Can someone walk me through this geometry problem? Please? Area?
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Rec first  8 X 4 = 32\n32 X 16 = 512  this is Area of rectangle.\n\nCircles pi X r^2 = Area\n3.14159 X 8 X 8 = 201\n\n512 - (2 X 201) = shaded area\n109.87 = shaded area\n\nOK ?
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
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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
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i wont answer.... log onto www.mathguru.com and discover solutions with explanations.
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Area of what?\n\nRectangle = b×h\nSquare = l^2\nTrapezium = ½(l+b)h\nEllipse = πlb\nCircle = πr2 \nTriangle = ½b×h
1015745
here goes again .. seems like im gona do the whole hw for u ;) to find the limit of the area that ur gona revolve around the x=-2 axis u have to draw the two equations again and the area enclosed is basicly the shape of a circle (not a perfect circle though) and its located right above the x-axis touching it from below and the center of it is on the y-axis .. now the limit is the intersection of these two euations y=x^2 and y=2-x^2 that is x^2= 2- x^2 ==> 2*x^2 = 2 ==> x^2=1 ==> | x | = 1 ==> x = +1 or x = -1 .. and thats ur interval limits
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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
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Surface Area = 4*PI*r^2\n\nr = SQRT(SA/4/PI)
175713
Your question is too complicated to appreciate the math in it, which is supposed to be the real beauty in the first place. Anyway, I have tried figuring it out and come with equations. I do not even want to try and begin substitution and end up with more complications. I just thought once you saw the relation, you would throw those 10 points my way.\n\nAssuming a to be sides of the squares of the prism and b to be the length of the rectangles, we know that the sum of all dimensions is 14a+5b. Now I do not even want to get started why you brought it the story of the equilateral triangle. You can just say the sum  of the dimensions is 61. Anyway, since any side of the trianle, as it is equiangular hence equilateral, is twice 14a+5b and 122, the sum of the dimensions will be 61. So your first equation is 14a+5b=61\n\nYou know the volume of the prism is the length X width X heiht = a x a x b = (a^2)b This is equal to the surface area. The surface area is the sum of all the areas on the prism and that is a^2+4ab. (How about you calculate how that came up?)\n\nSo there is your second equation, (a^2)b = a^2+4ab So then from 14a+5b=61, express a in terms of b or the other way around and plug it into the second equation and experience brutal complications and figure out what the variables are. I think you know it is too easy to calculate each dimension and area after that.
489663
Reviewing the facts:\n\n1.  You're constructing a pentagon and two trapezoids by using the center of the square.  Since the sides are 1, the short parallel side of the trapezoid is 1/2.\n\n2.  Since the two trapezoids are congruent, the short parallel side is midway in the square.  So the height of each trapezoid is 1/2.\n\n3.  Since there are three areas now, and each area is equal, the area of the trapezoid is 1/3.\n\nThe area of a trapezoid is A = (1/2)h(b1 + b2).  We have everything we need now to solve for b2, the longer parallel side:\n\nA = (1/2)h(b1 + b2)\n1/3 = (1/2)(1/2)(1/2 + b2)\n1/3 = (1/4)(1/2 + b2)\n4/3 = (1/2 + b2)\nb2 = 5/6
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Surface Area Formulas\nIn general, the surface area is the sum of all the areas of all the shapes that cover the surface of the object.\n\nCube | Rectangular Prism | Prism | Sphere | Cylinder | Units\n\n\nNote: "ab" means "a" multiplied by "b". "a2" means "a squared", which is the same as "a" times "a".\nBe careful!! Units count. Use the same units for all measurements. Examples\n  Surface Area of a Cube = 6 a 2  \n\n\n (a is the length of the side of each edge of the cube) \n\nIn words, the surface area of a cube is the area of the six squares that cover it. The area of one of them is a*a, or a 2 . Since these are all the same, you can multiply one of them by six, so the surface area of a cube is 6 times one of the sides squared. \n\n \n\nSurface Area of a Rectangular Prism = 2ab + 2bc + 2ac  \n\n (a, b, and c are the lengths of the 3 sides) \n\nIn words, the surface area of a rectangular prism is the are of the six rectangles that cover it. But we don't have to figure out all six because we know that the top and bottom are the same, the front and back are the same, and the left and right sides are the same. \n\nThe area of the top and bottom (side lengths a and c) = a*c. Since there are two of them, you get 2ac. The front and back have side lengths of b and c. The area of one of them is b*c, and there are two of them, so the surface area of those two is 2bc. The left and right side have side lengths of a and b, so the surface area of one of them is a*b. Again, there are two of them, so their combined surface area is 2ab. \n\n  Surface Area of Any Prism \n\n\n  (b is the shape of the ends) \n\nSurface Area = Lateral area + Area of two ends \n\n(Lateral area) = (perimeter of shape b) * L \n\nSurface Area = (perimeter of shape b) * L+ 2*(Area of shape b) \n\n  Surface Area of a Sphere = 4 pi r 2  \n\n\n (r is radius of circle) \n\n  Surface Area of a Cylinder = 2 pi r 2 + 2 pi r h \n\n\n (h is the height of the cylinder, r is the radius of the top) \n\nSurface Area = Areas of top and bottom +Area of the side \n\nSurface Area = 2(Area of top) + (perimeter of top)* height \n\nSurface Area = 2(pi r 2) + (2 pi r)* h \n\nIn words, the easiest way is to think of a can. The surface area is the areas of all the parts needed to cover the can. That's the top, the bottom, and the paper label that wraps around the middle. \n\nYou can find the area of the top (or the bottom). That's the formula for area of a circle (pi r2). Since there is both a top and a bottom, that gets multiplied by two. \n\nThe side is like the label of the can. If you peel it off and lay it flat it will be a rectangle. The area of a rectangle is the product of the two sides. One side is the height of the can, the other side is the perimeter of the circle, since the label wraps once around the can. So the area of the rectangle is (2 pi r)* h. \n\nAdd those two parts together and you have the formula for the surface area of a cylinder. \n\nSurface Area = 2(pi r 2) + (2 pi r)* h \n\n  \n\n\nTip! Don't forget the units.  \n\n\nThese equations will give you correct answers if you keep the units straight. For example - to find the surface area of a cube with sides of 5 inches, the equation is: \n\nSurface Area = 6*(5 inches)2 \n\n= 6*(25 square inches) \n\n= 150 sq. inches
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Assuming that A, B, C are all on the same line.\n\nLarger circle:\nRadius = AC = 6\nArea = PI * r^2 = 36*PI\n\nSmaller circle:\nRadius = AC - AB = 6 - 2 = 4\nArea = PI * r^2 = 16*PI\n\nShaded area:\nArea = (Large A) - (Small A)\n     =  36*PI - 16*PI = 20*PI = 62.8 units^2
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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.
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well you could split up the thing into triangles. you know what the base is (1m) and the height is unknown, so lets call it h for now. so for each side, the area of the triangle is h/2 (bh/2 and b=1 so h/2). so for every 2 sides, the area is h. so the total area is 18h. \n\nso to find h, you have to use some trig. the equation to find the internal angles is (n-2)*180/n. cuz the sum of the internal angles is (n-2)*180. and then to find each angle, you have to divide by number of angles (which is same as number of sides). so in this case 34*180/36 = 170 degrees each angle. so now for each side, if you make a triangle with the center of the polygon, u should notice that the triangle is isosceles and the base angle is half of the side angle. \nas in the bottom 2 angles are equal and they are equal to 170/2. which is um 85. you can try this with any polygon if you dont believe it. try a square, (n-2)*180 = 2*180 = 360 is the sum of the internal angles. so each angle is 360/4 = 90 degrees. so if u connect each vertex to the center of the circle, you get 4 isosceles triangles and the base angles are 45 degrees each. \nso you have 85 degrees as the angle. so now if you split the isosceles triangle into 2 right angle triangles, you have 3 sides in each right angle triangle: the h, 0.5m and the line from the vertex to center. so you have the angle 85 degrees and you need h and you have the adjacent side = 0.5m. so you can use tan 85 = h/0.5 and you get h. \nso the total area is 18h. \n\ni really hope you understand that. its kinda hard to explain without drawing it out and showing you. if you need more help, then message me or somethin. later
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the formula for the area of any regular shape is \n(pa)/2, so there is no multipling the base by 2 involved.\np=perimeter\na=apothem
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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.)
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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.