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Rectangular Conference
The suitability of a conference hall
I have been attending seminars and conferences in different types of hotels.
During one such seminar, I noticed the following in respect of the arrangement of the table and chairs for the members who had attended the seminar
The room was of rectangular type. A large sized rectangular table was in the middle of the room and on the one side of the table there were thirty chairs and on the other side there were thirty chairs. The chairman for the seminar has to sit on the far end of the table. Behind his seat a white board had been erected to display the statistical information through a projector.
I could witness that the room was very much spacious. There were proper lighting arrangements. Sufficient windows were available in the room so that uninterrupted air supply could move into the room comfortably in case of failure in airconditioning facilities
The hotel was of a moderate type located in the far end of the town.
During the course of the speech delivered by the Chairman, I could witness, the members sitting at the far end of the table were experiencing much trouble in bending their heads so that they could listen his speech clearly. They were very keen in listening to his speech which lasted for more than fortyfive minutes
The speech was over and the convenor announced a small break in the session for ten minutes. Six of the members sitting at the far end of the table started complaining neck pain, however, they were afraid of conveying the information to the convenor.
At this juncture, I offer my view points as follows:
Normally when this type of arrangement where a lengthy table is arranged for a seminar, members often find it difficult to sit in a comfortable position and listen to the speeches.
This type of arrangement is better when the seating capacity is not more than twenty – ten members sitting on both sides of the table. However, when more number of members are to be seated, it is better the hall can be arranged as follows:
The table can be of oval type so that the members can comfortably sit in their seats and listen to the speeches delivered by each member without any difficulty
Otherwise, the seating arrangement can be of the type of a class room – one behind the other. In this case as and when the member sitting on the back row starts talking, the members sitting in the front row has to turn back and listen to his speech.
These are the points to be noted by the organizers of the seminars and conferences in hotels and the seating arrangement should vary according to the capacity of the audience attending the meeting
About the Author
A. Gauri Sankar, a retired bank official from a nationalised bank in India. He is a trainer for banking subjects and he is also a trainer for soft skills. He has served in the bank for more than 39 years, He is a writer, orator and communicator. His qualifications are: M B A; BSc; CAIIB. He is a lead auditor for ISO 9001: 2000.
He can be contacted at: gausan51@gmail.com and gausanchennai@gmail.com
and his articles can be viewed at www.gaurisankars@blogspot.com
Finding the length and the width?
A large table for a conference room is to be constructed in the shape of a rectangle with two semicircles at the ends. The table is to have a perimeter of 40 ft, and the area of the rectangular portion is to be twice the sum of the areas of the two ends. Find the length and the width of the rectangular portion.
The required length is 10 ft, and the required width is 20/π ft.
If you analyze the appearance of the table, it will look like this: (top only)
http://welcome.com.au/shop/images/d%20end%20table.JPG
Hence, the width of the rectangle is not included in calculating the perimeter. The most effective method to solve this problem is to use a system of equations to represent the given information.
First equation:
2L + πW = 40
With the πW, remember that the perimeter of a circle is its circumference. (C = πD). However, we have a semicircle, and the formula for its circumference is (C = 1/2(πD)). Then again, since we have two semi-circles, and their circumferences are included in the perimeter, it's as if we have a circle, nevertheless. Their diameter is the width of the rectangle, so we have (1/2)πW + (1/2)πW = πW.
Second equation:
LW = πW² / 2
The problem states that the area of the rectangle is twice the area of the two semi-circles, or the circle. Remember that our diameter is the width of the rectangle, and since we need the radius to find the area, we need half of the width. Hence, we have 2π(W/2)² = πW² / 2.
Eliminate one of the variables to solve for the other.
2L + πW = 40
LW - πW² / 2 = 0
2L + πW = 40
2LW - πW² = 0
W(2L + πW = 40)
2LW - πW² = 0
2LW + πW² = 40W
2LW - πW² = 0
4LW = 40W
L = 10 ft
Finding the width:
2L + πW = 40
2(10) + πW = 40
πW = 40 - 20
πW = 20
W = 20/π ft
Dimensions: l = 10 ft; w = 20/π ft
Hope this helps!
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