Draw 5 rectangles or squares that have the same area but different perimeters.?
554315
First pick what you want your area to be.\nLet A = 100.\nThe first one is easy, just let it be a square with all side lengths of 10\nNow pick two other numbers that when you multiply them you get 100.\n4 times 25 = 100\nA rectangle with width 4 and length 25.\nKeep picking more\n2*50 = 100\nwidth = 2 and length = 50\nYou can find more.\nAlso, you don't have to use 100, just pick any number you want for the area.
789618
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.
630757
A pentagon has 5 sides of equal length if that is what you are asking.
844185
square-ish in shape\ncome in different sizes\nmany varieties
606368
Yes form a hexagon with 6 of the boards and then place the other six boards so they meet at the center. Each pig will then have an equilateral triangle as a pen.
397805
This problem is certainly tractable, you just need to\nfind a consistent/methodical way of counting the\nrectangles -- without missing any and without counting\nany of them multiple times.\n\nThe way I chose is to define each rectangle as being\ndefined by the square in the upper left corner and the\nsquare in the lower right corner.  Consider the following\nsimplified 2x3 case (where each square is labelled by\nits X and Y coordinates:\n\n0,0     0,1   0,2\n1,0     1,1   1,2\n\n\nThis 2X3 example has the following rectangles starting\nat 0,0.  Remember, each rectangle is defined by the\nsquare in its upper left and the square in the lower right:\n0,0   0,0\n0,0   0,1\n0,0   0,2\n0,0   1,0\n0,0   1,1\n0,0   1,2\n\nThen it also has the set anchored in the upper left by 0,1:\n0,1   0,1\n0,1   0,2\n0,1   1,1\n0,1   1,2\n\nThen it has the set anchored in the upper left by 0,2:\n0,2   0,2\n0,2   1,2\n\nThen those anchored by 1,0:\n1,0   1,0\n1,0   1,1\n1,0   1,2\n\nThose anchored by 1,1:\n1,1   1,1\n1,1   1,2\n\nAnd finally those anchored by 1,2:\n1,2   1,2\n\n\nRewriting the matrix another way, showing how many\nrectangles are anchored by each location in the matrix:\n\n2x3    2x2   2,1\n1x3    1x2   1x1\n\n\nand then doing the math:\n\n6      4     2\n3      2     1\n\nfor a total of 6+4+2+3+2+1 = 18\n\n\nDoing the same analysys for a 4x5 gives:\n\n4x5   4x4   4x3   4x2  4x1\n3x5   3x4   3x3   3x2  3x1\n2x5   2x4   2x3   2x2  2x1\n1x5   1x4   1x3   1x2  1x1\n\nor rewriting after doing the multiplication:\n\n20    16    12    8    4\n15    12     9    6    3\n10     8     6    4    2\n 5     4     3    2    1\n\nAdding up the columns gives:\n\n50    40    30   20   10\n\nFor a total of 50+40+30+20+10 = 150 total rectangles
1021945
The first five squares are easy:\n1, 4, 9, 16, 25.\n\nRectangular:\n2, 6, 12, 20, 30\n\nTriangular:\n1, 3, 6, 10, 15\n\nPentagonals are:\n1, 5, 12, 22, 35
899386
The x's will all be the same, and the y's will be the difference between 4 & the number:\n\nA=(2,5)\nB=(0,3)\nC=(5,2)\n\nAnother way to do this is to actually draw the triangle and the line y=4. This will give you a visual of what should happen.
614253
In a rectangle, the opposite sides are equal.  Therefore, two of the sides = 2x-5 and the other two = x.\n\nPerimeter is the total (add) of all the sides.  So, your equation is\n2x-5 + 2x-5 + x + x =50  (Then you combine like terms and solve)\n\n6x -10 = 50\n6x = 60\nx = 10\n\nTo find the length and width, plug in the number you got.\n\n2x-5 = 2 (10) -5 = 20 -5 = 15\nx = 10\n\nTo check:  Add the perimeter with your numbers.\n15+15+10+10 = 50
505748
If the perimeter is 36 and 1 side is 4, then you know by the rules of a rectangle that 2 of the sides are equal.  So if 2 sides are 4 each, that is a total of 8.  to find the length of the other two sides, you subtract 8 from the total perimeter 36 which gives you 28.  So the sum of the remaining two sides is 28 and by definition of a rectangle, they are equal.  So 28/2 is 14.  So 2 sides are 14 in length and 2 sides are 4 in length.  \n\nto find square footage, which is also the AREA of the rectangle, multiply LxW which is 14x4 which gives you 56.
978615
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
30239
You can't just draw one? It's 10 boxes across the top and down the side, starting with 0 and going to 9 (and no, the corner doesn't share both sides). All you need are a pen and a ruler.
411731
The perimeter might be 14.0625. \n3+4+5+6+7 = 25 and the pentagons are similar. It is just a matter of proportions. \n(the perimeter of given)/ (perimeter you r tryin to find) = (area of given)/ (area given for the second pentagon)\n\nso in numbers it is (25/x)=(48/27) and you just solve for x because the pentagons are similiar.
897940
perimeter = 2 * (length + width)\n\nfor square, length = width..\n\nso, perimeter = 2 * (2 * length) = 4 * length\n\ngiven, 4 * length = 104\nso, length = 26\n\nso, each side is 26 cm long.
756458
w1=width of 1st rect.\nL1=length of 1st rect\nA1=area of 1st rect\nw2=width of 2nd rect\nL2=length of 2nd rect\nA2=area of 2nd rect\nw1=x\nL1=4x\nA1=length*width\n=4*(square of x)\nw2=x+2\nL2=4x+5\nA2=(x+2)(4x+5)\n=x(4x+5)+2(4x+5)\n=4*(square of x)+5x+8x+10\n=4*(square of x)+13x+10\n\nAccording to the given condition\nA1+270=A2\n4*(square of x)+270=4*(square of x)+13x+10\n270=13x+10                //4*(square of x) gets cancelled\n13x=260\nx=20\ntherefore width of 1st rect=20cm\nlenght of 1st rect=20*4=80cm
359093
Square and Rectangle= lenght*height\ntriangle=base*height divided by 2
639826
1. Draw a square.\n2. Bisect the square top to bottom by drawing a line through its midpoint.\n3. Draw a diagonal line from the base of the bisecting line (at the bottom center of the square) to the outer, upper corner of the square. \n4. Using the diagonal line as the radius, drop an arc from the outer, upper corner of the square down to the baseline.\n5. Draw a line straight up from the point of intersection of the arc and the baseline. Extend the top edge of the square to meet this line and form a rectangle - it's golden.