Calculating Cone Properties
To solve the problems in this table, we'll need to use the formulas for the area of a circle (base of the cone), and volume of a cone. The formulas are:
- Area of a circle (A): A = πr², where r is the radius.
- Volume of a cone (V): V = (1/3)πr²h, where r is the radius and h is the height.
Let's fill in the table using the information given:
1. For the first row, we're given the diameter (4 units), so the radius is half the diameter, which is 2 units.
The area of the base is calculated by A = πr² = π(2)² = 4π (approximately 12.57 square units).
We're given the height of the cone (3 units), so we can calculate the volume using V = (1/3)πr²h = (1/3)π(2²)(3) = (1/3)π(4)(3) = 4π (approximately 12.57 cubic units).
For the second cone, we have partial information:
- The diameter is missing.
- The radius is 1.5 units.
- The area of the base is given as 144π square units.
- The height is missing.
- The volume of the cone is 200π cubic units.
We have the area, so we use the formula for the area of a circle to find the radius:
\[ A = πr² \]
\[ 144π = πr² \]
\[ r² = 144 \]
\[ r = 12 \text{ units (since we already know it's 1.5 units, this confirms the given area is correct)} \]
To find the height, we use the formula for the volume of a cone:
\[ V = \frac{1}{3}πr²h \]
\[ 200π = \frac{1}{3}π(1.5²)h \]
\[ 200π = \frac{1}{3}π(2.25)h \]
\[ 200π = \frac{π}{3}(2.25)h \]
\[ 200 = \frac{2.25}{3}h \]
\[ 200 = 0.75h \]
\[ h = \frac{200}{0.75} \]
\[ h = \frac{800}{3} \text{ or approximately } 266.67 \text{ units} \]
For the diameter, since the radius is 1.5 units, the diameter is twice the radius, which is 3 units.
For the third cone:
- The diameter is given as 20 units, so the radius is 10 units.
- The area of the base can be calculated using the formula A = πr² = π(10)² = 100π (which is approximately 314.16 square units).
- The height is missing.
- The volume of the cone is given as 200π cubic units.
Using the volume formula, we can solve for height:
\[ V = \frac{1}{3}πr²h \]
\[ 200π = \frac{1}{3}π(10)²h \]
\[ 200π = \frac{1}{3}π(100)h \]
\[ 200π = \frac{100π}{3}h \]
\[ 200 = \frac{100}{3}h \]
\[ h = \frac{200 * 3}{100} \]
\[ h = 6 \text{ units} \]
Lastly, for the fourth cone:
- The diameter is missing.
- The radius is missing.
- The area of the base is 3π square units.
- The height is given as 12 units.
- The volume of the cone is given as 64π cubic units.
Using the formula for the area of the circle, we can find the radius:
\[ A = πr² \]
\[ 3π = πr² \]
\[ r² = 3 \]
\[ r = \sqrt{3} \text{ units} \]
The diameter is twice the radius, so the diameter is \(2\sqrt{3}\) (approximately 3.46 units).
With the given information, the rows of the table should be completed as follows:
First row:
- Diameter: 4 units
- Radius: 2 units
- Area of the Base: \(4π\) square units
- Height: 3 units
- Volume of Cone: \(4π\) cubic units
Second row:
- Diameter: 3 units
- Radius: 1.5 units
- Area of the Base: \(144π\) square units
- Height: \(266.67\) units
- Volume of Cone: \(200π\) cubic units
Third row:
- Diameter: 20 units
- Radius: 10 units
- Area of the Base: \(100π\) square units
- Height: 6 units
- Volume of Cone: \(200π\) cubic units
Fourth row:
- Diameter: \(2\sqrt{3}\) units
- Radius: \(\sqrt{3}\) units
- Area of the Base: \(3π\) square units
- Height: 12 units
- Volume of Cone: \(64π\) cubic units
Please note that for simplicity, we are working with the symbol π as it is given in the problem where areas and volumes are represented with π. If you wanted an exact decimal value, you would have to use the value of π (approx. 3.14159) to get the decimal result.
