The cost to fill the building with road salt is $ 8591.04
<em><u>Solution:</u></em>
Given that, A cone-shaped building has a height of 11.4 meters and a base with a diameter of 12 meters
Therefore,
Height = 11.4 meters
diameter = 12 meters
![radius = \frac{diameter}{2} = \frac{12}{2} = 6](https://tex.z-dn.net/?f=radius%20%3D%20%5Cfrac%7Bdiameter%7D%7B2%7D%20%3D%20%5Cfrac%7B12%7D%7B2%7D%20%3D%206)
radius = 6 meters
<em><u>Let us first find the volume of cone</u></em>
<em><u>The volume of cone is given as:</u></em>
![V = \frac{\pi r^2h}{3}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B%5Cpi%20r%5E2h%7D%7B3%7D)
Where, "V" is the volume of cone
"h" and "r" are height and radius of cone
<em><u>Substituting the given values, we get</u></em>
![V = \frac{3.14 \times 6^2 \times 11.4}{3}\\\\V = 3.14 \times 12 \times 11.4\\\\V = 429.552](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B3.14%20%5Ctimes%206%5E2%20%5Ctimes%2011.4%7D%7B3%7D%5C%5C%5C%5CV%20%3D%203.14%20%5Ctimes%2012%20%5Ctimes%2011.4%5C%5C%5C%5CV%20%3D%20429.552)
Thus volume of cone is 429.552 cubic meters
<em><u>The building will be filled with road salt that costs $20 per cubic meter</u></em>
1 cubic meter = $ 20
<em><u>Therefore, for 429.552 cubic meters, we get,</u></em>
![cost = 20 \times 429.552\\\\cost = 8591.04](https://tex.z-dn.net/?f=cost%20%3D%2020%20%5Ctimes%20429.552%5C%5C%5C%5Ccost%20%3D%208591.04)
Thus cost to fill the building with road salt is $ 8591.04