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

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

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All Common Core: High School - Geometry Resources

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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:

8100.0

22902.2

25446.9

254.5

81.0

Correct answer:

254.5

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 .

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

Now we round our answer to the nearest tenth

A = 254.5

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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 9764 find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Possible Answers:

48.3

213.0

1.6

13.3

175.8

Correct answer:

175.8

Explanation:

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

A= 175.8

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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 8472 find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Possible Answers:

159.9

1.6

45.0

12.6

193.7

Correct answer:

159.9

Explanation:

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

A= 159.9

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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 5124 find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Possible Answers:

114.4

138.6

1.6

35.0

10.7

Correct answer:

114.4

Explanation:

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

A= 114.4

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

Possible Answers:

10.2

104.0

1.6

32.6

125.9

Correct answer:

104.0

Explanation:

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

A= 104.0

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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 6254 find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Possible Answers:

1.6

38.6

158.2

11.4

130.6

Correct answer:

130.6

Explanation:

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

A= 130.6

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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 4276 find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.

Possible Answers:

101.4

32.0

1.6

122.8

10.1

Correct answer:

101.4

Explanation:

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

A= 101.4

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

Possible Answers:

27.8

84.3

9.2

102.1

1.6

Correct answer:

84.3

Explanation:

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

A= 84.3

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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:

530.9

36.0

169.0

1470.3

113.1

Correct answer:

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 .

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

Now we round our answer to the nearest tenth

A = 113.1

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Example Question #10 : 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:

201.1

5929.0

15481.8

18626.5

64.0

Correct answer:

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 .

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

Now we round our answer to the nearest tenth

A = 201.1

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

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

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

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

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

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

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

Example Question #6 : Identify Shapes In 2 D Cross Sections And Rotations In 3 D: Ccss.Math.Content.Hsg Gmd.B.4

Example Question #7 : Identify Shapes In 2 D Cross Sections And Rotations In 3 D: Ccss.Math.Content.Hsg Gmd.B.4

Example Question #8 : Identify Shapes In 2 D Cross Sections And Rotations In 3 D: Ccss.Math.Content.Hsg Gmd.B.4

Example Question #9 : Identify Shapes In 2 D Cross Sections And Rotations In 3 D: Ccss.Math.Content.Hsg Gmd.B.4

Example Question #10 : Identify Shapes In 2 D Cross Sections And Rotations In 3 D: Ccss.Math.Content.Hsg Gmd.B.4

All Common Core: High School - Geometry Resources

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