Can you find a plane shape, with total surface of 34 m^2 having minimal perimeter and surrounding a 5x5 square
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So am I answering a homework question for you? lol\n\nCircles have the minimal perimeter for the area contained.\nArea of a circle is pi x Radius x Radius. (pi R squared)\n\nNext best would be something close to a circle, so consider a hexagon or octagon, with the corners of the square intersecting the vertices of the outer figure.\n\nLet's try a circle first...\n\nDiameter of the circle would be the diagonal of the square, or five times the square root of two.  5 x 1.414 = 7.07.  So the Radius is 3.535 meters.\n\nPi x 3.535 x 3.535 = 3.14 x 12.5 = 39.34\n\nSo the circle solution is too big.\n\nNow work out on paper what an octagon or hexagon surrounding the square would look like.  One of those should be about 34 m square.  Remember the area of the square is already 25 m2, so you are looking for only 9 more m2.\n\nI played around for minute or so in paint, and came up with a parallelogram that worked too.  Finally, a smooth curving perimeter shape get you benefits similar to the "circle" solution, so an ellipse would be a good comparison solution.  Try that, and compare the perimeter result against the perimeter of the parallelogram, which has a perimeter of 24.23 meters.\n\nHave fun!\nOC
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a) 24 in.\n\nSince a square has four sides...  You must add six for each side (i.e. 6 + 6 + 6 + 6) to get perimeter.
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a polygon with 5 sides is a pentagon.\n\nthis has two right angles so it would look like the picture of a basic drawing of a house.   see the pic of the house attached.\n\nhttp://www.thestationerystudio.com/index.cfm/fuseaction/catalog.viewProduct/productId/5214
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7.3m (width) by 2.4m (height)
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A shape with minimum surface area for a given volume!...a round shaped ball-like mass would have minmum area of 4 pi r square for a volume 4/3 pi r cube !
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width = 5 \nlength = 12
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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)
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I am trying to figure how large a .745" round wire can be shaped into a rectangle shape wire. \n\nAssuming the wire is malleable, find the area of a circle with diameter .745 or radius = .3725\n\n(.3725)squared times pi = .4359156156 sq in.\n\nFor a square to have this area, each side would be the square root of that number which is .6602390595.\n\nFor a rectangle that is not a square, pick any 2 numbers whose product is .4359156156.  Choose one dimension and divide by it to get the other.
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You never mentioned that there are smaller squares that make up this 5x5 shape. Therefore, the technical answer is one. For all we know...5x5 could be referring to inches or some other measurement.
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Why do you think this isn't possible?  Perimeter is 2 times the sum of width and length.  Any combination of width and length that sum up to 18 cm will satisfy this requirement
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I am only using visual techniques and not math theorems here:\n\nThe first plane will divide space into 2 parts.\n\nIf totally random the second plane will probably not be parallel with the first.  This will divide the 2 parts into 4.  If it did happen to be parallel with the first, then the space is divided into 3 parts.\n\nThe third plane will most likely not be parallel with either of the other two.  It would divide the space into 8ths.  If it was parallel to either of the other two then the minimum parts is 4 and the maximum is 6 parts.\n\nThe fourth plane will most likely not be parallel with any of the first 3, and can divide the space into maximum of 15 parts.  The fifth will make that 26.  Parallel planes will reduce this number.\n\nThe space will be divided into a minimum of 6 and a maximum of 26 parts with 5 planes.\n\nI just found a great description at the site below:
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Yes, it is possible.\n\nDenote the number of black dots belonging to a red dot by N.\n\nThe present problem is formulated for N=10.\n\nLet us first solve a simpler problem with N=2.\n\nStep 1. \nTake a square, that is a four-sided regular polygon, with the side equal to R. Put the two red dots at the opposite [i.e., connected by a diagonal] vertices of the square. Put the two black dots at the remaining two vertices. If you draw a circle of radius R around each of the two red dots, you will find exactly 2 black dots lying precisely on the circle line. The problem is almost solved, but the number of red dots (2) is equal to the number of black dots (2).\n\nStep 2. \nTake another square and repeat the above procedure. \nNow we have the two square compounds. In total we have 4 red dots and 4 black dots.\n\nStep 3. \nPut the 5th red dot anywhere in the plane and draw a circle of radius R around it. \nDrag each of the above compounds as a whole closer to this circle. \nPlace one of the black dots belonging to the first compound and one of the black dots belonging to the second compound on the circle line surrounding the 5th red dot.\n\nWhile doing so, make sure that you move and rotate each compound as a whole, keeping all the sides of the squares equal to R. \n\nThe problem with N=2 has been solved: in total we have 5 red dots and 4 black dots satisfying the condition of the problem, 5>4. \n\nLet us solve the same problem for N=3.\n\nIn this case, one has to extend the Step 1.\n\nStep 1a. \nHaving constructed the square compound according to the Step 1 above, draw parallel vectors of length R starting at the vertices of the square. Make sure that the vectors do NOT coincide with the sides of the square. \nIf the vector starts at the red vertex, put the black dot at the end of the vector. If the vector starts at the black vertex, put the red dot at the end of the vector. \n\nNow we have one compound consisting of the two shifted squares containing 4 red dots and 4 black dots. If you draw a circle of R around each of the four red dots, you will find exactly 3 black dots lying precisely on the circle line.\n\nStep 2. Repeat the procedure 1+1a.\nStep 2a. Repeat the procedure 1+1a.\n\nNow that we have 3 compounds with equal numbers of red and black dots (in total 12 red dots and 12 black dots), put the 13th red dot anywhere in the plane and follow the instructions from Step 3 above.\n\nThe problem with N=3 has been solved: in total we have 13 red dots and 12 black dots satisfying the condition of the problem, 13>12. \n\nFor N=4, take the compound constructed for N=3 at Step 1a and translate ALL 8 vertices by a vector of length R.\nAgain, if the vector starts at the red vertex, put the black dot at the end of the vector and vice versa. You will have 8 red dots and 8 black dots, etc.\n\nThe above procedure applies to an arbitrary N, in particular to N=10.\n\n\nAn interesting question is: what is the minimal number of dots needed to solve the problem for a given N.
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This might help --> http://www.uwm.edu/~ericskey/TANOTES/Geometry/node15.html
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now, if you add that it has 4 sides and all right angles, you probably can pass that off as the definition of a rectangle.\n\notherwise you just have a parallelogram.
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Finding a runway close by.
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Surface area. Seriously. 8D
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Diamond-Square is a great height-map generation algorithm. It's kind of cool that something that simple could create realistic landscapes.