### All Common Core: High School - Geometry Resources

## Example Questions

### Example Question #1 : Identify Shapes In 2 D Cross Sections And Rotations In 3 D: Ccss.Math.Content.Hsg Gmd.B.4

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.

**Possible Answers:**

**Correct 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 .

Now we round our answer to the nearest tenth

### Example Question #2 : Identify Shapes In 2 D Cross Sections And Rotations In 3 D: Ccss.Math.Content.Hsg Gmd.B.4

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.

**Possible Answers:**

**Correct answer:**

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

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

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

Now we solve for

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

Now we simply substitute the radius we just found for .

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

### Example Question #3 : Identify Shapes In 2 D Cross Sections And Rotations In 3 D: Ccss.Math.Content.Hsg Gmd.B.4

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.

**Possible Answers:**

**Correct answer:**

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

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

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

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

Now we simply substitute the radius we just found for .

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

### Example Question #4 : Identify Shapes In 2 D Cross Sections And Rotations In 3 D: Ccss.Math.Content.Hsg Gmd.B.4

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.

**Possible Answers:**

**Correct answer:**

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

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

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

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

Now we simply substitute the radius we just found for .

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

### Example Question #5 : Identify Shapes In 2 D Cross Sections And Rotations In 3 D: Ccss.Math.Content.Hsg Gmd.B.4

**Possible Answers:**

**Correct answer:**

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

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

Now we solve for r

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

Now we simply substitute the radius we just found for .

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

### Example Question #431 : High School: Geometry

**Possible Answers:**

**Correct answer:**

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

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

Now we solve for r

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

Now we simply substitute the radius we just found for .

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

### Example Question #432 : High School: Geometry

**Possible Answers:**

**Correct answer:**

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

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

Now we solve for r

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

Now we simply substitute the radius we just found for .

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

### Example Question #433 : High School: Geometry

**Possible Answers:**

**Correct answer:**

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

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

Now we solve for r

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

Now we simply substitute the radius we just found for .

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

### Example Question #1 : Identify Shapes In 2 D Cross Sections And Rotations In 3 D: Ccss.Math.Content.Hsg Gmd.B.4

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.

**Possible Answers:**

**Correct 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 .

Now we round our answer to the nearest tenth

### Example Question #1 : Identify Shapes In 2 D Cross Sections And Rotations In 3 D: Ccss.Math.Content.Hsg Gmd.B.4

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.

**Possible Answers:**

**Correct 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 .

Now we round our answer to the nearest tenth

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