Dimensions- Table

[Guest]

A conference room table is constructed, rectangular in shape, with semicircles at the ends.

Perimeter- 40 feet

Area of rectangle- Twice the sum of areas of ends

Calculate length and width of rectangular part.

[Guest]

Assume you mean area of rect bit = twice sum of the ends
You need to attempt this same as the ones posted before
Let x and y equal the unknowns
Set up equations using the info given
This time you will have twounknowns and two equation....so simultaneous equations
Rearrange one of the equations to isolate one of the unknowns
Substitute it into the other or any otherprocess of elimination to solveone of the unknowns then substitute it in to the other equation to get the other unknown
The answer I get is the rectangular bit of the table is 10 x 6.37

[Guest]

Perimeter- 2L + 2W

Area- Pi * r^2 / 2

P = 2L + 2W = 40
L + W = 20
20 - L = W

Area- 3.14 * r^2 / 2

I don't know how to proceed from here.

[Guest]

All you have done is list a few generic formulae
Not related to the question in hand
I assume perimeter in the question means perimeter of the table
You give 2L + 2W that is perimeter of a rectangle and the question says the table is not rectangular
Also you introduced r for radius that now means 3 unknowns
I think you are messing about and not giving it any real thought based on what was done for previous posted questions. You need to show how your equations are related and how they relate to the actual question

[Guest]

I am not "messing about" using your words. I don't know how to solve the problem.

Perimeter- 40 feet

Area of semicircle- pi * r^2 / 2 (ends are semicircles)

Area of rectangle- Length x Width

Area of rectangle - twice sum of areas of ends

Let L = Length of rectangle
Let W = Width of rectangle

I don't know how to set up the equations.

[Guest]

OK you are not messing about.....but are you giving the problem any "real thought"

Compare the table problem to the earlier one posted about selling fruit.....

It is using algebra to provide a solution.
and it is using simultaneous equations in particular.
In the Fruit question we are told 2 things about selling fruit on week1 and week2.
1......how many pieces were sold in week1.....108
2......how many times more of each fruit was sold in week2 also the total of fruit sold in week2......5 time apples and 3 times oranges and total of 452 fruit...apples+oranges.

So if we.... Let x=apples and y=oranges sold in week1

x + y = 108 .....this is what we are told for week1 ....eqn1

5x + 3y = 452 .....this is what we are told for week2 ....eqn2

So follow the same type of procedure for the table problem and think about what you are being told and what you have to find out...

....In the table problem we are told several things....
1....its construction..Rectangular shape with semicircle ends.
2....Its perimeter is 40 feet...that is the perimeter of the whole table with the curved ends...not the perimeter of the rectangular bit in the middle between the 2 curved ends. ...so the perimeter of the whole table will not be 2L + 2W.
3....We are told the area of the rectangular bit in the middle is equal to twice the area of the curved ends.
4....We are asked to find the dimensions of the rectangular bit in the middle. The question calls it the length and width of the rectangular part...but we dont know at this stage is it a narrow rectangle with the length across the width of the table or is it a large rectangle with the width across the width of the table.....it can be either way...
5...These are the unknown quantities you have to find. but if it is a small rectangle bit the length of it may be the width of the whole table or it may be the other way round. So as a start lets call one dimension "x" and the other "y" and let "Y" be the dimension across the width of the whole table.

From the information given you should be able to setup equations as was done in the fruit problem...follow what is done above...
There should be 2 simultaneous equations and we have to find a value for x and y that satisfies both.

[Guest]

Let x = length
Let y = width

x * y = area of rectangle


xy = (2) pi * r^2 / 2

The problem states the area of the rectangle is equal to twice the sums of the ends (semicircles). I don't know what other information to use.

I don't know what to use for the other equation.

I don't understand how the two problems are related.

[Guest]

OK from point 5 in my last post I suggested we let "x" and "y" be the dimensions of the rectangular bit in the middle and also that "y" be the dimension that corresponds to the width of the whole table....so "x" then is along length ways on the "whole table.
And the table has semicircle ends so see if you can answer the following questions....in terms of "x" and "y", don't worry about actual figures.
1...What is the table straight side distance between the 2 semicircular ends?.
2...What is the width of the "whole" table?.
3...What is the diameter of the semicircular ends?.
4...What is the radius of the semicircular ends?.
5...What is the overall length of the "whole table?.
6...What is the area of the rectangular bit in the middle?.
7...What is the area of each semicircular end?.
8...What is the length of the curved edge of one of the semicircular ends?.
9...What is the distance around the edge of the whole table?

Now after answering the above questions see if you can put equations together to represent the statements 2 and 3 in my last post.
These equations should in terms of "x" and "y" and can be solved to get values for "x" and "y"

[Guest]

Questions 1- 8:
No idea

Question 9:
Perimeter- 40 feet

[Guest]

I wanted you to answer in terms of "x" and "y" not with real figures. The question told you the perimeter was 40 feet so that did not take much thinking to answer that, but I wanted the answer in terms of "x" and "y" and then you could show it equalled 40 for point 9.
Just as we did in the fruit question......
x + y = 108 .....this is what we are told for week1 ....eqn1

This is the type of equation you need to do for the table question but using the information given in the table question.

1 to 9 are questions about the individual pieces of information either given in the question or can be derived simply from the fact that we let x and y be the dimensions of the rectangular bit in the middle and the fact the ends of the table are semicircular.
Example .... if the side of a triangle is "x" inches. What is its perimeter.
You could say Perimeter is the distance around the outside OR the sum of the 3 sides.....
So it would be x + x + x = P OR 3x = P
Example2....A object is formed by a square with an equal sided triangle (equilateral triangle) attached to one side. (makes the shape of an open envelope). The perimeter of the object is 20 inches. Find the length of a side of the square.
Answer.......
We know a square has 4 equal sides and an equilateral triangle has 3 equal sides. The triangle is attached to the side of the square so the side of the triangle must be the same length as the side of the square. Both are attached so they share a common side. So the perimeter will be 3 sides of the square plus 2 sides of the triangle. That gives 5 sides and we are told they add up to 20 inches. This is how we analyse the information given in the question and establish what are the unknowns.
From an algebra point of view we can say.......
All the sides are the same length and there are 5 of them....Let s = the length of a side.
Then.... 5 times s is the perimeter and we are told it is 20 inches.....
So......5 x s = 20
gives us s = 20/5
So...... s = 4
So the length of one side is 4 inches......

This is the type of reasoning you need to do for the table question ans compare the method with the Fruit question.
If you can try an answer questions 1 to 9.....these are simple question on different bits of the table and do not involve equations...Just seeing if you understand what you are being told in the question.

[Guest]

Question 6:

Area of rectangle - x * y

Question 7:

pi * r^2 / 2 (2) = xy


"these are simple question on different bits of the table and do not involve equations...Just seeing if you understand what you are being told in the question."

Not simple to me.

[Guest]

Ok
Have you agreed that the dimensions of the rectangular bit in the middle is "x" by "y"......we said that we would let x and y be the dimensions of the bit in the middle.......so it must be.....do you agree....
If yes then answer 1 to 9 in terms of x and y.....

Answer from question 1 down......

1.....What is the distance (length) of the straight sides between the 2 semicircular ends, bearing in mind that the rectangular bit is in the middle between the 2 semicircular ends
2.....and from that what is the width of the whole table...

You need to answer these before you can go any further.....otherwise you don't understand what the question is telling you

[Guest]

I agree on x and y.

I know what each question is asking. I don't know how to solve them.

[Guest]

The questions in my last post do not need any solving. They just need answered....stating facts you are already told in the question or have said we would let equal x or y. If you cannot say what x and y are then you must not understand the statement "Let "x" and "y" equal the dimensions of the rectangular bit in the middle". We are told in the original question that the bit in the middle is rectangular and the ends are semicircular.

A similar example problem would be.....

I have a round ball that is painted black....
Question 1....What is the colour of the ball?.
Question 2....What is the shape of the ball?.
Question 3....Who has the ball?.
See if you can answer these 3 questions and then follow the same logic to answer the questions in my last post.

[Guest]

Question 1- Let B= Black

Question 2- Let R = Round

Question 3- Let U = You

If that is not correct I don't know.


Let L= Length of table

Let W = Width of table


I don't know.

[Guest]

You are not understanding the question as it is written....
Question 1....What is the colour of the ball?.
The question simply asked what is the colour of the ball and the question had actually told you it was painted black. That means it was painted with paint that is coloured black so the the answer to Q1 is Black. It did not ask you to let some variable B = Black. The answer required was simply it colour....Black.
Same with Q2...answer is Round......and the question told you the ball was round so Round must be the answer.
Same with Q3....answer is "I have" the question told you so.

Now referring to the table problem.....
We said we would let "x" and "y" be the dimensions of the rectangular bit in the middle....The question told us there was a rectangular bit in the middle. A rectangle has 2 long sides and 2 short sides. We don't know at this stage if the long sides of the rectangle are across the width of the table or along the length of the table. So we cannot say if "x" is longer than "y" or if "y" is longer than "x". But we said we would let the "x" dimension be along the straight sides of the table and then the "y" dimension would be across the width of the table. But the "y" dimension may be the length of the rectangular bit and the "x" dimension may the the width of the rectangular bit....we don't know at this stage. So you cannot say which is the length until we find a value for "x" and for "y"....

Based on this explanation .....Can you now answer the Questions 1 to 9.....in terms of "x" and "y"....

1...What is the table straight side distance between the 2 semicircular ends?.
2...What is the width of the "whole" table?.
3...What is the diameter of the semicircular ends?.
4...What is the radius of the semicircular ends?.
5...What is the overall length of the "whole table?.
6...What is the area of the rectangular bit in the middle?.
7...What is the area of each semicircular end?.
8...What is the length of the curved edge of one of the semicircular ends?.
9...What is the distance around the edge of the whole table?

[Guest]

I can't answer them. I have wasted too much of your time. I posted the question but I still can't solve.

I have been out of school over 30 years and am trying to re-learn math/algebra on my own. I am not making much progress.

[Guest]

I have answered below each question copied from last post.....

Answers the Questions 1 to 9.....in terms of "x" and "y"....

1...What is the table straight side distance between the 2 semicircular ends?.
The straight sides of the table between the semicircular ends will be "x" because we said we would let "x" be the dimension along the length of the overall table.

2...What is the width of the "whole" table?.
And that means that "y" will be the dimension across the width of the table. We dont know wich is the bigger "x" or "y".

3...What is the diameter of the semicircular ends?.
The semicircular ends will be the full width of the whole table so the diameter will also by "y"

4...What is the radius of the semicircular ends?.
And the radius will be the half of the diameter....y/2

5...What is the overall length of the "whole table?.
The overall length will be the length of the straight sides plus the radius if the curved ends (2 off)
so will be x + 2(y/2) or simply (x + y)

6...What is the area of the rectangular bit in the middle?.
The area of the rectangular bit in the middle will be x multiplied by y OR xy.

7...What is the area of each semicircular end?.
The area of the semicircle at each end will be Pi times (y/2)^2 divided by 2 OR Pi.(y^2)/8

8...What is the length of the curved edge of one of the semicircular ends?.
The length of the curved edge of one semicircular end is Pi.y/2

9...What is the distance around the edge of the whole table?
The distance round the whole table is two straight sides plus two curved ends
and two curved ends makes same as one circle, so whole perimeter is 2x + (Pi.y)

This should give you a better understanding of the problem details and let you build the equations based on the facts given in the question. When you get the equations you can then sole for values for "x" and "y"

[Guest]

I still can't solve the problem. I don't know what information to use for the equations.

[Guest]

You should see by now that we have broken the "big" problem down into a collection of "little" problems.
And provided answers to specific bits of the information given in the question. The question attempts to describe the problem in "english" telling us various facts about the objects and items relating to the problem. It also describes in "english" what unknown values we are to find.

When we try to solve this problem mathematically instead of writing "english" sentences to describe it we have to write mathematical sentences or "equations or mathematical expressions" to describe it in terms of numbers and variables known or unknown. We state the facts given in the problem as a mathematical sentence.

Now to get back to the original problem....copied below...

====================
A conference room table is constructed, rectangular in shape, with semicircles at the ends.

Perimeter- 40 feet

Area of rectangle- Twice the sum of areas of ends

Calculate length and width of rectangular part.
=====================

What are the facts.......
1.....It is rectangular in shape with semicircular ends....so combination of simple shapes....
2.....Perimeter of the whole table is 40 feet......
3.....Area of rectangular bit = twice sum of the areas of the ends.....This fact is ambiguous....does it mean twice the sum of the area of one end or twice the sum of the area of both ends added together...???.We will take twice the sum of both ends added together....so that is two full circles of diameter "y"

We have worked out most of these facts individually before in the previous more detailed posts.....

Perimeter is two straight edges plus (Pi times diameter of curved ends) and this equals 40 feet....
So..... 2x + Pi.y = 40...... this is a mathematical equation describing the perimeter of the whole table....Eqn.1...

Area... rectangle bit = twice the are of both ends
So.........xy = 2.Pi.(y/2)^2 .....mathematical equation describing the facts relating the curved area to the rectangular area.

So........xy = 2.Pi.((y^2)/4).........(y/2 squared equals y squared over 4)
So........xy = Pi.y^2/2
So......divide across by y gives.....x = Pi.y/2........Eqn.2....

We now have 2 simultaneous equations and we need to solve for "x" and "y" that satisfies both......

So....multiply across by 2....2x = Pi.y......and notice we have a Pi.y is in eqn1 so substitute in eqn.1 to eliminate Pi.y...
Gives 2x + 2x = 40 OR 4x = 40 ..... so .... x = 10 feet..........solution.1......

Substitute in eqn.1 again to eliminate "x"......
So.....Pi.y + Pi.y = 40 OR 2.Pi.y = 40.....so y = 40/(2.PI) ...So y = 40/6.28 = 6.37 feet......solution.2......

So the dimensions of the rectangular bit is 10 feet by 6.37 feet.
Now we can say x = length of 10 feet and y = width of 6.37 feet.
you can now test if these dimension are correct by trying them out on the original statements in the problem