The first step is see what the cross section parallel to the base looks like.

Since we are dealing with a cylinder, the cross section that we are dealing with parallel to the base is a circle.

So all we need to do is recall the area of a circle equation, and substitute the radius given for $\displaystyle r$.

$\displaystyle \\A = \pi r^{2} \\A= \pi \cdot 9 ^2 \\A = 81 \pi \\A = 254.469004940773$

Now we round our answer to the nearest tenth

$\displaystyle A = 254.5$

The first step is to recall the volume equation of a sphere.

$\displaystyle V = \frac{4 \pi}{3} r^{3}$

Since we are given the volume, we can plug it in for V

$\displaystyle 9764 = \frac{4 \pi}{3} r^{3}$

Since we want to find the area of the perpendicular cross section, we need to find what the radius is.

Now we solve for $\displaystyle r$

$\displaystyle \\7323 = \pi r^{3} \\\\\frac{7323}{\pi} = r^{3} \\r=\sqrt[3]{ \frac{7323}{\pi} }$

Now we can use the radius we found to find the area of the cross section.

The cross section in this case is a circle, so now we will use the area of a circle formula, which is

$\displaystyle A = \pi r^{2}$

Now we simply substitute the radius we just found for $\displaystyle r$.

$\displaystyle \\A= \pi \cdot 13.259069706343505 ^2 \\A= 175.80292947767606$

Now we round our answer to the nearest tenth, which is

$\displaystyle A= 175.8$

The first step is to recall the volume equation of a sphere.

$\displaystyle V = \frac{4 \pi}{3} r^{3}$

Since we are given the volume, we can plug it in for V

$\displaystyle 8472 = \frac{4 \pi}{3} r^{3}$

Since we want to find the area of the perpendicular cross section, we need to find what the radius is.

Now we solve for r

$\displaystyle \\6354 = \pi r^{3} \\\\\frac{6354}{\pi} = r^{3} \\r=\sqrt[3]{ \frac{6354}{\pi} }$

Now we can use the radius we found to find the area of the cross section.

The cross section in this case is a circle, so now we will use the area of a circle formula, which is

$\displaystyle A = \pi r^{2}$

Now we simply substitute the radius we just found for $\displaystyle r$.

$\displaystyle \\A= \pi \cdot 12.646366950388915 ^2 \\A= 159.930597043889$

Now we round our answer to the nearest tenth, which is

$\displaystyle A= 159.9$

The first step is to recall the volume equation of a sphere.

$\displaystyle V = \frac{4 \pi}{3} r^{3}$

Since we are given the volume, we can plug it in for V

$\displaystyle 5124 = \frac{4 \pi}{3} r^{3}$

Since we want to find the area of the perpendicular cross section, we need to find what the radius is.

Now we solve for r

$\displaystyle \\3843 = \pi r^{3} \\\\\frac{3843}{\pi} = r^{3} \\r=\sqrt[3]{ \frac{3843}{\pi} }$

Now we can use the radius we found to find the area of the cross section.

The cross section in this case is a circle, so now we will use the area of a circle formula, which is

$\displaystyle A = \pi r^{2}$

Now we simply substitute the radius we just found for $\displaystyle r$.

$\displaystyle \\A= \pi \cdot 10.694820600410134 ^2 \\A= 114.379187674957$

Now we round our answer to the nearest tenth, which is

$\displaystyle A= 114.4$

The first step is to recall the volume equation of a sphere.

$\displaystyle V = \frac{4 \pi}{3} r^{3}$

Since we are given the volume, we can plug it in for V

$\displaystyle 4440 = \frac{4 \pi}{3} r^{3}$

Since we want to find the area of the perpendicular cross section, we need to find what the radius is.

Now we solve for r

$\displaystyle \\3330 = \pi r^{3} \\\\\frac{3330}{\pi} = r^{3} \\r=\sqrt[3]{ \frac{3330}{\pi} }$

Now we can use the radius we found to find the area of the cross section.

The cross section in this case is a circle, so now we will use the area of a circle formula, which is

$\displaystyle A = \pi r^{2}$

Now we simply substitute the radius we just found for $\displaystyle r$.

$\displaystyle \\A= \pi \cdot 10.196038192871926 ^2 \\A= 103.95919483050302$

Now we round our answer to the nearest tenth, which is

$\displaystyle A= 104.0$

The first step is to recall the volume equation of a sphere.

$\displaystyle V = \frac{4 \pi}{3} r^{3}$

Since we are given the volume, we can plug it in for V

$\displaystyle 6254 = \frac{4 \pi}{3} r^{3}$

Since we want to find the area of the perpendicular cross section, we need to find what the radius is.

Now we solve for r

$\displaystyle \\\frac{9381}{2} = \pi r^{3} \\\\\frac{9381}{2 \pi} = r^{3} \\r=\sqrt[3]{ \frac{9381}{2 \pi} }$

Now we can use the radius we found to find the area of the cross section.

The cross section in this case is a circle, so now we will use the area of a circle formula, which is

$\displaystyle A = \pi r^{2}$

Now we simply substitute the radius we just found for $\displaystyle r$.

$\displaystyle \\A= \pi \cdot 11.429390973615462 ^2 \\A= 130.6309780277626$

Now we round our answer to the nearest tenth, which is

$\displaystyle A= 130.6$

The first step is to recall the volume equation of a sphere.

$\displaystyle V = \frac{4 \pi}{3} r^{3}$

Since we are given the volume, we can plug it in for V

$\displaystyle 4276 = \frac{4 \pi}{3} r^{3}$

Since we want to find the area of the perpendicular cross section, we need to find what the radius is.

Now we solve for r

$\displaystyle \\3207 = \pi r^{3} \\\\\frac{3207}{\pi} = r^{3} \\r=\sqrt[3]{ \frac{3207}{\pi} }$

Now we can use the radius we found to find the area of the cross section.

The cross section in this case is a circle, so now we will use the area of a circle formula, which is

$\displaystyle A = \pi r^{2}$

Now we simply substitute the radius we just found for $\displaystyle r$.

$\displaystyle \\A= \pi \cdot 10.06892321760049 ^2 \\A= 101.38321476193423$

Now we round our answer to the nearest tenth, which is

$\displaystyle A= 101.4$

The first step is to recall the volume equation of a sphere.

$\displaystyle V = \frac{4 \pi}{3} r^{3}$

Since we are given the volume, we can plug it in for V

$\displaystyle 3242 = \frac{4 \pi}{3} r^{3}$

Since we want to find the area of the perpendicular cross section, we need to find what the radius is.

Now we solve for r

$\displaystyle \\\frac{4863}{2} = \pi r^{3} \\\\\frac{4863}{2 \pi} = r^{3} \\r=\sqrt[3]{ \frac{4863}{2 \pi} }$

Now we can use the radius we found to find the area of the cross section.

The cross section in this case is a circle, so now we will use the area of a circle formula, which is

$\displaystyle A = \pi r^{2}$

Now we simply substitute the radius we just found for $\displaystyle r$.

$\displaystyle \\A= \pi \cdot 9.18138362235783 ^2 \\A= 84.29780522090057$

Now we round our answer to the nearest tenth, which is

$\displaystyle A= 84.3$

The first step is see what the cross section parallel to the base looks like.

Since we are dealing with a cylinder, the cross section that we are dealing with parallel to the base is a circle.

So all we need to do is recall the area of a circle equation, and substitute the radius given for $\displaystyle r$.

$\displaystyle \\A = \pi r^{2} \\A= \pi \cdot 6 ^2 \\A = 36 \pi \\A = 113.097335529233$

Now we round our answer to the nearest tenth

$\displaystyle A = 113.1$

The first step is see what the cross section parallel to the base looks like.

Since we are dealing with a cylinder, the cross section that we are dealing with parallel to the base is a circle.

So all we need to do is recall the area of a circle equation, and substitute the radius given for $\displaystyle r$.

$\displaystyle \\A = \pi r^{2} \\A= \pi \cdot 8 ^2 \\A = 64 \pi \\A = 201.061929829747$

Now we round our answer to the nearest tenth

$\displaystyle A = 201.1$