The area of a garden, given perimeter is 50 feet, width is 7 feet

[eddy2017]

Hi, dear friends, I am having trouble solving this problem. i need your help

A rectangular garden has a perimeter of 50 feet. The width of the garden is 7 feet. What is the area of the garden?.

This is what I have been able to do.

the formula for the area of a rectangle is A = l *w
You are given that the width of the garden is 7 feet...
W=7
but I do not know the length of the garden.I will need to find the length of the garden before I can find its area.

perimeter of this rectangle>50 feet
if the width of one side is 7 the other side is also 7=14
The length of the garden is twice the width.
L= 7+7 =14
Now that you know length and width
I go ahead and find the area of the garden using the formula A = l.w

A= l x w
A=14 x 7
A= 98

from here on out, i don't know what else to do.
I have been told the answer is wrong.
Thank you so much in advance.

 

[Steven G]

Hi, dear friends, I am having trouble solving this problem. i need your help

A rectangular garden has a perimeter of 50 feet. The width of the garden is 7 feet. What is the area of the garden?.

This is what I have been able to do.

the formula for the area of a rectangle is A = l *w
You are given that the width of the garden is 7 feet...
W=7
but I do not know the length of the garden.I will need to find the length of the garden before I can find its area.

perimeter of this rectangle>50 feet
if the width of one side is 7 the other side is also 7=14
The length of the garden is twice the width.
L= 7+7 =14
Now that you know length and width
I go ahead and find the area of the garden using the formula A = l.w

A= l x w
A=14 x 7
A= 98

from here on out, i don't know what else to do.
I have been told the answer is wrong.
Thank you so much in advance.

The problem you are having is when you say that the length is twice the width.

In the rectangle you have two sides (opposite one another) that are called the width. This uses 7ft + 7ft = 14ft.

Now there are two other sides that make up the length.

How many feet are left after using 14ft to be used for the two sides which we call length? How long is the length? What is the area?

 

[eddy2017]

The problem you are having is when you say that the length is twice the width.

In the rectangle you have two sides (opposite one another) that are called the width. This uses 7ft + 7ft = 14ft.

Now there are two other sides that make up the length.

How many feet are left after using 14ft to be used for the two sides which we call length? How long is the length? What is the area?

jomo, thank you very much for helping me out.
let me go this route and you tell me what you think
the perimeter is 50 feet. if the width is 14 feet, then i will have to find a number that when added to 15 gives me 50. right?.
I found that 18 + 18 = 36 and 36 + 14 = 50
so, the length of the two sides amount to 36.

the problem is asking about the area of the rectangle
so i have the length and the width and if the formula for area is
A = L X W

should i plug in these values?
i'm stuck here.
thank you for your help

ok, i think i am able now to go a little further

i can find the length like this:
P=2(l+w)
50= 2(l+7)
50= 2l + 14
distribute to make it easier to solve
2l+14 = 50
sub 14 from both sides
2l+ 14 -14 = 50 -14
2l = 36
divide 2 by both sides to isolate the l
2l = 36
2 2
l=18 so the entire length of the rectangle is 36
so we have
w= 7
l= 18

now i have the length and the width and the perimeter, but i dont know what operation i should perform to get 126 f² which they have told me is the correct answer.

a friend offer me this path

A=P w/2- w^2= 50 * 7/2 -7^2
I have found that i get the result 126 following it, but why so complicated. i asked him why w^2? and he told me it was the omega constant.
why is it used here?
please, if anybody can help deciper for me the above equation is going to do me a great favor.
thank you

 

[Steven G]

jomo, thank you very much for helping me out.
let me go this route and you tell me what you think are getting there but still a bit confused.

The width is 7 NOT 14. The length is 18 NOT 36.

There are two sides that you can call length and 2 sides that you can call width. If you and I are both 30 years old, you can NOT say we are each 60 years old. Do NOT add the two lengths together. Do not add the two width together.

Now can you find the area?

 

[eddy2017]

reply

thank you, I was adding the two sides. oh my goodness!. yes, I can find it now. i did this
P=2(l+w)
50= 2(l+7)
50= 2l + 14
distribute
2l+14 = 50
sub 14 from both sides
2l+ 14 -14 = 50 -14
2l = 36
divide 2 by both sides
2l = 36
2 2
l=18 so the length of the rectangle is 18
so we have
w= 7
l=18

A = w. l
A= 18 x 7= 126 f²

Now,could clear up for me this other oath i was given but i did not understand?

a friend gave me this path which yielded the same result but i dont understand the steps he took.

this is the formula he used. for me very complicated. Why did he use this one?, and he kept telling me that w^2 was the constant omega. when do I use the constant omega?, why did he use it here, when getting the area was so easy?.
i know you will be able to decipher this for me, mojo. if you can't, it is ok, big help you have given me. thank you

A=P w/2- w²

=50 x 7/2 - 7²

A= pw/2 - W²

A=50X7 - 7²
2

A= 350 - 49
2
= 175 - 49
=126

 

[Steven G]

thank you, I was adding the two sides. oh my goodness!. yes, I can find it now. i did this
P=2(l+w)
50= 2(l+7)
50= 2l + 14
distribute
2l+14 = 50
sub 14 from both sides
2l+ 14 -14 = 50 -14
2l = 36
divide 2 by both sides
2l = 36
2 2
l=18 so the length of the rectangle is 18
so we have
w= 7
l=18

A = w. l
A= 18 x 7= 126 f²

Now,could clear up for me this other oath i was given but i did not understand?

a friend gave me this path which yielded the same result but i dont understand the steps he took.

this is the formula he used. for me very complicated. Why did he use this one?, and he kept telling me that w^2 was the constant omega. when do I use the constant omega?, why did he use it here, when getting the area was so easy?.
i know you will be able to decipher this for me, mojo. if you can't, it is ok, big help you have given me. thank you

A=P w/2- w²

=50 x 7/2 - 7²

A= pw/2 - W²

A=50X7 - 7²
2

A= 350 - 49
2
= 175 - 49
=126

A=P w/2- w² = 2(l+w)w/2 -w² = lw + w² - w² = lw = A
I would not use that formula.

I would not even use the formula P = 2(l+w). Personally I would use the formula P/2 = l+w.

In your case P=50, so P/2 = 25. Then 25 = 7 + l, so l=18 and A=lw=7*18=126

 

[Dr.Peterson]

this is the formula he used. for me very complicated. Why did he use this one?, and he kept telling me that w^2 was the constant omega. when do I use the constant omega?, why did he use it here, when getting the area was so easy?

A=P w/2- w²

=50 x 7/2 - 7²

A= pw/2 - W²

A=50X7 - 7²
2

A= 350 - 49
2
= 175 - 49
=126

As far as I can tell, "w" here is just the width. The Greek letter omega looks like w, but I see no reason to call it that here.

It appears that he has memorized a formula for solving this particular kind of problem, which is the worst way to learn math. The formula can be derived by solving for L:

2W + 2L = P
2L = P - 2W
L = P/2 - W

Then put this into the formula for area:

A = LW = (P/2 - W)W = PW/2 - W^2

I agree, it's much easier to just do the work you did (eventually!) than to try to memorize and use such an unpleasant formula.

 

[JeffM]

I go beyond Dr. Peterson.

Memorizing lots of formulas is not a way to learn math at all. Math is a form of reasoning. Memorization is not reasoning.

Moreover, if you go down the memorization route, you need to remember not only the formula but when the formula applies. All that memorization gradually fades away so that when you actually need to use math, you have nothing to fall back on. Even if you manage to retain in memory thousands of formulas and when each applies, you will not be able to apply math outside of those formulas. Admittedly, it saves time to memorize a few basic formulas, and you will have to memorize some additional formulas temporarily to pass a closed-book test. But math beyond basic arithmetic is primarily about logical thinking rather than memorized formulas. I have a book on my desk that is over 500 pages long filled with tables and formulas: I don't memorize; I look up after I have thought about what I need.

 

[eddy2017]

I go beyond Dr. Peterson.

Memorizing lots of formulas is not a way to learn math at all. Math is a form of reasoning. Memorization is not reasoning.

Moreover, if you go down the memorization route, you need to remember not only the formula but when the formula applies. All that memorization gradually fades away so that when you actually need to use math, you have nothing to fall back on. Even if you manage to retain in memory thousands of formulas and when each applies, you will not be able to apply math outside of those formulas. Admittedly, it saves time to memorize a few basic formulas, and you will have to memorize some additional formulas temporarily to pass a closed-book test. But math beyond basic arithmetic is primarily about logical thinking rather than memorized formulas. I have a book on my desk that is over 500 pages long filled with tables and formulas: I don't memorize; I look up after I have thought about what I need.

A big thnak you to all of you who contributed. I can't appreciate it enough. And thank you for guiding me theough the steps by asking questions to prod me along. Excellent way to learn. Dennis, thank you for the link. Wonderful!