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January 17th, 2013, 04:45 AM  #1 
Newbie Joined: Jan 2013 Posts: 3 Thanks: 0  Finding the Area
Hallo Can someone please help me to calculate the inside area of the drawing. The drawing is not to scale. I am really struggling with this. [attachment=0:818mtdbp]Bloemfontein2013011600428.jpg[/attachment:818mtdbp] Thank you 
January 17th, 2013, 01:17 PM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  Re: Finding the Area
Line under the 2 lines of 9200: 2200 or 3200? You show a line that seems split 8100: 3700; is that a straight line? WHY the split mark? 
January 17th, 2013, 07:47 PM  #3 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications  Re: Finding the Area
Hi [color=#0000FF]narkie[/color], and welcome to the forums. I think you are struggling because it is impossible to accurately calculate the area without either a scale drawing or measures of the angles. Visualize the following: Let the joints on the top part of the diagram be fixed at right angles starting with the intersection of the sides of lengths 3600 and 4000, proceeding clockwise and ending with the intersection of the sides of lengths 1400 and 2900. Let the other joints be flexible. If you take the tick mark at the intersection of the sides of length 8100 and 3700 and pull it into the interior of the figure, even if you keep the side of length 9300 vertical and the side of length 20800 horizontal (each of which is implied), the angles of the sides with lengths of 12100 and 10100 are free to change. Obviously, the angles of the sides of lengths 8100 and 3700 will also change as you pull. You could also pull the tick mark the opposite way. This results in shapes with the lengths of sides as indicated, but with different areas. To illustrate this, let's even fix the angle of the side of length 12100 at +135 degrees. Since the next side (of length 9300) is taken to be vertical, this fixes the points from the intersection of the sides of lengths 8100 and 3600 to the point of intersection of the sides of lengths 9300 and 10100. In the first diagram below, the tick mark is manipulated so that the sides of lengths 8100 and 3700 form a straight line of length 11800. The coordinates of the relevant points are given and the area is 928,810,981. In the second diagram, the tick mark is pulled into the interior by an arbitrary amount. Again, the coordinates of the relevant points are given and the area is 795,527,987. Note that I took the length of the side between the sides of lengths 9200 as 3200. As a crude check, the figure is approximately 40,000*20,000=800,000,000 which is in the ballpark. I used a Ruby script that I wrote for this problem so if you have figures for the angles, I can plug them in and get a result very easily. I will post the script if anyone is interested. As explained in the referenced post, the shoelace formula is used to calculate the area. 
January 17th, 2013, 11:15 PM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  Re: Finding the Area
Well, if "that line" is fixed at 11800, and your 4700 line is changed to 4760, and your 3750 line changed to 3740 (are you sure of your lengths?), then your diagram can be fit into a rectangle width 38320 by height 31680. The 11800 becomes the hypotenuse of a right triangle with legs 7080, 9440, the 10100 becomes the hypotenuse of a right triangle with legs 6060, 8080 , the 12100 becomes the hypotenuse of a right triangle with legs 7260, 9680. Top width line: 4000 + 4760 + 3200 + 17700 + 8660 = 38320 Bottom width line: 9440 + 20800 + 8080 = 38320 Left height line: 21000 + 3600 + 7080 = 31680 Right height line: 3740 + 2900 + 9680 + 9300 + 6060 = 31680 Your area would be: 38320 * 31680  21000 * 4000  9200 * 3200  8660 * 3740  2900 * 7260  7260 * 9680 / 2  6060 * 8080 / 2  7080 * 9440 / 2 = 954,056,800 : YIKES! Pretty close to jks' 928,810,981....probably wrong :P 
January 17th, 2013, 11:20 PM  #5 
Newbie Joined: Jan 2013 Posts: 3 Thanks: 0  Re: Finding the Area
Hey jks and Denis Thanks for the help. No wonder I couldn't figure it out. I did not have all the dimensions. I made a new drawing and entered the missing dimensions and also divided all the numbers by 1000 to make it easier to write on the drawing. [attachment=0:1o4nza9j]Bloemfontein20130118004291.jpg[/attachment:1o4nza9j] Thanks 
January 17th, 2013, 11:41 PM  #6 
Newbie Joined: Jan 2013 Posts: 3 Thanks: 0  Re: Finding the Area
The answer I got was : 698.3485 squares. 
January 18th, 2013, 08:54 AM  #7  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  Re: Finding the Area Quote:
What size rectangle do you get if you insert your diagram in a rectangle?  
January 18th, 2013, 10:14 AM  #8 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications  Re: Finding the Area
I do not think that all of the dimensions and angles are correct. The dimensions in the x direction add up to be the same along the top and bottom, but there is a discrepancy of 2.73 in the y direction between the left and right sides. The following is the output of my script. The last point should be very close to (0,0), but instead it is (0,2.73). The last number is the area, 779.56 but I cannot really trust it. x1 2.2042914368802792e16 y1 3.6 x2 4.0 y2 3.6 x3 4.000000000000001 y3 24.6 x4 8.700000000000001 y4 24.6 x5 8.700000000000001 y5 15.400000000000002 x6 11.900000000000002 y6 15.400000000000002 x7 11.900000000000002 y7 24.6 x8 29.6 y8 24.6 x9 29.6 y9 20.85 x10 31.0 y10 20.85 x11 31.0 y11 17.950000000000003 x12 41.79846598554391 y12 12.490775480065778 x13 41.79846598554391 y13 3.1907754800657777 x14 32.2492429380759 y14 0.09895690870169771 x15 11.449242938075901 y15 0.09895690870169771 x16 7.89847771750891 y16 0.9412672731032843 x17 0.0012754557813288159 y17 2.7312113507467153 779.563391683334 
January 18th, 2013, 04:55 PM  #9  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038  Re: Finding the Area Quote:
Only way some sense can be made is forget about the bottom left corner slants of 3.7 and 8.1; make that a straight horizontal line equal to 11.45 top horizontal line: 4 + 4.7 + 3.2 + 17.7 + 1.4 + 10.8 = 41.8 bottom horizontal line: 9.55 + 20.8 + 11.45 = 41.8 right vertical line: 3.75 + 2.9 + 5.46 + 9.3 + 3.29 = 24.7 left vertical line: 21 + 3.7 = 24.7 (the 3.6 slightly extended to 3.7) 41.8 * 24.7 = 1042.56 (area of rectangle) 21*4 + 9.2*3.2 + 12.2*3.75 + 10.8*2.9 + 10.8*5.46/2 + 9.55*3.29/2 = 235.70 (area to be removed) 1042.56  235.70 = 796.76 ; fairly close to your 779.56, jks. To Narkie: you keep throwing "not to scale" stuff at us: are you too busy to sit down with pencil and paper and have an actual look at what you're trying to do? All you've actually succeeded to do is make us lose our time  

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