# Patio Dimensions

• Jan 12th 2007, 01:16 PM
symmetry
Patio Dimensions
A contractor orders 8 cubic yards of premixed cement, all of which is to be used to pour a rectangular patio that will be 4 inches thick. If the length of the patio is specified to be TWICE the width, what will be the patio dimensions?

NOTE: 1 cubic yard = 27 cubic feet
• Jan 12th 2007, 01:25 PM
CaptainBlack
Quote:

Originally Posted by symmetry
A contractor orders 8 cubic yards of premixed cement, all of which is to be used to pour a rectangular patio that will be 4 inches thick. If the length of the patio is specified to be TWICE the width, what will be the patio dimensions?

NOTE: 1 cubic yard = 27 cubic feet

Let the width of the patio be W (ft), then the volume will be:

V=W*(2W)*(1/3) cubic ft,

so:

(2/3)W^2 = 8*27 cubic ft,

so:

W=sqrt(8*27*3/2)=18 ft

so the patio is 18 by 36 ft.

RonL
• Jan 12th 2007, 04:22 PM
symmetry
ok
Below is your answer followed by my reply in brackets.

Let the width of the patio be W (ft), then the volume will be:

V=W*(2W)*(1/3) cubic ft,[What part of the question indicated that we are dealing with volume?]

so:

(2/3)W^2 = 8*27 cubic ft,

so:

W=sqrt(8*27*3/2)=18 ft{Why did you take the square root?]

so the patio is 18 by 36 ft.
• Jan 12th 2007, 10:40 PM
CaptainBlack
Quote:

Originally Posted by symmetry
Below is your answer followed by my reply in brackets.

Let the width of the patio be W (ft), then the volume will be:

V=W*(2W)*(1/3) cubic ft,[What part of the question indicated that we are dealing with volume?]

We are given the amount of cement ordered in cubic yards, which is a
volume, and this will be the the volume of the patio when laid.

Quote:

so:

(2/3)W^2 = 8*27 cubic ft,

so:

W=sqrt(8*27*3/2)=18 ft{Why did you take the square root?]
we have from the previous equation:

(2/3)W^2 = 8*27,

so:

W^2=8*27*3/2,

the term on the left of this is W squared, so to get W we need to take
square roots, so:

W=sqrt(8*27*3/2)=18

Quote:

so the patio is 18 by 36 ft.
RonL
• Jan 13th 2007, 02:06 AM
symmetry
ok
I can see from your reply that it's super important to pay attention to detail in terms of word problems, right? Every word leads to a clue.

Of course, this does NOT take away the fact that there are word problems that are simply complicated in nature.

I thank you very much.
• Jan 13th 2007, 02:24 AM
CaptainBlack
Quote:

Originally Posted by symmetry
I can see from your reply that it's super important to pay attention to detail in terms of word problems, right? Every word leads to a clue.

Of course, this does NOT take away the fact that there are word problems that are simply complicated in nature.

I thank you very much.

You gave to look out for problems with redundant information (information
not needed to solve the problem, but just there to confuse things)

RonL