What is the area of a cone that is 10" H and 4"to the center ?
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
29575
well, the formula for the volume of a cone is 1/3 * base * height.\n\nthe trick to doing this is to find the volume of the large cone, then subtract the volume of the small cone.  what's leftover will be the tumbler-shaped volume.  plug in numbers for the height and base for both the large and small cones, and then subtract. voila!
952353
The curved surface area is 2*pi*r*h\n\nor 2*3.142*27*12 sq cms\n\nand 2*3.142*17*35 sq mms\n\nyou can do the rest
801034
Here's the answer: (I'm assuming the top of the box is counted in the area)---------   x = side of square base, h = height, S = surface area, P = perimeter length\n\nS = 2x^2 + 4xh\n\nP = 8x + 4h ---->  h = (P-8x)/4  Substituting,\n\nS = 2x^2 + 4*x*(P-8x)/4 = Px - 6x^2\n\nFor your particular case where P = 120 cm,\n\nS = (120x - 6x^2) cm^2
22742
4 inches. the radius is half of the diameter.
543816
Yes it is !!\n\nFor proof of volume of cone, pls go to :\n\nhttp://mathforum.org/library/drmath/view/53646.html\n\n\nThe following site has got the proof for area of circle :-\n\n\nhttp://64.233.161.104/search?q=cache:FbBn10NweQkJ:www.artofproblemsolving.com/LaTeX/Examples/AreaOfACircle.pdf+area%2Bcircle%2Bproof&hl=en&gl=in&ct=clnk&cd=1
783277
let the height be x\nthe cross section will be (40-2x)*x\ndA/dx=40-4x\nsetting this to 0\nx=10\nso the length of the sides will be 10 cm each for max area of x section
841638
4.00751601 × 10^9 centimeters
976121
Let us assume that we require a cone with d as bottom circle dia, h as vertical height which are the known factors. Let l be the slant height of the cone which is equal to Square root of (h^2+(d/2)^2). To arrive the cone we have to make a sector with l as radius and A as angle of the sector. \nl is already calculated. A is calculated as follows:\nA = pi()*d/l which is obtained in radians; \nTo convert it to degrees, multiply A with 180 and divide by pi();\nSo we get A in degrees and l, from which we can get a sector. By connecting the two slant heights together we can get a cone with dia d and height h.
854741
The above answerer is on the right track but 10^-4 kg seemed too large for me.  10-4 g is more reasonable and I see is part of the problem.  Multiply his/her answer by 10^3, and you might want to do the calculation yourself to double check.
528597
Total Area = Base Area + Lateral Area\n\nLateral Area= pi x radius x height of the slant.\nHeigh of slant = sqrt(cone height^2 + radius^2)\n\nLateral Area = (pi)(r) * sqrt(cone height^2 + radius^2)\nBase Area = (pi)(r^2)\n\nTotal area = (pi)(r) * sqrt(cone height^2 + radius^2) + (pi)(r^2)
311471
lateral surface area of a cone = pi*r*l\n\nwhere \n         pi = 3.14( a mathematical constant)\n         r = radius of base of cone\n         l = slant height of cone\n           =  square root of sum of squares of base radius  and height of cone\n         h = height of cone from centre of base circle to apex of cone\n          l = sq rt(r*r + h*h)
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.
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
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.
816108
dA = ½ (b.dh + h.db)\nSo δA ≈ ½ (b.δh + h.δb)\n= ½ (100 x ±1 + 10 x ±2)\n= ½  * ±120\n= ±60ft²\n\nCheck:\nMin area = ½ * 98 * 9 = 441ft²\nMax area = ½ * 102 * 11 = 561ft²\n\nNow 500±60 ft² = (440 to 560)ft²\n\nSo estimate is pretty close!!!
577740
Imagin 2 circles. \n\nTake the first which is the 10'diameter center and add 6ft (3ft walk twice) for a total od 16' diameter. So that radius is half the diameter or 8'.\n\nSo that area is A=3.14(r^2) or 3.14(8^2)=200.96\n\nThen find the area of the inside foundation that is 10' in diameter or 5' radius.\n\nso that area A=3.14(r^2) or 3.14(5^2)=78.5\n\nnow subtract the two areas to get the sidewalk area.  \n\n200.96-78.5=122.46\n\nIt may help if you draw it out. Then you can see how subtracting the foundation area from the entire area will give you the answer.