How can the area increase cutting a square into 4 parts and rearranging it into a rectangle?
396811
This a variation on the hidden parallelogram trick that relies on the viewer trusting his eyes (seeing a straight line) over objective measurement (explicitly measuring the disagreeing slopes of the hypotenuse). The original trick can be found in the source.
99265
I asked that same question to my teacher last time. She told me that I have to Seperated the shape into different common shapes. (for example like lets say I wanted to seperate a "square" into some common shapes I could turn it into two seperate triangles or four different triangles. Well Just try to use the formulas you know for the areas of the different disected figures you get.
529015
The old area is found through 2lh + 2wh.  There are 4 rectangles of the same height - one wall is l*h and the other is w*h.  Since there's two each, the total area to be painted will be 2lh + 2wh.\n\nNow if the new space has a length 2l and a breadth w/2, then one wall becomes 2lh and the other becomes wh/2.  Again, since there's two each, the new area will be 4lh + wh.\n\nHow does this affect cost?  It's probably easier to just plug in numbers to see.  Using the formula 2lh + 2wh and varying the length and breadth for each scenario, we have:\n\nOld space (l = 3, w = 2, h = 2) = 2(3)(2) + (2)(2)(2)\n= 12 + 8\n= 20\n\nNew space (l = 6, w = 1, h = 2) = 2(6)(2) + (2)(1)(2)\n= 24 + 4\n= 28\n\n% change = 40%\n--------------------------------------------------\nOld space (l = 4, w = 2, h = 2) = 2(4)(2) + (2)(2)(2)\n= 16 + 8\n= 24\n\nNew space (l = 8, w = 1, h = 2) = 2(8)(2) + (2)(1)(2)\n= 32 + 4\n= 36\n\n% change = 50%\n--------------------------------------------------\nOld space (l = 5, w = 2, h = 2) = 2(5)(2) + (2)(2)(2)\n= 20 + 8\n= 28\n\nNew space (l = 10, w = 1, h = 2) = 2(10)(2) + (2)(1)(2)\n= 40 + 4\n= 44\n\n% change = 57%\n\nThe total area increases every time.
681891
Sub-divide it into regular polygons, and then add up their areas.
1050164
Divide the L-shape into two rectangles, and compute the area of each rectangle separately. Then add the two areas.
237985
Cut a 2x2 piece off. Cut the remaining 2x3 piece in half lengthwise.  Put the two resulting 1x3 pieces one atop the other and cut the stack diagonally.  You now have 4 triangles with sides = 1, 3 and sqrt(10).  Arrange them with their right angles touching the corners of the 2x2 square piece, 3-meter sides parallel to and touching the sides of the square piece, and all "pointing" in the same direction, clockwise or counterclockwise.  The result is a square that is sqrt(10) on a side.  Area = 2x5 = sqrt(10)^2 = 10 square meters.  Three cuts were used to make five pieces.\nTo solve this I started with the requirement of the large square being sqrt(10) on a side.  Sqrt(anything) suggests the hypotenuse of a right triangle.  A little mental fiddling produced the sides of 1 and 3 meters.  Luckily a 2x2 piece remained.
35229
Start with a multi-cut-out prototype & tape.  Once you've figured the sizes for your parts, disassemble and use a prototype 1-piece pattern with tabs & slots to hold it together.  Work it till it works  :-)
328495
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.
771669
Cause squares would look weird
518004
You would need to disect the polygon into two shapes.  A triangle and a rectangle.  Figure the volume of those individual pieces and then add them together.
766270
im assumuing you mean a rectangle? If it was a square itd be 400, if its a rectangle its 80l-l^2
953741
Just break the pentagon up into 5 triangles.  You can then add together the area of the 5 triangles to get the total area.
593727
I can remember upto 4 to 5 ways of doing it and not more than that.  btw you are talking about halving a square (diagram)! Right ?
71096
let area be A\nwidth be W\nlength be L\nnow area of rectangle is = length * breadth \n    A = L * B\n  L = A/B\n L = 3x (squared) +14x+8 / x+4\n   = (3x+2)(x+4) / (x+4)\n   = 3x+2
1150117
Start with the upper left square.  Using that square, you can make 14^2 = 196 rectangles (of course a square counts as a rectangle) since any other square can be used as the bottom right end of the rectangle.  Then move to the next square to the right.  Using that square, there are 14*13 = 182 rectangles that do not include the first square.  Then move to the next square to the right; there are 14*12 = 168 rectangles that do not include the previous two squares.  And so on:  14*11 = 154, 14*10 = 140... down to 14*1 = 14 with the top right square.  So, in total, there are 14*(14+13+12+...+1) possible rectangles that can be made from the squares in the top row.\n\nThen move to the next row.  Starting with the leftmost square in the second row, you can make 13*14 = 182 rectangles starting with that square that does not include any squares in the first row.  Then move to the right one square.  With that square, you can make 13^2 = 169 rectangles that don't include the first row or the first square in the second row.  And so on: 13*12 = 156, 13*11= 143, and so on down to 13*1 = 13.  So there are a total of 13*(14+13+12+...+1) rectangles that can be made from the squares in the second row that do not include the squares in the first row.\n\nSo, in general, if you are in row k, then there are k*(14+13+...+1) rectangles that can be made that do not include the previous rows.  The last row will have 1*(14+13+12+...+1).\n\nSo, in total, there are \n14*(14+13+12+...+1) + 13*(14+13+12+...+1) + ... + \n1*(14+13+12+...+1) \n= (14+13+...+1)(14+13+...+1)\n = (14+13+...+1)^2 rectangles.\n\nYou may remember that the sum of the first n whole numbers is equal to n(n+1)/2.  So the sum of the first 14 whole numbers is 14+13+...+1 = 14(14+1)/2 = 105.  So, in total, there are 105^2 = 11,025 rectangles that can be formed from the squares of a 14 by 14 grid.\n\nIn general, if you have an n by n grid, there are {n(n+1)/2}^2 rectangles that can be made.
805184
It depends on where the 3x3 square was cut out.\nIf cut out of a corner, then the total perimeter remains the same, 39".\nIf cut out of a side, such that a side of the cut out square is on one of the sides of the paper, it increases by 6", or 45"\nIf cut out of a side, such that a corner of the 3x3 piece is on one of the sides of the paper, it increases by 12", or 51"\nsee:\nhttp://mudandmuck.com/str2/cutout.JPG\n\nAlso, I assume that she doesn't cut off a corner.  But if she did, the perimeter could increase anywhere between 0" and (9-3√2)"
1045508
click on the picture. there should be some boxes on the corner of the picture. you have to click on these squares and drag it to the size you want