Question 316969
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I highly suspect you mean that you need the explanation in layman's terms.  If not, I don't know what you are talking about because I don't understand the word "lamins"


12 feet 6 inches is 12.5 feet. *[tex \LARGE 12\frac{1}{2}\ =\ \frac{25}{2}]


Likewise:


10 feet 4 inches is *[tex \LARGE 10.\overline{3}] feet.  *[tex \LARGE 10\frac{1}{3}\ =\ \frac{31}{3}]


The area of a rectangle is simply the length times the width.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{25}{2}\,\cdot\,\frac{31}{3}\ =\ \frac{775}{6}\text{ ft^2}]



Each tile is 5 inches by 5 inches, hence the area of a tile is *[tex \Large 25\text{ in^2}]


A square foot is 12 inches by 12 inches, hence the area of a square foot in square inches is *[tex \Large 144\text{ in^2}]


Therefore each tile is *[tex \Large \frac{25}{144}\text{ ft^2}]


To determine the number of tiles required divide the total number of square feet required by the number of square feet per tile:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{775}{6}\right)\ \div\ \left(\frac{25}{144}\right)\ =\ \left(\frac{775}{6}\right)\ \times\ \left(\frac{144}{25}\right)\ =\ 31\ \times\ 24\ =\ 744]


In real terms, nothing like this comes out exactly.  In the first place, even though 12'6" divides evenly into 5" pieces, the distance from the left edge of one tile to the left edge of the adjacent tile is more than the measurement of the tile because the grout takes up some space.  10'4" does NOT divide evenly into 5" pieces, and you still have the grout line spacing to contend with.


Next, you will have to contend with the fact that you probably won't be able to buy 744 tiles.  The store  will want to sell them in boxes.  If they come in boxes of 12 or 24 you are in luck (remember you got to 744 by multiplying 31 times 24) but if they come in boxes of 10 or 20 or 25, you are going to have to buy more than you need in order to have enough.  Furthermore, you are going to want to buy extra anyway.  Some will break during shipping and handling.  You will break some when installing it.  You will make mistakes cutting it.  You need 10% to 15% more than you calculated in real life.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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