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Geometry [#permalink] New post 02 Oct 2004, 22:06
Attached. Please try to explain your answer.
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 [#permalink] New post 02 Oct 2004, 22:50
E. 4 ft 9 inches

The rectangle has been divided into four equal parallelograms -- 2 of them inverted.

Draw a perpendilar line to 120 inch line from the end of slant line at C. This line makes a 90-45-45 triangle at C. Since one side opposite to 45 degree is 6 inches, the other side will also be 6 inches.

AB + BC = 120 inches
AB + (AB - 6) = 120
AB = 57 inches = 4ft 9inches.

(let me know, if this is not clear; I will come up with a diagram.)
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 [#permalink] New post 02 Oct 2004, 22:59
4ft 9". Same approach.
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 [#permalink] New post 02 Oct 2004, 23:03
4 feet 9 inches.
Divide the rectangle into two and consider the symmetry. Sorry can't attach a figure as i am on linux, but the explanation given above is a good enough one.
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 [#permalink] New post 02 Oct 2004, 23:06
Answer given is C..thanks for your effort but I could not understand your approach .
AB + BC =120 CM..HOW?
hOW IS BC = AB -6
Also, shouldnt we draw perpendicular from end of slant line which is starting from A..
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 [#permalink] New post 02 Oct 2004, 23:10
C. 5 ft 3 in

First draw a perpendicular line on AB and on BC. Suppose the perpendicular line inersect on AB at point D and perpendicular line inersect on BC at point E. As all the parts are identical, So DB=BC.
Now BC = (240-12)/4 = 57 inches.
Now AB = AD+DB = 6+57= 63 inches = 5 ft 3 inches.
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 [#permalink] New post 02 Oct 2004, 23:18
Yep one of the slly mistakes... BC is 57 inches but AB is 63 inches.
Remember the lenght of rectangle is 20 feet which implies 240 inches.
Half of that is 120.
Consider half the rectangle, BC + the small segment obtained by dropping a perpendicular from the point of intersection of slant line in this half with the top line..to the bottom line + the rest of the right side line = 120 inches
since x = 45 this middle segment is 6 inches.
By symmetry the first abd last segment r equal thus BC = 57 inches . From C to the end of the right side bottom vertex of the rectangle the distance is 63 inches...
This is the same as asked for by symmetry again.
I know its not so clear but that's all i could help with...without a figure.
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 [#permalink] New post 02 Oct 2004, 23:36
Ok tried something :)....nto elegant at all though :(

A E = 20 foot = 240 inches
BE = 120 inches
BC + CD + DE = 120
CD = 6 as x = 45 degrees and given that the two sides of this right angles triangle must be same.
Look at the figure and see the symmetry
BC = DE . Hence BC + DE = 120-6 = 114
BC = DE = 57 inches
Now AB is corrsponding to CE as per symmetry , the mistake we committed in haste was to correspond AB with BC.
CE = DE + CD = 63 inches = 5 foot 3 inches.
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 [#permalink] New post 03 Oct 2004, 01:29
hardworker_indian wrote:
E. 4 ft 9 inches

The rectangle has been divided into four equal parallelograms -- 2 of them inverted.

Draw a perpendilar line to 120 inch line from the end of slant line at C. This line makes a 90-45-45 triangle at C. Since one side opposite to 45 degree is 6 inches, the other side will also be 6 inches.

AB + BC = 120 inches
AB + (AB - 6) = 120
AB = 57 inches = 4ft 9inches.


(let me know, if this is not clear; I will come up with a diagram.)


Wow, instead of adding 6 on the other side, I had subtracted. :beat
AB + (AB - 6) = 120
AB = 63 inches = 5ft 3inches.
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 [#permalink] New post 03 Oct 2004, 04:22
Don't be so brutal to yurself
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 [#permalink] New post 03 Oct 2004, 08:44
Franky..thanks a lot for your effort. But I must say that this question remains difficult for me..i mean under time pressure, i cannot be sure of my symmetric sense..and this question requires a bit of it. :(
  [#permalink] 03 Oct 2004, 08:44
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