All Common Core: High School - Geometry Resources
Example Questions
Example Question #43 : Geometric Measurement & Dimension
Find the radius of a hemisphere, where the volume is
Round your answer to digits.
Before we find the radius of a hemisphere, we need to recall the equation.
Since we are given the volume (), we can plug it into the equation.
Now we can solve for .
Now we divide by on each side.
Now our equation is
To solve for , we need to take the cube root on each side.
Here is a picture representation of the hemisphere.
Example Question #5 : Cylinders, Pyramids, Cones, And Spheres Volume Formulas: Ccss.Math.Content.Hsg Gmd.A.3
Find the radius of a hemisphere, where the volume is .
Round your answer to digits.
Before we find the radius of a hemisphere, we need to recall the equation.
Since we are given the volume (), we can plug it into the equation.
Now we can solve for .
Now we divide by on each side.
Now our equation is
To solve for , we need to take the cube root on each side.
Here is a picture of what the hemisphere look like.
Example Question #44 : Geometric Measurement & Dimension
Find the radius of a hemisphere, where the volume is .
Round your answer to digits.
Before we find the radius of a hemisphere, we need to recall the equation.
Since we are given the volume (), we can plug it into the equation.
Now we can solve for .
Now we divide by on each side.
Now our equation is
To solve for , we need to take the cube root on each side.
Here is a picture of the hemisphere.
Example Question #45 : Geometric Measurement & Dimension
Find the radius of a sphere, where the volume is .
Round your answer to digits.
Before we find the radius of a sphere, we need to recall the equation.
Since we are given the volume (), we can plug it into the equation.
Now we can solve for .
Now we divide by on each side.
Now our equation is
To solve for , we need to take the cube root on each side.
Here is a picture of the sphere.
Example Question #46 : Geometric Measurement & Dimension
Find the radius of a sphere, where the volume is .
Round your answer to digits.
Before we find the radius of a sphere, we need to recall the equation.
Since we are given the volume (), we can plug it into the equation.
Now we can solve for .
Now we divide by on each side.
Now our equation is
To solve for , we need to take the cube root on each side.
Here is a picture of the sphere.
Example Question #42 : Geometric Measurement & Dimension
Find the radius of a hemisphere, where the volume is .
Round your answer to digits.
Before we find the radius of a hemisphere, we need to recall the equation.
Since we are given the volume (), we can plug it into the equation.
Now we can solve for .
Now we divide by on each side.
Now our equation is
To solve for , we need to take the cube root on each side.
Here is a picture of the hemisphere.
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.
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 sphere with a volume of find the area of the perpendicular cross section right through its center. Round your answer to the nearest tenth.
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 #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.
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 #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.
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