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# Geometry

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Senior Manager
Joined: 22 Feb 2004
Posts: 347

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02 Oct 2004, 23:06
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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Director
Joined: 20 Jul 2004
Posts: 592

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02 Oct 2004, 23:50
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.)
Director
Joined: 16 Jun 2004
Posts: 891

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02 Oct 2004, 23:59
4ft 9". Same approach.
Manager
Joined: 26 Sep 2004
Posts: 137

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03 Oct 2004, 00:03
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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Franky
http://franky4gmat.blogspot.com

Senior Manager
Joined: 22 Feb 2004
Posts: 347

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03 Oct 2004, 00:06
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..
Intern
Joined: 24 Sep 2003
Posts: 38
Location: India

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03 Oct 2004, 00:10
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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Vipin Gupta

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Joined: 26 Sep 2004
Posts: 137

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03 Oct 2004, 00:18
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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Franky
http://franky4gmat.blogspot.com

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Joined: 26 Sep 2004
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03 Oct 2004, 00:36
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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Franky
http://franky4gmat.blogspot.com

Director
Joined: 20 Jul 2004
Posts: 592

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03 Oct 2004, 02:29
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.)

AB + (AB - 6) = 120
AB = 63 inches = 5ft 3inches.
Manager
Joined: 26 Sep 2004
Posts: 137

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03 Oct 2004, 05:22
Don't be so brutal to yurself
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Franky
http://franky4gmat.blogspot.com

Senior Manager
Joined: 22 Feb 2004
Posts: 347

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03 Oct 2004, 09:44
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.
03 Oct 2004, 09:44
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