Question

Can someone help please?

Answer 1

Answer:

Part 1) The area of the structure is $2,453.25\ ft^{2}$

Part 2) The slab will contain  $1,226.625\ ft^{3}$ 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

$A=LW$

we have

$L=70\ ft$

$W=30\ ft$

substitute

$A=(70)(30)$

$A=2,100\ ft^{2}$

Find the area of semicircle

The area of semicircle is

$A=\frac{1}{2}\pi r^{2}$

we have

$\pi =3.14$

$r=15\ ft$

substitute

$A=\frac{1}{2}(3.14)(15)^{2}$

$A=353.25\ ft^{2}$

Find the area of the structure

Adds the areas

$A=2,100+353.25=2,453.25\ ft^{2}$

Part 2) If the structure is 6 in thick, how many cubic feet of cement will the slab contain?

Remember that

$1\ foot=12\ inches$

Convert inches to feet

$6\ in=6/12=0.5\ ft$

Find the volume of the structure

To obtain the volume , multiply the area by 0.5 ft (thick)

so

$V=2,453.25*0.5=1,226.625\ ft^{3}$