I don't know the formulas for Lateral Area and Surface areas. I need one for Cylinder, Trangular Prism, Cone, and Regular pyramid.
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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
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1) Area of a trap = (a+b)h/2\n88=(12+b)(8)/2\n2*88=(12+b)(8)\n2*88/8=(12+b)\nb=2*88/8 - 12\nb=10\n\n2) lateral surface area S = p*s/2\np = perimeter, s=slant height\nperimeter of a pentagon of base length 3 = 5*3=15\nS=15*5/2\nS=37.5
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Let z be the amount to multiply each dimension by:\nLet A be the original area\nLet A' be the new area\n\nA = W * L\nA' = zW * zL\nA' = z²(W * L)\nA' = z²A\n\nWe want the new area (A') to be 4/9 the old area (A).\n\nThat means z² = 4/9\nAnd therefore:\nz = sqrt(4/9)\nz = 2/3\n\nIf the original dimensions are 18 cm * 24 cm.  The new dimensions will each be 2/3 of that (12 cm * 16 cm).\n\nThe old area will be 432 cm², the new are will be 192 cm² which is 4/9 the original.\n\nThus the new dimensions should be 2/3 the old or 12 cm * 16 cm.
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http://www.johnstonsarchive.net/spaceart/cylmaps.html
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A rectangular prism is a box.\n\nSA (surface area) = 2 * l * w + 2 * h * w + 2 * l * h\n24 cm^2 = = 2 * l * w + 2 * h * w + 2 * l * h\n12 cm^2 = lw + hw + lh\n\nThe simplest prism that fits this would be a cube of 2 cm on a side:\n12 cm^2 = 2 * 2 + 2 * 2 + 2 * 2\n12 cm^2 = 4 + 4 + 4\n12 = 12
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to find the surface area of a pyramid take 1/2 of the perimeter of the base times the slant height and add on the area of the base.  First lets find the area of the base with 3 sides of 17.32 each. You need more information for this, unless you have been taught some simple trig functions. To get the area of the base we need to know the height of this base(if it were standing up).If I use the tangent function I find that the height of the triangle is approximately 15 inches. So the area of the base is 1/2 times 17.32 times 15. I did the tangent of 60 degrees = the height of the triangle divided by 1/2 the side of the triangle.  I hope this helps solve the rest of the problem.
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to calculate the area of a 2-dimensional space you always multiply length x width, or in this case height\n\nC
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1. volume = h*2*pi*r^2 =1000 so r^2= 1000/2/pi/h\n\nsurface area = 2*pi*(h+2*r^2)\nsurface area / volume = (h+2*r^2)/h/r^2 \n=1/r^2+2/h  = 2*pi*h/1000+2/h  which is a minimum when h is about 17.84  [(1000/pi)^0.5]
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There are two ways to measure the area of a shaded figure. The third is by integral calculus.\n\nMethod 1. \n\nStep 1. Draw the figure on a graph paper by choosing suitable scales in x and y axes.\nStep 2. Count the small squares that are shaded. If a square is not whole, then count it as one if it is greater than half a square. Skip it if it is less than half. And of course, two halves make a whole.\nStep 3. Convert it to area by multiplying the number of squares with area of 1 square.\n\nMethod 2.\nCase 1:\nIf the figure is a polygon, draw diagonals from from corner points and break it down to triangles.\nNow, the area of triangles can be calculated mainly by two formulae.\n\nArea of a triangle = (1/2) base * altitude\n= square root of [s(s-a)(s-b)(s-c)] \n\nThe second formula is very useful and practised in surveying. Here, a,b, and c are the three sides of a triangle; and s is the semi-perimeter i.e. s=(1/2)(a+b+c)\n\nCase 2:\nIf the boundary of the figure is curved, the the matter is a bit harder. If it is a circle or a part of a circle, use formula: area= pi*r squared; pi is approximately 3.1416..  Or else, you have to replace the curve by a straight line such that the area remains the same by eye.\n\nBy integral calculus, area can be found by using table of formula.
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The total area or the can is give by:\nArea of curved surface = 2π r h\nArea of top of can = π r²\nArea of bottom of can = π r²\n-----------------------------------------\nTotal area (A) of can = 2π r h + 2 π r²\n A = 2π r h + 2 π r²\n A - 2 π r² = 2π r h\n (A - 2 π r²) / 2π r = h\n h = (A / 2π r) - r\n\nIt's the best I can do for you based on the information given.
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Not a site, but I suggest you invest in something called Mathcad.\nTeach, please note, it is a site not a sight! Unless you were 'looking' for the answer... sorry!
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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!
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V = area of base times height\nheight = 14\nbase = 3*3*pi = 9pi\nV = 14*9pi\nV = 126pi\nV = 395.8
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Take the formula for the surface area(S) and solve for r\nS=4*pi*r^2\nr^2 = S/(4pi)\nr =squareroot(S/4pi) or (S/4pi)^(1/2)\n\nNow take the formula for the volume\n4/3*pi*r^3\nand substitute the formula for r into the equation\n4/3*pi*(squareroot(S/4pi))^3\n4/3*pi(S/4pi)^(3/2)
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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)
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Yes, the answer is h = 2r\n\nThis is a classic optimization problem. You want the minimum surface area(SA) for a cylindrical shaped tin for a fixed Volume(V)\n\n(1) SA = 2pi r^2 + 2pi r h\n\nV = pi r^2 h (for simplicity set V = 1)\n\nso (2)    h = 1/(pi r^2)\n\nSubstituting this value for h into (1) gives:\n\nSA = 2pi r^2 + 2/r\n\nd/dr SA = 4pi r - 2/r^2\n\nd/dr SA = 0 when 2/r^2 = 4pi r\n\nthat is: (3)   r = 1/(2 pi r^2)\n\nTo get the value of the ratio h/r divide equation (2) by equation (3), which yields a value of 2 to 1.\n\nh = 2r
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rare gem's answer is right..given the radius you'll just need to compute for the area using the formula right?\n\nArea=(radius^2)(pi)\n    = (5)(5)(pi)