Chikis Posted October 12, 2014 Share Posted October 12, 2014 A rectangle is twice as long as it is wide. If the lengh and width are decreased by 2cm and 3cm respectively, the area is decreased by 30 cm^2. Find the original dimension of the rectangle. We have that 2L=w where L = lenght of rectangle and w = width of rectangle. (L-2)(w-3)-30 = area How do I form the equations? Link to comment Share on other sites More sharing options...
Robittybob1 Posted October 13, 2014 Share Posted October 13, 2014 A rectangle is twice as long as it is wide. If the lengh and width are decreased by 2cm and 3cm respectively, the area is decreased by 30 cm^2. Find the original dimension of the rectangle. We have that 2L=w where L = lenght of rectangle and w = width of rectangle. (L-2)(w-3)-30 = area How do I form the equations? If a rectangle is twice as long as it is wide? An example of this rectangle would be one that was 8 long and 4 wide so the first equation should be L = 2W (8 = 2*4), not what you've written 2L=W (2*8 <> 4). You have to get that bit right to begin with. Link to comment Share on other sites More sharing options...
studiot Posted October 13, 2014 Share Posted October 13, 2014 (edited) You asked in your previous question if you can reduce the number of equations. This problem is the ideal situation for this, you do not need L = 2W, You can simply use 2W, whereever you have written L. So original area= A = W (2W) = 2W2 Can you now write the second equation so that (A - 30) = ?? Edited October 13, 2014 by studiot Link to comment Share on other sites More sharing options...
Robittybob1 Posted October 13, 2014 Share Posted October 13, 2014 You asked in your previous question if you can reduce the number of equations. This problem is the ideal situation for this, you do not need L = 2W, You can simply use 2W, whereever you have written L. So original area= A = W (2W) = 2W2 Can you now write the second equation so that (A - 30) = ?? The only reason you can "simply use 2W, whereever you have written L" because L = 2W. Link to comment Share on other sites More sharing options...
studiot Posted October 13, 2014 Share Posted October 13, 2014 The only reason you can "simply use 2W, whereever you have written L" because L = 2W. Well yes that is true, but that is what we are told. Link to comment Share on other sites More sharing options...
Robittybob1 Posted October 13, 2014 Share Posted October 13, 2014 (edited) Well yes that is true, but that is what we are told. OK but where do you go from there? I couldn't see the point of going "(A - 30) = ??" I thought you should write the dimensions of the bits cut off the main area. 2w and 2(L-2) and they must add up to 30. Edited October 13, 2014 by Robittybob1 Link to comment Share on other sites More sharing options...
studiot Posted October 13, 2014 Share Posted October 13, 2014 (edited) OK but where do you go from there? I couldn't see the point of going "(A - 30) = ??" I can't yet answer this, until Chikis has had a chance to have a go, as this is homework help. But think about the one piece (or pair of pieces) of information in the problem statement, not yet used. As a matter of interest I make W = 36/8 this way. Edited October 13, 2014 by studiot Link to comment Share on other sites More sharing options...
Chikis Posted October 13, 2014 Author Share Posted October 13, 2014 Am not in a hurry to solve the problem. I would like us to look at the way some problems on rectangle are phrased and study them. A rectangle is twice as long as it is wide. Mean L = 2W where L and W are lenght and width of the rectangle. It also means that the lenght is two times the width. Can the statement be written also as a lenght is two times as long as it wide? If the statement is, a rectangle is twice as wide as it is long, then the equation W = 2L. The same statement also mean that the width is twice it lenght. Hope my understanding concerning the following statement are very correct. Link to comment Share on other sites More sharing options...
Robittybob1 Posted October 13, 2014 Share Posted October 13, 2014 Am not in a hurry to solve the problem. I would like us to look at the way some problems on rectangle are phrased and study them. A rectangle is twice as long as it is wide. Mean L = 2W where L and W are lenght and width of the rectangle. It also means that the lenght is two times the width. Can the statement be written also as a lenght is two times as long as it wide? If the statement is, a rectangle is twice as wide as it is long, then the equation W = 2L. The same statement also mean that the width is twice it lenght. Hope my understanding concerning the following statement are very correct. OK so what is the situation in the problem you are wanting to solve? Link to comment Share on other sites More sharing options...
Chikis Posted October 14, 2014 Author Share Posted October 14, 2014 (edited) If a rectangle is twice as long as it is wide? An example of this rectangle would be one that was 8 long and 4 wide so the first equation should be L = 2W (8 = 2*4), not what you've written 2L=W (2*8 <> 4). You have to get that bit right to begin with. Am not in a hurry to solve the problem.I would like us to look at the way some problems on rectangle are phrased and study them. A rectangle is twice as long as it is wide. Mean L = 2W where L and W are lenght and width of the rectangle. It also means that the lenght is two times the width. Can the statement be written also as a lenght is two times as long as it wide? If the statement is, a rectangle is twice as wide as it is long, then the equation W = 2L. The same statement also mean that the width is twice it lenght. Hope my understanding concerning the following statement are very correct. [math](L-2)(W-3) = 2W^2-30[/math] W = 4.5cm L = 9cm Edited October 14, 2014 by Chikis Link to comment Share on other sites More sharing options...
Robittybob1 Posted October 14, 2014 Share Posted October 14, 2014 OK but where do you go from there? I couldn't see the point of going "(A - 30) = ??" I thought you should write the dimensions of the bits cut off the main area. 2w and 2(L-2) and they must add up to 30. Read the problem incorrectly the width was reduced by 3 So it should be 2w and 3(L-2) and they must add up to 30 Am not in a hurry to solve the problem. I would like us to look at the way some problems on rectangle are phrased and study them. A rectangle is twice as long as it is wide. Mean L = 2W where L and W are lenght and width of the rectangle. It also means that the lenght is two times the width. Can the statement be written also as a lenght is two times as long as it wide? If the statement is, a rectangle is twice as wide as it is long, then the equation W = 2L. The same statement also mean that the width is twice it lenght. Hope my understanding concerning the following statement are very correct. [math](L-2)(W-3) = 2W^2-30[/math] W = 4.5cm L = 9cm I think you haven't drawn the problem on a bit of paper. If you had you would notice that you have taken away some squares twice. Show me the workings of your maths please. You seem to have the right answer but how do you go from "[math](L-2)(W-3) = 2W^2-30[/math]" to the answer? . Link to comment Share on other sites More sharing options...
Chikis Posted October 14, 2014 Author Share Posted October 14, 2014 Here is the work. Since L= 2w, the area of the triangle would be LW. Thus [math]LW= 2W(W) = 2W^2[/math] Based on the condition of the problem, we have [math](L-2)(w-3)=2w^2-30[/math] If expanded, we have [math]2w^2-6w-2w+6=2w^2-30[/math] [math]\to[/math] -8w = -36 [math]\therefore[/math] W = -36/-8 = 4.5cm L = 2(4.5) = 9cm Link to comment Share on other sites More sharing options...
Robittybob1 Posted October 14, 2014 Share Posted October 14, 2014 Here is the work. Since L= 2w, the area of the triangle would be LW. Thus [math]LW= 2W(W) = 2W^2[/math] Based on the condition of the problem, we have [math](L-2)(w-3)=2w^2-30[/math] If expanded, we have [math]2w^2-6w-2w+6=2w^2-30[/math] [math]\to[/math] -8w = -36 [math]\therefore[/math] W = -36/-8 = 4.5cm L = 2(4.5) = 9cm Thanks Link to comment Share on other sites More sharing options...
studiot Posted October 14, 2014 Share Posted October 14, 2014 Of course having got values for L and W you can check to see if they correctly match the conditions of the problem. Link to comment Share on other sites More sharing options...
Chikis Posted October 15, 2014 Author Share Posted October 15, 2014 L = 9 W = 4.5 (9-2)(4.5-3) = 10.5 (4.5)9 = 40.5 40.5 - 10.5 = 30 1 Link to comment Share on other sites More sharing options...
studiot Posted October 15, 2014 Share Posted October 15, 2014 Doesn't that give you a nice warm feeling. +1 Link to comment Share on other sites More sharing options...
Chikis Posted October 15, 2014 Author Share Posted October 15, 2014 It does. Thank you. Link to comment Share on other sites More sharing options...
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