Problem 30 - Contemporary Precalculus

Question

What is the radius of the base of a cone whose surface area is \(18 \pi\) square centimeters and whose height is \(4 \mathrm{cm} ?\)

Step-by-Step Solution

Answer

Answer: The radius of the base of the cone is 2 cm.

Step 1: Write down the given information

The exercise provides us with information: - Surface area (A) \(= 18\pi \,\text{cm}^2\) - Height (h) \(= 4\, \mathrm{cm}\)

Step 2: Find the slant height using the Pythagorean theorem

We'll start by finding the slant height using the relationship between height, radius, and slant height. In this right-angled triangle, height (h) forms one leg, radius (r) forms another leg, and slant height (l) forms the hypotenuse. Using the Pythagorean theorem, we have: \(l^2 = r^2 + h^2\) Now plug in the value of h: \(l^2 = r^2 + (4)^2\)

Step 3: Set up the surface area formula

The surface area (A) formula for a cone can be written as: \(A = \pi r (r + l)\) And given that A = \(18\pi\): \(18\pi = \pi r (r + l)\)

Step 4: Solve for radius (r)

Now, we will use the above equation to find the value of r: \(\frac{18\pi}{\pi} = r (r + l)\) \(18 = r^2 + rl\) As we found in Step 2: \(l^2 = r^2 + 16\) Now, we can solve for l: \(l^2 - r^2 = 16\) Plug this expression into the equation from Step 4, and we get: \(18 = r^2 + r(\sqrt{l^2 - r^2})\) Now, square both sides: \(324 = r^4 + 2r^3\sqrt{l^2 - r^2} + r^2(l^2 - r^2)\) But, we know that \(l^2 - r^2 = 16\). So we can substitute it into our equation: \(324 = r^4 + 2r^3\sqrt{16} + r^2(16)\) \(324 = r^4 + 32r^3 + 16r^2\) Now, solve for r: \(r^4+ 32r^3 + 16r^2 - 324 = 0\) This equation can be factored further: \((r^2+18)(r+2)^2 = 0\) So, either \(r^2+18=0\) or \((r+2)^2 = 0\). The first equation gives us a non-real answer, which we discard. Therefore, the second equation \((r+2)^2 = 0\) gives us the answer. \(r = -2\) However, since we are dealing with lengths and the radius cannot be negative, it's implied that we were actually looking for the absolute value of r: \(r = | -2 |\)

Step 5: Determine the radius

The radius of the base of the cone is: \(r = 2 \, \mathrm{cm}\)

Key Concepts

Surface Area of a Cone

Understanding the surface area of a cone is crucial in many geometry problems. A cone's surface area is composed of two parts: the base and the lateral surface. The formula to find the total surface area, A, of a cone is given by:

• Total Surface Area: \[ A = \pi r (r + l) \]

where:

• \( r \) is the radius of the base of the cone
• \( l \) is the slant height, which is the diagonal distance from the base to the apex of the cone.

The slant height is not the same as the perpendicular height of the cone, making it important to calculate it separately if not directly provided. To solve problems involving surface area, make sure you know these components and can substitute the correct values into the formula. This will unveil the required measurements, such as the radius if it is unknown.

Pythagorean Theorem

The Pythagorean Theorem is pivotal when dealing with triangles, especially in precalculus. It's especially useful for right-angled triangles, where one must understand that:

• \( a^2 + b^2 = c^2 \)

Here:\

• \( a \) and \( b \) are the lengths of the triangle's legs
• \( c \) is the length of the hypotenuse (the longest side)

In the context of cones, the theorem helps find the slant height (\( l \)), acting as the hypotenuse of a right triangle whose legs are the cone's height (\( h \)) and the cone base's radius (\( r \)). By rearranging \( l = \sqrt{r^2 + h^2} \), you can uncover missing dimensions from given parameters. This makes the theorem a powerful tool in geometry within precalculus.

Solving Quadratic Equations

Quadratic equations appear frequently in geometry and precalculus, often in problems requiring us to find missing dimensions. A standard quadratic equation looks like this: \[ ax^2 + bx + c = 0 \]To solve it, several methods can be employed:

• Factoring: Try to express the equation in the form \((x - p)(x - q) = 0\)
• Completing the Square: Alter the equation to make one side a perfect square trinomial
• Quadratic Formula: \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]

In the exercise, we derived a quadratic equation in the process of finding the cone's radius. Once formed, these equations yield two potential solutions, where one might not make sense in the real-world context, such as a negative length, which we discard.

Geometry in Precalculus

Geometry integrates with calculus to form the foundation of precalculus, where understanding shape properties helps solve complex math problems. This branch of mathematics often involves:

• Geometric Shapes: Understanding properties of cones, cylinders, spheres, etc.
• Angle Calculations: Determining angles within different shapes, involving trigonometry
• Measurement Calculations: Using formulas to find area, volume, and surface area

By combining algebraic techniques, geometry in precalculus helps you establish relationships between different mathematical concepts. It's essential to solve for variables like lengths, areas, and angles, often requiring tools like the Pythagorean Theorem, surface area formulas, and quadratic equations. This makes geometry both challenging and rewarding as you decode complex problems.