Problem 98
Question
Solve for the indicated variable. Surface Area of a Cone Solve for \(l\) in \(S=\pi r^{2}+\pi r l\).
Step-by-Step Solution
Verified Answer
The slant height \(l\) can be solved from the surface area equation of a cone as \(l = \frac{S - \pi r^{2}}{\pi r}\
1Step 1: Understand the equation
The given equation \(S=\pi r^{2}+\pi r l\) is the formula for the surface area of a cone. Here, \(S\) is the total surface area, \(r\) is the radius of the base, and \(l\) is the slant height.
2Step 2: Isolate the term with \(l\)
In order to solve for \(l\), we need to isolate the term that includes \(l\). We can do this by subtracting \( \pi r^{2}\) from both sides of the equation, resulting in \( S - \pi r^{2} = \pi r l \)
3Step 3: Solve for \(l\)
Finally, we divide both sides of the equation by \( \pi r\), this gives us: \(l = \frac{S - \pi r^{2}}{\pi r}\)
Key Concepts
GeometrySurface Area of a ConeSolving Equations
Geometry
Geometry is a branch of mathematics that deals with shapes, sizes, and the properties of space. It connects to our everyday life, from architecture to nature. In geometry, we study various forms such as circles, triangles, and, importantly for this topic, cones. Understanding geometric shapes involves visualizing both their flat and 3D forms, which is crucial for more advanced math concepts.
Cones are 3-dimensional figures with a circular base and a single vertex. Think of them like traffic cones or ice-cream cones. The study of cones brings together circular shapes and the ideas of height and slant height, which give it its unique structure.
In this exercise, we specifically look at the surface area of a cone, linking geometry with algebra to solve the problem at hand.
Cones are 3-dimensional figures with a circular base and a single vertex. Think of them like traffic cones or ice-cream cones. The study of cones brings together circular shapes and the ideas of height and slant height, which give it its unique structure.
In this exercise, we specifically look at the surface area of a cone, linking geometry with algebra to solve the problem at hand.
Surface Area of a Cone
The surface area of a cone is a measure of the total area covered by its outer surface. It consists of two parts: the base and the lateral (side) surface. The base is simply a circle, and its area is found using the formula:
\[ S = \pi r^2 + \pi r l \]
Where \(S\) is the total surface area, \(r\) is the radius of the base, and \(l\) is the slant height. Understanding this formula is crucial as it allows us to solve for different variables, such as \(l\) in this exercise.
- Base Area: \( ext{Base Area} = \pi r^2\)
- Lateral Surface Area: \( ext{Lateral Surface Area} = \pi r l\)
\[ S = \pi r^2 + \pi r l \]
Where \(S\) is the total surface area, \(r\) is the radius of the base, and \(l\) is the slant height. Understanding this formula is crucial as it allows us to solve for different variables, such as \(l\) in this exercise.
Solving Equations
Solving equations is about finding the unknown value that makes the equation true. In algebra, it's like working with a puzzle where each piece fits snugly into place. Here, we focus on solving for the slant height, \(l\), in the surface area equation of a cone.
The skills involved include:
Then, we divided every term by \(\pi r\) to solve for \(l\), giving the formula: \[ l = \frac{S - \pi r^2}{\pi r} \]
Solving equations like this is foundational in algebra and geometry, honing one's ability to manipulate and understand mathematical relationships.
The skills involved include:
- Rearranging terms to isolate the variable of interest, which often involves adding or subtracting terms on both sides of the equation.
- Dividing or multiplying both sides to solve for the unknown variable, ensuring that the equation remains balanced.
Then, we divided every term by \(\pi r\) to solve for \(l\), giving the formula: \[ l = \frac{S - \pi r^2}{\pi r} \]
Solving equations like this is foundational in algebra and geometry, honing one's ability to manipulate and understand mathematical relationships.
Other exercises in this chapter
Problem 97
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