Area of Lawn

[Guest]

A square garden has a square lawn in the center. Outside of this is a gravel walkway 4 ft. wide, then a border of flowers 6 ft. wide. The border and lawn combined has an area of 721 sq. ft. Determine area of lawn.

Area = length x width.

A square has equal sides.

Area of square = s^2

How do I proceed ?

[Guest]

Please respond. Thanks.

[Guest]

Outside of this is a gravel walkway 4 ft. wide

Do you mean "outside the lawn in the centre" that is around the outside edge of the lawn is a gravel walkway 4 feet wide and a 6 feet wide flower border around the gravel walkway. ????

[Guest]

Yes, that is the way I understand it.

[Guest]

Well then,

Let "s" equal the side of the lawn

You can now find the areas of all the bits in terms of "s" and the numbers given

Work out the area of the flower border and add on the area of the lawn, all in terms of "s" and put this equal to 721 which the question says is the area of the lawn plus the flower bed. Now solve for "s"...... it will be a quadratic equation, so you will have to pick the "reasonable" root

[Guest]

Area of lawn = s x s = s^2

The walkway and flower border is 10 ft. wide, so each side is 20 ft. longer than the lawn.

s + 20 = side of garden.

(s + 20 )^2 - s^2 = 721

(s + 20) (s + 20) = s^2 + 20s + 20s + 20 x 20 = s^2 + 40s + 400 - s^2 = 721

40s = 321

40s / 40 = 321 / 40 = 8.025 ft. = side of lawn

I am not sure.

[Guest]

s + 20 = side of garden.

(s + 20 )^2 - s^2 = 721

You have taken the area of the garden minus the area of the lawn and put it equal to 721. The question does not say that...??


This is what it says in the question..??
The border and lawn combined has an area of 721 sq. ft

[Guest]

ax^2 + bx + c = 0 (quadratic form)

a = 2 , b = 40 , c = -321

x = - b + - sq. rt. b^2 - 4ac / 2a (quadratic formula)

x = -(40) + - sq. rt. 40^2 - 4(2) (-321) / 2(2)

x = -(40) + - sq. rt. 1600 + 2568 / 4

x = -(40) + - sq. rt. 4168 / 4

x = -(40) + - 64.56 / 4

x = - (40) + - 16.14

x = -40 - 16.14 = -56.14

x = - 40 + 16.14 = -22.86

Where am I correct ?

[Guest]

I assume the last post is some different problem...?????

What you need to do for the original lawn question is as follows......

Let "s" equal the side of the lawn

You can now find the areas of all the bits in terms of "s" and the numbers given

Work out the area of the flower border and add on the area of the lawn, all in terms of "s" and put this equal to 721 which the question says is the area of the lawn plus the flower bed. Now solve for "s"...... it will be a quadratic equation, so you will have to pick the "reasonable" root

[Guest]

I used information from the original lawn problem for the quadratic:

x^2 + x + 20x + 20 x + 20 X 20 = 2x^2 + 40x + 400 = 721 (I used x for the sides)

2x^2 + 40x + 400 - 721 = 0

2x^2 + 40x - 321 = 0

Then I used the steps in last post.

[Guest]

But I told you this was the wrong quadratic.... as copied here again
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
s + 20 = side of garden.

(s + 20 )^2 - s^2 = 721

You have taken the area of the garden minus the area of the lawn and put it equal to 721. The question does not say that...??


This is what it says in the question..??
The border and lawn combined has an area of 721 sq. ft

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

as posted in an earlier post, you need to do this.....

What you need to do for the original lawn question is as follows......

Let "s" equal the side of the lawn

You can now find the areas of all the bits in terms of "s" and the numbers given

Work out the area of the flower border and add on the area of the lawn, all in terms of "s" and put this equal to 721 which the question says is the area of the lawn plus the flower bed. Now solve for "s"...... it will be a quadratic equation, so you will have to pick the "reasonable" root

[Guest]

I don't know how.

[Guest]

You have most of it worked out already , You need to put it into logical steps so that you can see how you are progressing through the problem.


What you need to do for the original lawn question is as follows......

Let "s" equal the side of the lawn ........So.....If "s" is the length of the side of the lawn, then the lawn area is "s^2" which you have already got.


Work out the area of the flower border and add on the area of the lawn, all in terms of "s" and put this equal to 721 which the question says is the area of the lawn plus the flower bed. Now solve for "s"...... it will be a quadratic equation, so you will have to pick the "reasonable" root

You can now find the areas of all the bits in terms of "s" and the numbers given.
You are given a lot more information in the question so make use of it.
The flower border is 6 ft wide and the gravel walkway is 4 ft wide.
So.... Work out the area of the flower border ....this will be a 6 ft wide strip around the outside of the gravel path.
The overall size of the garden ..... s + 20 = side of garden....which you had already got....and the overall size of the gravel is 8 + s. From this you can cut the flower border into pieces and work out its area......in terms of "s"
Then the question say add on the area of the lawn and put this equal to 721.......and then all you have to do is solve the equation

[Guest]

I don't know how.

[Guest]

You can now find the areas of all the bits in terms of "s" and the numbers given.
You are given a lot more information in the question so make use of it.
The flower border is 6 ft wide and the gravel walkway is 4 ft wide.
So.... Work out the area of the flower border ....this will be a 6 ft wide strip around the outside of the gravel path.
The overall size of the garden ..... s + 20 = side of garden....which you had already got....and the overall size of the gravel is 8 + s. From this you can cut the flower border into pieces and work out its area......in terms of "s"

If you draw a diagram of the garden and lawn you will see how it all fits together.....but I cannot draw here. You will see how the flowerbed can be cut into rectangle strips 6 ft wide and added together....OR....easier to describe here is.....

To get the area of the flower border....take the overall area (s + 20)^2 and subtract from it the overall area of the gravel and the lawn (8 + s)^2. That will leave the area of the flower border in terms of "s".

Then the question says.....the area of the flower border plus the lawn area "s^2" is equal to 721......and that is your equation to solve.

[Guest]

(s + 20)^2

(s + 20) (s + 20) = s^2 + s + 20 + 20s + 400

(8 + s)^2

(8 + s) (8 + s)

64 + 8s + 8s +2s

s^2 + 21S + 420 + 64 + 8s + 8s + 2s

s^2 + 395 + 484 = 721

s^2 + 395 + 484 - 721 = 0

s^2 + 395 - 237 = 0

x = -(1) + - sq. rt. 39^2 - 4(1)(-237) / 2

x = -(1) + - sq. rt. 1521 + 948 / 2

x = -(1) + - sq. rt. 2469 / 2

x = -(1) + - 49.68 / 2

x = -(1) + - 24.84 = 23.84

s = 23.84

I don't know.

x = -(1) - 24.84 = -25.84

[Guest]

s + 20)^2 .........This is area of the overall garden

(s + 20) (s + 20) = s^2 + s + 20 + 20s + 400 ....error in multiplying out brackets

(8 + s)^2 ......this is area of the gravelpath plus the lawn within it.

(8 + s) (8 + s)

64 + 8s + 8s +2s .....error in this as well...multiplying out brackets

s^2 + 21S + 420 + 64 + 8s + 8s + 2s....this is wrong because of above errors
But you have added the areas....remember (s+ 20)^2 is area of whole garden including the (8 + s)^2 so why would you want to add it on.


What you need to do.............
Sort out the problems you have in understanding the question. Read over the descriptive posts that tell you exactly what to do. Do each step and understand it before going on to the next. That way you can progress steadily through it. Your first step is to multiply out the two squared expressions and get expressions for the the (whole garden) and the (gravel + lawn). Then subtract these to get an expression for the flower border. Forget about solving the quadratic equations until you get these expressions sorted first.



The rest below is wrong for various reasons as above.
s^2 + 395 + 484 = 721

s^2 + 395 + 484 - 721 = 0

s^2 + 395 - 237 = 0

x = -(1) + - sq. rt. 39^2 - 4(1)(-237) / 2

x = -(1) + - sq. rt. 1521 + 948 / 2

x = -(1) + - sq. rt. 2469 / 2

x = -(1) + - 49.68 / 2

x = -(1) + - 24.84 = 23.84

s = 23.84

I don't know.

x = -(1) - 24.84 = -25.84
0

[Guest]

(s + 20) ^2

(s + 20) (s + 20) = s^2 + 20s + 20s + 400

--------------------------------------------------
(8 + s) ^2

(8 + s) (8 + s)

64 + 8s + 8s + s^2

s^2 + 20s + 20s + 400 - 64 + 8s + 8s + s^2

56s + 336 = 721

56s + 336 - 721

56s - 385

56s /56s = -385/56

s = - 6.875

[Guest]

(s + 20) ^2

(s + 20) (s + 20) = s^2 + 20s + 20s + 400 .....yes, correct expr. for o/all area

but you could have written it as..........s^2 + 40s + 400 ........OK

--------------------------------------------------
(8 + s) ^2

(8 + s) (8 + s)

64 + 8s + 8s + s^2.....yes, correct expr. for area of gravel+lawn
but could have been written as....... s^2 + 16s + 64 .......OK

s^2 + 20s + 20s + 400 - 64 + 8s + 8s + s^2 ....you have only subtracted the 64 and added everything else on..?????

Put brackets around each if you cannot see what you are doing
(s^2 + 20s + 20s + 400) - (64 + 8s + 8s + s^2)

This is (overall area) - (gravel + lawn) ....the brackets show what is what.

The rest of this is wrong due to the errors above

56s + 336 = 721

56s + 336 - 721

56s - 385

56s /56s = -385/56

s = - 6.875

[Guest]

(s^2 + 20s + 20s + 400) - (64 + 8s + 8s + s^2)

40s + 400 - 64 - 16s

24s + 336 = 721

24s + 336 - 721

24s = -385

24s / 24 = -385 / 24

s = -385/24 = -16.04