Question

Match each spherical volume to the largest cross sectional area of that sphere

Answer 1

Answer: The above answer is correct.

Step-by-step explanation: I got this right on Edmentum.

Answer 2

Part 1)
`324\pi\ units^(2)` ------>
`7,776\pi\ units^(3)`

Part 2)
`36\pi\ units^(2)` ------>
`288\pi\ units^(3)`

Part 3)
`81\pi\ units^(2)` ------>
`972\pi\ units^(3)`

Part 4)
`144\pi\ units^(2)` ------>
`2,304\pi\ units^(3)`

Explanation:

we know that

The largest cross sectional area of that sphere is equal to the area of a circle with the same radius of the sphere

Part 1) we have

`A=324\pi\ units^(2)`

The area of the circle is equal to

`A=\pi r^(2)`

so

`324\pi=\pi r^(2)`

Solve for r

`r^(2)=324`

`r=18\ units`

Find the volume of the sphere

The volume of the sphere is

`V=(4)/(3)\pi r^(3)`

For
`r=18\ units`

substitute

`V=(4)/(3)\pi (18)^(3)`

`V=7,776\pi\ units^(3)`

Part 2) we have

`A=36\pi\ units^(2)`

The area of the circle is equal to

`A=\pi r^(2)`

so

`36\pi=\pi r^(2)`

Solve for r

`r^(2)=36`

`r=6\ units`

Find the volume of the sphere

The volume of the sphere is

`V=(4)/(3)\pi r^(3)`

For
`r=6\ units`

substitute

`V=(4)/(3)\pi (6)^(3)`

`V=288\pi\ units^(3)`

Part 3) we have

`A=81\pi\ units^(2)`

The area of the circle is equal to

`A=\pi r^(2)`

so

`81\pi=\pi r^(2)`

Solve for r

`r^(2)=81`

`r=9\ units`

Find the volume of the sphere

The volume of the sphere is

`V=(4)/(3)\pi r^(3)`

For
`r=9\ units`

substitute

`V=(4)/(3)\pi (9)^(3)`

`V=972\pi\ units^(3)`

Part 4) we have

`A=144\pi\ units^(2)`

The area of the circle is equal to

`A=\pi r^(2)`

so

`144\pi=\pi r^(2)`

Solve for r

`r^(2)=144`

`r=12\ units`

Find the volume of the sphere

The volume of the sphere is

`V=(4)/(3)\pi r^(3)`

For
`r=12\ units`

substitute

`V=(4)/(3)\pi (12)^(3)`

`V=2,304\pi\ units^(3)`