Answer:
Part 1) The area of the structure is ![2,453.25\ ft^{2}](https://tex.z-dn.net/?f=2%2C453.25%5C%20ft%5E%7B2%7D)
Part 2) The slab will contain
of cement
Step-by-step explanation:
Part 1) Find the area of the structure
we know that
The area of the structure is equal to the area of rectangle plus the area of semicircle
<u><em>Find the area of rectangle</em></u>
The area of rectangle is equal to
![A=LW](https://tex.z-dn.net/?f=A%3DLW)
we have
![L=70\ ft](https://tex.z-dn.net/?f=L%3D70%5C%20ft)
![W=30\ ft](https://tex.z-dn.net/?f=W%3D30%5C%20ft)
substitute
![A=(70)(30)](https://tex.z-dn.net/?f=A%3D%2870%29%2830%29)
![A=2,100\ ft^{2}](https://tex.z-dn.net/?f=A%3D2%2C100%5C%20ft%5E%7B2%7D)
<u><em>Find the area of semicircle</em></u>
The area of semicircle is
![A=\frac{1}{2}\pi r^{2}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1%7D%7B2%7D%5Cpi%20r%5E%7B2%7D)
we have
![\pi =3.14](https://tex.z-dn.net/?f=%5Cpi%20%3D3.14)
![r=15\ ft](https://tex.z-dn.net/?f=r%3D15%5C%20ft)
substitute
![A=\frac{1}{2}(3.14)(15)^{2}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1%7D%7B2%7D%283.14%29%2815%29%5E%7B2%7D)
![A=353.25\ ft^{2}](https://tex.z-dn.net/?f=A%3D353.25%5C%20ft%5E%7B2%7D)
<u><em>Find the area of the structure</em></u>
Adds the areas
![A=2,100+353.25=2,453.25\ ft^{2}](https://tex.z-dn.net/?f=A%3D2%2C100%2B353.25%3D2%2C453.25%5C%20ft%5E%7B2%7D)
Part 2) If the structure is 6 in thick, how many cubic feet of cement will the slab contain?
Remember that
![1\ foot=12\ inches](https://tex.z-dn.net/?f=1%5C%20foot%3D12%5C%20inches)
Convert inches to feet
![6\ in=6/12=0.5\ ft](https://tex.z-dn.net/?f=6%5C%20in%3D6%2F12%3D0.5%5C%20ft)
<u><em>Find the volume of the structure</em></u>
To obtain the volume , multiply the area by 0.5 ft (thick)
so
![V=2,453.25*0.5=1,226.625\ ft^{3}](https://tex.z-dn.net/?f=V%3D2%2C453.25%2A0.5%3D1%2C226.625%5C%20ft%5E%7B3%7D)