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 \displaystyle 4440 find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

 

Possible Answers:

\displaystyle 32.6

\displaystyle 104.0

\displaystyle 10.2

\displaystyle 125.9

\displaystyle 1.6

Correct answer:

\displaystyle 104.0

Explanation:

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

Example Question #6 : 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 \displaystyle 6254 find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

 

Possible Answers:

\displaystyle 158.2

\displaystyle 11.4

\displaystyle 1.6

\displaystyle 38.6

\displaystyle 130.6

Correct answer:

\displaystyle 130.6

Explanation:

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

Example Question #7 : 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 \displaystyle 4276 find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

 

Possible Answers:

\displaystyle 10.1

\displaystyle 1.6

\displaystyle 32.0

\displaystyle 122.8

\displaystyle 101.4

Correct answer:

\displaystyle 101.4

Explanation:

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

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 \displaystyle 3242 find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Possible Answers:

\displaystyle 27.8

\displaystyle 84.3

\displaystyle 9.2

\displaystyle 102.1

\displaystyle 1.6

Correct answer:

\displaystyle 84.3

Explanation:

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

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 \displaystyle 6 and height \displaystyle 13 find the area of a cross section that's parallel to its base. Round your answer to the nearest tenth.

Possible Answers:

\displaystyle 530.9

\displaystyle 36.0

\displaystyle 113.1

\displaystyle 1470.3

\displaystyle 169.0

Correct answer:

\displaystyle 113.1

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 \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

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 \displaystyle 8 and height \displaystyle 77 find the area of a cross section that's parallel to its base. Round your answer to the nearest tenth.

Possible Answers:

\displaystyle 64.0

\displaystyle 201.1

\displaystyle 15481.8

\displaystyle 5929.0

\displaystyle 18626.5

Correct answer:

\displaystyle 201.1

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 \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

Example Question #11 : Identify Shapes In 2 D Cross Sections And Rotations In 3 D: Ccss.Math.Content.Hsg Gmd.B.4

Given a cylinder with radius \displaystyle 16 and height \displaystyle 31 find the area of a cross section that's parallel to its base. Round your answer to the nearest tenth.

Possible Answers:

\displaystyle 804.2

\displaystyle 3019.1

\displaystyle 961.0

\displaystyle 256.0

\displaystyle 24931.7

Correct answer:

\displaystyle 804.2

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 \displaystyle r.

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

Now we round our answer to the nearest tenth

\displaystyle A = 804.2

Example Question #52 : Geometric Measurement & Dimension

Given a sphere with a volume of \displaystyle 6850 find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Possible Answers:

\displaystyle 1.6

\displaystyle 40.4

\displaystyle 168.1

\displaystyle 138.8

\displaystyle 11.8

Correct answer:

\displaystyle 138.8

Explanation:

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 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

\displaystyle \\\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

\displaystyle A = \pi r^{2}

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

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

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

\displaystyle A= 138.8

Example Question #423 : High School: Geometry

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

Triangle abc

Possible Answers:

Screen shot 2020 07 01 at 6.11.14 pm

Screen shot 2020 07 01 at 6.16.55 pm

Triangle surface area triangular prism png favpng swy7w2myrhyx3esxq6kyf7333 removebg preview

Cone

Screen shot 2020 07 01 at 6 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:

Jhmfu

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

Trapezoid, Equilateral Triangle

Square, Trapezoid, Triangle

Polygon, Rectangle

Trapezoid, Triangle, 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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