n\n(Drawing a diagram while slogging through the exercise and looping around the separate triangles in the same direction can yield a shortcut whereby all of the internal perimeters eliminate each other, leaving only the perimeter of the original figure.)