Example Question - cone dimensions calculation
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Calculating Dimensions of Cones
This image shows a table to be filled in with information about a cone's dimensions. Let's fill it out row by row: 1st row: Given: - Radius = 4 units Unknowns: - Diameter = 2 * Radius = 2 * 4 units = 8 units - Area of the Base = π * Radius^2 = π * 4^2 square units = π * 16 square units ≈ 50.27 square units (Using π ≈ 3.14159) - Height = 3 units - Volume of Cone = (1/3) * Area of Base * Height = (1/3) * π * 16 * 3 cubic units ≈ 50.27 cubic units 2nd row: Given: - Area of the Base = 144π square units To find the radius, we know that Area of the Base = π * Radius^2, which means Radius = sqrt(Area of Base / π) = sqrt(144) = 12 units Now use this to find the unknowns: - Diameter = 2 * Radius = 2 * 12 units = 24 units - Height = 6 units - Volume of Cone = (1/3) * Area of Base * Height = (1/3) * 144π * 6 = 288π cubic units ≈ 904.78 cubic units (Using π ≈ 3.14159) 3rd row: Given: - Volume of Cone = 200π cubic units - Diameter = 20 units Unknowns: - Radius = Diameter / 2 = 20 units / 2 = 10 units - Area of the Base = π * Radius^2 = π * 10^2 square units = 100π square units - To find the Height, use the formula Volume of Cone = (1/3) * Area of Base * Height. Rearrange this formula to solve for Height as Volume of Cone / (Area of Base / 3): Height = 200π cubic units / (100π square units / 3) = 6 units 4th row: Given: - Volume of Cone = 64π cubic units - Height = 12 units Unknowns: - To find the Area of the Base, we rearrange the volume formula to solve for Area of Base as (Volume of Cone / Height) * 3: Area of the Base = (64π cubic units / 12 units) * 3 = 16π square units - Radius = sqrt(Area of Base / π) = sqrt(16) = 4 units - Diameter = 2 * Radius = 2 * 4 units = 8 units 5th row: Given: - Height = 3 units - Volume of Cone = 3.14 cubic units (assuming π ≈ 3.14 for calculation simplicity) Unknowns: - Use the volume formula to solve for the Area of the Base: Area of the Base = (Volume of Cone / Height) * 3 = (3.14 cubic units / 3 units) * 3 ≈ 3.14 square units - Radius = sqrt(Area of Base / π) ≈ sqrt(3.14 / 3.14) = 1 unit - Diameter = 2 * Radius = 2 * 1 unit = 2 units Please note that when filling out the table, I only provided approximate values for areas and volumes involving π, as I assumed π ≈ 3.14159 for calculation purposes.
Calculating Missing Dimensions for Cones in a Worksheet
The image shows a worksheet with a cone and a table where we need to fill in the missing dimensions. The volume of a cone is given by the formula \( V = \frac{1}{3} \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. Let's fill in the missing dimensions for each row: 1. For the first row with missing height: We have the Diameter (8 units), Radius (4 units), Area of the Base \( (16\pi \) square units), and Volume of Cone \( (16\pi \) cubic units). The volume is given by the formula: \[ V = \frac{1}{3} \pi r^2 h \] Now, we solve for \( h \): \[ 16\pi = \frac{1}{3} \pi \times 4^2 \times h \] \[ 16\pi = \frac{1}{3} \pi \times 16 \times h \] \[ 16\pi = \frac{16}{3} \pi \times h \] \[ h = 16\pi / (\frac{16}{3} \pi) \] \[ h = 3 \] So, the missing height is 3 units. 2. For the second row with missing radius: We have the Diameter (20 units), so the Radius is half of the diameter: \[ r = 20/2 \] \[ r = 10 \] units 3. For the third row with missing diameter and area of the base: We have the Radius \( (5 \frac{1}{2} \) units) and the Volume of Cone \( (200\pi \) cubic units), but no height. First, let’s find the height. We can rearrange the volume formula to solve for \( h \): \[ 200\pi = \frac{1}{3} \pi \times (5 \frac{1}{2})^2 \times h \] \[ 200\pi = \frac{1}{3} \pi \times \frac{121}{4} \times h \] \[ 200\pi = \frac{121}{12} \pi \times h \] \[ h = \frac{200\pi}{\frac{121}{12}\pi} \] \[ h = \frac{200 \times 12}{121} \] \[ h = \frac{2400}{121} \] \[ h = 19.83 \] (approx.) Now, let’s find the Diameter and the area of the base: Since the Radius is \( 5 \frac{1}{2} \), the Diameter is: \[ d = 2r \] \[ d = 2 \times 5 \frac{1}{2} \] \[ d = 2 \times \frac{11}{2} \] \[ d = 11 \] The area of the base is: \[ A = \pi r^2 \] \[ A = \pi \times (5 \frac{1}{2})^2 \] \[ A = \pi \times \frac{121}{4} \] \[ A = 30.25\pi \] square units So, the missing diameter is 11 units, and the area of the base is \( 30.25\pi \) square units. 4. For the fourth row with missing volume: We have the Radius (3 units), the Area of the Base \( (9\pi \) square units), and the Height (\( 12 \) units). The volume is calculated as follows: \[ V = \frac{1}{3} \pi r^2 h \] \[ V = \frac{1}{3} \pi \times 3^2 \times 12 \] \[ V = \frac{1}{3} \pi \times 9 \times 12 \] \[ V = 36\pi \] So, the missing volume is \( 36\pi \) cubic units. Fill the table with these missing dimensions: 1. Height: 3 units 2. Area of the base for row 3: \( 30.25\pi \) square units 3. Diameter for row 3: 11 units 4. Volume for row 4: \( 36\pi \) cubic units
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