Math Problem Statement
a right circular cone has a slant height of 13 ft and a lateral area of 91𝜋 ft2. Find its volume in cubic feet.
Solution
We are given the following information about the right circular cone:
- Slant height () = 13 ft
- Lateral area () =
We need to find the volume of the cone. Let's go step by step:
Step 1: Find the radius () of the base
The formula for the lateral area of a cone is: where:
- is the lateral area
- is the radius of the base
- is the slant height
We are given and . Substituting these into the formula: Now, divide both sides by : Solve for :
Step 2: Find the height () of the cone
To find the volume, we need the height of the cone. We can use the Pythagorean theorem, since we have a right triangle formed by the radius, the slant height, and the height of the cone. The relation is: Substitute the known values and : Solve for : Now, take the square root of both sides:
Step 3: Find the volume () of the cone
The formula for the volume of a cone is: Substitute the values and : Approximating as 3.1416:
Thus, the volume of the cone is approximately 561.56 cubic feet.
Would you like more details on any of the steps, or do you have any further questions? Here are some related questions to explore:
- How would the volume change if the slant height were increased to 15 ft?
- What would the lateral area be if the radius were 10 ft?
- How does the height relate to the slant height in different types of cones?
- How would you derive the formula for the lateral area of a cone from its surface area formula?
- How can you use the volume of a cone to calculate the amount of material needed to construct it?
Tip: For cones, knowing any two of the following — radius, slant height, lateral area, or height — can help you find the rest!
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Math Problem Analysis
Mathematical Concepts
Geometry
Volume of a Cone
Lateral Surface Area
Pythagorean Theorem
Formulas
Lateral Area of a Cone: A_L = πrl
Pythagorean Theorem: l² = r² + h²
Volume of a Cone: V = (1/3)πr²h
Theorems
Pythagorean Theorem
Suitable Grade Level
Grades 9-12
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