Question

“An architect is planning to incorporate several stone spheres of different sizes into the landscaping of a public park, and workers who will be applying a finish to the exterior of the spheres need to know the surface area of each sphere. The finishing process costs $92 per square meter. The surface area of a sphere is equal to 4πr2, where r is the radius of the sphere.“In the table, select the value that is closest to the cost of finishing a sphere with a 5.50-meter circumference as well as the cost of finishing a sphere with a 7.85-meter circumference. Make only two selections, one in each column.”Circumference 5.50 m Circumference 7.85 m Finishing cost$900$1,200$1,800$2,800$3,200$4,500

Answer 1

Part 1) The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Part 2) The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

Explanation:

Step 1

Find the radius of each sphere

we know that

The circumference of a circle is equal to

`C=2\pi r`

Find the radius of the sphere with a 5.50-meter circumference

For
`C=5.50\ m`

assume

`\pi =3.14`

substitute and solve for r

`5.50=2(3.14)r`

`r=5.50/[2(3.14)]=0.88\ m`

Find the radius of the sphere with a 7.85-meter circumference

For
`C=7.85\ m`

assume

`\pi =3.14`

substitute and solve for r

`7.85=2(3.14)r`

`r=7.85/[2(3.14)]=1.25\ m`

step 2

Find the surface area of each sphere

The surface area of sphere is equal to

`SA=4\pi r^(2)`

Find the surface area of sphere with a 5.50-meter circumference

For
`r=0.88\ m`

assume

`\pi =3.14`

substitute

`SA=4(3.14)(0.88)^(2)`

`SA=9.73\ m^(2)`

Find the surface area of sphere with a 7.85-meter circumference

For
`r=1.25\ m`

assume

`\pi =3.14`

substitute

`SA=4(3.14)(1.25)^(2)`

`SA=19.63\ m^(2)`

step 3

Find the cost of finishing each sphere

we know that

To find out the cost , multiply the surface area by $92 per square meter

Find the cost of sphere with a 5.50-meter circumference

`9.73*(92)=\$895.16`

therefore

The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Find the cost of sphere with a 7.85-meter circumference

`19.63*(92)=\$1,805.96`

therefore

The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800