Answer:
Part 1) The area of the structure is [tex]2,453.25\ ft^{2}[/tex]
Part 2) The slab will contain [tex]1,226.625\ ft^{3}[/tex] 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
Find the area of rectangle
The area of rectangle is equal to
[tex]A=LW[/tex]
we have
[tex]L=70\ ft[/tex]
[tex]W=30\ ft[/tex]
substitute
[tex]A=(70)(30)[/tex]
[tex]A=2,100\ ft^{2}[/tex]
Find the area of semicircle
The area of semicircle is
[tex]A=\frac{1}{2}\pi r^{2}[/tex]
we have
[tex]\pi =3.14[/tex]
[tex]r=15\ ft[/tex]
substitute
[tex]A=\frac{1}{2}(3.14)(15)^{2}[/tex]
[tex]A=353.25\ ft^{2}[/tex]
Find the area of the structure
Adds the areas
[tex]A=2,100+353.25=2,453.25\ ft^{2}[/tex]
Part 2) If the structure is 6 in thick, how many cubic feet of cement will the slab contain?
Remember that
[tex]1\ foot=12\ inches[/tex]
Convert inches to feet
[tex]6\ in=6/12=0.5\ ft[/tex]
Find the volume of the structure
To obtain the volume , multiply the area by 0.5 ft (thick)
so
[tex]V=2,453.25*0.5=1,226.625\ ft^{3}[/tex]
