Common Core: High School - Geometry : High School: Geometry

Study concepts, example questions & explanations for Common Core: High School - Geometry

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All Common Core: High School - Geometry Resources

6 Diagnostic Tests 114 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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:

Explanation:

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 #1 : 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:

Explanation:

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 #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:

Explanation:

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 #8 : 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:

Explanation:

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 #8 : 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:

Explanation:

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 #9 : 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:

Explanation:

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 #51 : Geometric Measurement & Dimension

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:

Explanation:

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 #11 : 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:

Explanation:

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 #61 : Geometric Measurement & Dimension

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

Triangle abc

Possible Answers:

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Screen shot 2020 07 01 at 6.16.55 pm

Cone

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Triangle surface area triangular prism png favpng swy7w2myrhyx3esxq6kyf7333 removebg preview

Correct answer:

Cone

Explanation:

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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Example Question #1 : Modeling With Geometry

Identify all of the different shapes that make up the following figure of a two-dimensional boat.

Boat

Possible Answers:

Triangle, Rectangle

Square, Trapezoid, Triangle

Trapezoid, Triangle, Rectangle

Trapezoid, Equilateral Triangle

Polygon, Rectangle

Correct answer:

Trapezoid, Triangle, Rectangle

Explanation:

In the world around us, all figures can be looked at in geometric shapes.

Looking at the two-dimensional boat,

Boat

the boat can be broken down into three geometric shapes. The bottom of the boat is a trapezoid, the pole in the center of the boat is a rectangle and the sail is a triangle.

All Common Core: High School - Geometry Resources

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