Example Question - calculate cone volume
Here are examples of questions we've helped users solve.
Calculating the Volume of a Cone
The volume of a cone can be calculated using the formula: \[ \text{Volume} = \frac{1}{3} \pi r^2 h \] Where \( r \) is the radius of the base of the cone and \( h \) is the height of the cone. For a cone with a height \( h = 20 \) cm and a base radius \( r = 10 \) cm, we can plug these values into the formula to get the volume: \[ \text{Volume} = \frac{1}{3} \pi \times (10 \text{ cm})^2 \times 20 \text{ cm} \] The correct formula to use for the calculation, based on the given information and the options provided in the image, is the third one: \[ \boxed{\frac{1}{3} \pi \times (10 \times 10) \times 20} \]
Calculating Volume of a Cone with Known Dimensions
The image contains a question related to a cone with the given dimensions: the radius is 5.5 cm and the height is 12.5 cm. However, the specific question to be solved is not visible in the image. I can provide assistance with common questions related to these dimensions such as calculating the volume or the surface area if that's what you need. For a cone, the volume (V) can be calculated using the formula: \[ V = \frac{1}{3}\pi r^2 h \] Where: - \( r \) is the radius, - \( h \) is the height, and - \( \pi \) is Pi, approximately 3.14159. Given that \( r = 5.5 \) cm and \( h = 12.5 \) cm, the volume would be: \[ V = \frac{1}{3} \times \pi \times (5.5)^2 \times 12.5 \] \[ V = \frac{1}{3} \times \pi \times 30.25 \times 12.5 \] \[ V = \frac{1}{3} \times \pi \times 378.125 \] \[ V = \pi \times 126.0416667 \] \[ V ≈ 3.14159 \times 126.0416667 \] \[ V ≈ 396.85 \text{ cm}^3 \] The volume of the cone is approximately 396.85 cubic centimeters. If you need the surface area or any other calculation, please provide the specific question or let me know which calculation you require.
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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