# Find area of lawn if whole park is 5m longer than wide.

#### sandyk109

##### New member
A small city park consists of a rectangular laawn surrounded on all sides by a 330 m2 border of flowrs 2.5m wide. Find the area of the lawn if the entire park is 5 m longer than it is wide.

x(x +5) -(x -10)(x - 15) = x +330

xsquared + 5x - x squared + 15x + 10x + 150 + x + 330
5x + 15x + 10x + 150 = x+ 30
30X + 150 = x + 330 - 150
30x = x + 280

after this I have absolutely no idea what to do.

Please show work and explain. Thanks

#### Subhotosh Khan

##### Super Moderator
Staff member
Re: Beginning Algebra

sandyk109 said:
A small city park consists of a rectangular laawn surrounded on all sides by a 330 m2 border of flowrs 2.5m wide. Find the area of the lawn if the entire park is 5 m longer than it is wide.

Deleted the solution below - read the problem wrong

x(x +5) -(x -10)(x - 15) = x +330

xsquared + 5x - x squared + 15x + 10x + 150 + x + 330
5x + 15x + 10x + 150 = x+ 30
30X + 150 = x + 330 - 150
30x = x + 280

after this I have absolutely no idea what to do.

Please show work and explain. Thanks

#### scoresofsteel

##### New member
Hi Sandyk109!

This is how I would solve:

Code:
A ____________B
|    _____    |
|  a|     |b  |
|  d|_____|c  |
|_____________|
D             C
ABCD is the Park boundary; abcd is the Lawn boundary.

It has been given that "entire park is 5m longer than it is wide".

The width of the park is BC (in the drawing); let's name that value as 'X'

Then, due to statement above, DC = X + 5

Since width of the flower border on either side is 2.5, then:

. . .dc = DC - 2.5 - 2.5 = DC - 5 = (X + 5) - 5 = X

Similarily:

. . .bc = BC - 2.5 - 2.5 = X - 5

We have been given that area of flower path is 330m, which means that ABCD - abcd = 330

This can also be written as:

. . .X(X + 5) - X(X - 5) = 330

Take the common factor of X out front, and simplify:

. . .X((X + 5) - (X - 5)) = 330

. . .X((X + 5 - (X - 5)) = 330

. . .X(X + 5 - X + 5) = 330

. . .X(X - X + 5 + 5) = 330

. . .X(10) = 330

. . .10X = 330

. . .=> X = 33, and X - 5 = 33 - 5 = 28

So the area of the lawn abcd = X(X - 5) = 33 * 28 = 924 m^2
______________________________
Edited by stapel -- Reason for edit: fixing formatting.

#### scoresofsteel

##### New member
The equation you have wrritten is :

x(x +5) -(x -10)(x - 15) = x +330

however there if you look closely, right hand side should only be 330 and not (x+330)

Second mistake is that you have substracted '10' when you should substract only '5' as width of the flower path is 2.5.

it should be

X(X+5) - (X-5)(X) = 330

#### TchrWill

##### Full Member
]A small city park consists of a rectangular laawn surrounded on all sides by a 330 m2 border of flowrs 2.5m wide. Find the area of the lawn if the entire park is 5 m longer than it is wide.

Boy, did I miss the boat.

x = width and x + 5 = length

Then, x(x + 5) - (x - 5)x = 330

......x^2 - 5x - x^2 + =5x = 330

......10x = 330 making x = 33.

Check: 33(38) - 28(33) = 330

The lawn is then 28 by 33 with an area of 924 sq.m