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All Common Core: High School - Geometry Resources

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Flashcards: Identify Shapes in 2D Cross-Sections and Rotations in 3D: 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.

254.5

81.0

8100.0

25446.9

22902.2

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Flashcards: Identify Shapes in 2D Cross-Sections and Rotations in 3D: 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.

Your answer:

81.0

Your answer:

8100.0

Your answer:

25446.9

Your answer:

22902.2

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

Flashcards: Identify Shapes in 2D Cross-Sections and Rotations in 3D: 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.

13.3

48.3

1.6

213.0

175.8

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Flashcards: Identify Shapes in 2D Cross-Sections and Rotations in 3D: 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.

Your answer:

13.3

Your answer:

48.3

Your answer:

1.6

Your answer:

213.0

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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All Common Core: High School - Geometry Resources

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

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

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

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

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