Identify Shapes in 2D Cross-Sections and Rotations in 3D: CCSS.Math.Content.HSG-GMD.B.4 - Geometry

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Question

Given a cylinder with radius and height find the area of a cross section that's parallel to its base. Round your answer to the nearest tenth.

Answer

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 .

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

Now we round our answer to the nearest tenth

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Question

Given a sphere with a volume of find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Answer

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

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

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

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

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

Now we solve for

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

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

$A = \pi r^{2}$

Now we simply substitute the radius we just found for .

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

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

Compare your answer with the correct one above

Question

Given a sphere with a volume of find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Answer

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

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

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

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

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

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

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

$A = \pi r^{2}$

Now we simply substitute the radius we just found for .

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

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

Compare your answer with the correct one above

Question

Given a sphere with a volume of find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Answer

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

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

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

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

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

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

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

$A = \pi r^{2}$

Now we simply substitute the radius we just found for .

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

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

Compare your answer with the correct one above

Question

Given a sphere with a volume of find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Answer

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

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

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

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

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

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

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

$A = \pi r^{2}$

Now we simply substitute the radius we just found for .

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

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

Compare your answer with the correct one above

Question

Given a sphere with a volume of find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Answer

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

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

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

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

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

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

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

$A = \pi r^{2}$

Now we simply substitute the radius we just found for .

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

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

Compare your answer with the correct one above

Question

Given a sphere with a volume of find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Answer

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

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

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

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

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

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

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

$A = \pi r^{2}$

Now we simply substitute the radius we just found for .

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

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

Compare your answer with the correct one above

Question

Given a sphere with a volume of find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Answer

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

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

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

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

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

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

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

$A = \pi r^{2}$

Now we simply substitute the radius we just found for .

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

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

Compare your answer with the correct one above

Question

Given a cylinder with radius and height find the area of a cross section that's parallel to its base. Round your answer to the nearest tenth.

Answer

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 .

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

Now we round our answer to the nearest tenth

Compare your answer with the correct one above

Question

Given a cylinder with radius and height find the area of a cross section that's parallel to its base. Round your answer to the nearest tenth.

Answer

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 .

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

Now we round our answer to the nearest tenth

Compare your answer with the correct one above

Question

Given a cylinder with radius and height find the area of a cross section that's parallel to its base. Round your answer to the nearest tenth.

Answer

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 .

$\A = \pi r^{2} \A= \pi \cdot 16 ^2 \A = 256 \pi \A = 804.247719318987$

Now we round our answer to the nearest tenth

Compare your answer with the correct one above

Question

Given a sphere with a volume of find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Answer

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

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

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

$6850 = \frac{4 \pi}{3} r^{3}$

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

$\\frac{10275}{2} = \pi r^{3} \\\frac{10275}{2 \pi} = r^{3} \r=\sqrt[3]{ \frac{10275}{2 \pi} }$

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

$A = \pi r^{2}$

Now we simply substitute the radius we just found for .

$\A= \pi \cdot 11.78150180338734 ^2 \A= 138.80378474321913$

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

Compare your answer with the correct one above

Question

What is the resulting image when you rotate Triangle ABC around the point C?

Answer

When you rotate a triangle around one of its points, the resulting image is a cone. The following image helps illustrate this:

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Question

Given a cylinder with radius and height find the area of a cross section that's parallel to its base. Round your answer to the nearest tenth.

Answer

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 .

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

Now we round our answer to the nearest tenth

Compare your answer with the correct one above

Question

Given a sphere with a volume of find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Answer

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

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

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

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

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

Now we solve for

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

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

$A = \pi r^{2}$

Now we simply substitute the radius we just found for .

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

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

Compare your answer with the correct one above

Question

Given a sphere with a volume of find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Answer

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

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

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

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

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

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

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

$A = \pi r^{2}$

Now we simply substitute the radius we just found for .

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

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

Compare your answer with the correct one above

Question

Given a sphere with a volume of find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Answer

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

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

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

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

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

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

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

$A = \pi r^{2}$

Now we simply substitute the radius we just found for .

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

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

Compare your answer with the correct one above

Question

Given a sphere with a volume of find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Answer

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

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

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

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

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

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

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

$A = \pi r^{2}$

Now we simply substitute the radius we just found for .

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

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

Compare your answer with the correct one above

Question

Given a sphere with a volume of find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Answer

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

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

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

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

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

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

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

$A = \pi r^{2}$

Now we simply substitute the radius we just found for .

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

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

Compare your answer with the correct one above

Question

Given a sphere with a volume of find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Answer

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

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

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

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

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

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

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

$A = \pi r^{2}$

Now we simply substitute the radius we just found for .

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

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

Compare your answer with the correct one above

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