# Question #064a7

Aug 3, 2017

$4422.6$ pounds

#### Explanation:

"The concrete slab will be 12 feet 9 inches long, 9 feet 3 inches wide, and 3 inches thick"

Laying out the information

• 12 feet 9 inches is $\left(12 \setminus \cdot 12\right) + 9$ since each foot is 12 inches. This gets you $144 + 9 = 153$ inches for your length.
• 9 feet 3 inches is $\left(9 \setminus \cdot 12\right) + 3$ for the same reason. This gets you $108 + 3 = 111$ inches for your width.
• Your height (thickness of slab) is $3$ inches.

Summarizing:
$l = 153$ inches
$w = 111$ inches
$h = 3$ inches

Now find the volume of the slab by using the formula $l \setminus \cdot w \setminus \cdot h$, or $153 \times 111 \times 3 = 50949$ cubic inches.

Converting
But the weight is 150 pounds per cubic foot, so you have to turn the inches into feet. Again, since each foot is 12 inches:
$50949 \setminus {\textrm{\in}}^{3} \times \frac{1 \setminus {\textrm{f t}}^{3}}{12 \times 12 \times 12 \setminus {\textrm{\in}}^{3}} = \frac{50949}{1728} \setminus {\textrm{f t}}^{3} = 29.484 \setminus {\textrm{f t}}^{3}$

Final steps
Now to solve for weight, multiply the cubic feet weight by 150 to get your answer.
$29.484 \setminus {\textrm{f t}}^{3} \times \frac{150 \setminus \textrm{p o u n \mathrm{ds}}}{1 \setminus {\textrm{f t}}^{3}} = 4422.6$ pounds

Aug 4, 2017

We will need to find the volume of the slab in cubic feet ($f {t}^{3}$), to compare it with the weight of the concrete which is also given in $f {t}^{3}$

${V}_{s l a b} = l \times w \times h \to$ and $9 i = .75 f t$ and $3 i = .25 f t$

Then: $l = 12.75 f t , w = 9.25 f t , h = .25 f t$

Now: ${V}_{s l a b} = 12.75 f t l \times 9.25 w \times .25 h$

${V}_{s l a b} = 29.484 f {t}^{3}$

The weight of the slab is then: $\left({V}_{s l a b} = 29.484 \cancel{f {t}^{3}}\right) \times \left(\frac{150 l b}{\cancel{f {t}^{3}}}\right) = 4422.6 l b$