One vertex of the square is located at (3, 5) on the coordinate grid.

[eddy2017]

Hi, I am not understanding well how to analize this exercise

A square has a perimeter of 36 units.
One vertex of the square is located at (3, 5) on the coordinate grid.
What could be the x- and y-coordinates of another vertex of the square?

x=
y=

Thanks for any hint,
eddy
I have a suspicion they are giving me the perimeter of the square for something, but nothin' comes to mind

 

[lev888]

Hi, I am not understanding well how to analize this exercise

A square has a perimeter of 36 units.
One vertex of the square is located at (3, 5) on the coordinate grid.
What could be the x- and y-coordinates of another vertex of the square?

x=
y=

Thanks for any hint,
eddy
I have a suspicion they are giving me the perimeter of the square for something, but nothin' comes to mind

Draw a coordinate plane. Cut out a small paper square. Pin one of its vertices to (3, 5). Where can other vertices be? What measurement of the square do you need to know to describe those locations exactly?

 

[eddy2017]

On it

 

[pka]

A square has a perimeter of 36 units.
One vertex of the square is located at (3, 5) on the coordinate grid.
What could be the x- and y-coordinates of another vertex of the square?

[imath]\dfrac{36}{4}=~?[/imath] what does that have to do with this question?
Is [imath](12,14)[/imath] a nother vertex of the square? What not? or Why?
[imath][/imath][imath][/imath]

 

[eddy2017]

[imath]\dfrac{36}{4}=~?[/imath] what does that have to do with this question?
Is [imath](12,14)[/imath] a nother vertex of the square? What not? or Why?
[imath][/imath][imath][/imath]

63/4 =9 ..........................incorrect ..............................36/4 =9
P = S + S + S + S
P= 9 + 9 + 9 + 9
P=36

Each side of the square measures 9 units

 

[Steven G]

Go up 9, go down 9, go to the left 9, go to the right 9, go on a diagonal 9 units. Stop asking so much for help and do your problems more on your own.

 

[eddy2017]

Go up 9, go down 9, go to the left 9, go to the right 9, go on a diagonal 9 units. Stop asking so much for help and do your problems more on your own.

I'm at work Steve. I haven't even had time to take a look at it. I work for a living.

 

[eddy2017]

I'm at work Steve. I haven't even had time to take a look at it. I work for a living.

And with all due respect ? this free math help . The rules say that I don't understand something I should ask. Or is it not like that that?.

 

[Deleted member 4993]

Hi, I am not understanding well how to analize this exercise

A square has a perimeter of 36 units.
One vertex of the square is located at (3, 5) on the coordinate grid.
What could be the x- and y-coordinates of another vertex of the square?

x=
y=

Thanks for any hint,
eddy
I have a suspicion they are giving me the perimeter of the square for something, but nothin' comes to mind

Is that complete problem statement?

As stated, there could be infinite number of pairs of (x,y) coordinate.

 

[Harry_the_cat]

I'm at work Steve. I haven't even had time to take a look at it. I work for a living.

With all due respect Eddy, you really shouldn't post a question until you have taken the time to have a look at it and give it a go yourself. We are happy to answer your questions after you have done that.

 

[Deleted member 4993]

And with all due respect ? this free math help . The rules say that I don't understand something I should ask. Or is it not like that that?.

I have a suspicion they are giving me the perimeter of the square for something, but nothin' comes to mind

You had not shown ANY work for this problem - which provoked Steve's comment. I have suspicion that you post problems before you "think through".

 

[eddy2017]

You had not shown ANY work for this problem - which provoked Steve's comment. I have suspicion that you post problems before you "think through".

No, I was looking for a clue to initiate it. There has been many times in which I have posted part of the problem. Many times in which I have posted the whole solution
I think your assumption is less than fair.

 

[eddy2017]

This is my work.

 

[eddy2017]

I posted the problem and I said I needed a clue. Then made a comment about the perimeter that I was given. the negative comments are uncalled for. I thought you knew me better. And I am at work until 6 o'clock pm, get home around 7.
Once you even told I did not need to hurry, you all asked to take my time. Well I posted the problem because I knew I needed a clue once I reflected on it and saw no way to start.

With all due respect Eddy, you really shouldn't post a question until you have taken the time to have a look at it and give it a go yourself. We are happy to answer your questions after you have done that.

Harry, I did give it a look, a hard one
It is an easy problem once I have been given a clue, some of them are. Others are not. But if you see no work from me it is because even after rummaging thru the net I found nothing that helps.

Harry, I did give it a look, a hard one
It is an easy problem once I have been given a clue, some of them are. Others are not. But if you see no work from me it is because even after rummaging thru the net I found nothing that helps.

 

[Deleted member 4993]

I posted the problem and I said I needed a clue. Then made a comment about the perimeter that I was given. the negative comments are uncalled for. I thought you knew me better. And I am at work until 6 o'clock pm, get home around 7.
Once you even told I did not need to hurry, you all asked to take my time. Well I posted the problem because I knew I needed a clue once I reflected on it and saw no way to start.

The Picture that you had posted in response #13 has incorrect orientation. The correct orientation should be:



You should have noticed it and posted a correction.

 

[eddy2017]

Yes, you're right.

Yes, you're right.

That is the correct orientation.

Draw a coordinate plane. Cut out a small paper square. Pin one of its vertices to (3, 5). Where can other vertices be? What measurement of the square do you need to know to describe those locations exactly?

Lev, thank you for your tip. It led me to the solution.

Go up 9, go down 9, go to the left 9, go to the right 9, go on a diagonal 9 units. Stop asking so much for help and do your problems more on your own.

Thank you Steve for the tip.

 

[Harry_the_cat]

Harry, I did give it a look, a hard one
It is an easy problem once I have been given a clue, some of them are. Others are not. But if you see no work from me it is because even after rummaging thru the net I found nothing that helps.

But you said and I quote "I haven't even had time to take a look at it". See your post #7. I'm confused.

 

[eddy2017]

But you said and I quote "I haven't even had time to take a look at it". See your post #7. I'm confused.

I was in class. I took a look. Knew I had to ask for a clue.

 

[eddy2017]

Saw lev's and thought I had it. But needed time to print out the coordinate plane and do it. Just that. Then got Steve's lead that came with a very negative comment that hurt. Don't ask too much and do your exercises on your own. You all have given me hints and help but no one has never solved an exercise for me. Sometimes, as tutors, who are not
one on one with the students, nor on any other virtual platform for that matter just the forum, it would do a great service if the negative comments or things that could be considered rude were toned down a bit. There is no need for that notcwith students that do their best. As for me, any mathematical problem is a new discovery. Most of the exercises is a new challenge for me
and all I am asking is a clue. I have seen other post where posters get almost half the solution.

At the very least you should considered that there are a lot bad things that I could be doing instead of asking for help in a math forum. Mine is not help for homework is help for personal knowledge and development. There is a difference with schoolers here who post here looking for a quick math solution to a problem.

 

[Steven G]

I'm at work Steve. I haven't even had time to take a look at it. I work for a living.

If you haven't looked at the problem why did you post it? This is a help forum!