Question

The volume of a right circular cylinder with a height of 6 cm is 150\(\pi\) cubic centimeters. What is the lateral surface area, in square centimeters, of this cylinder? (Note: For a right circular cylinder with radius \(r\) and height \(h,\) the lateral surface area is 2\(\pi r h\) and the volume is \(\pi r^{2} h .\) ) A. 25\(\pi\) B. 36\(\pi\) C. 50\(\pi\) D. 60\(\pi\) E. 110\(\pi\)

Short answer

Answer: The lateral surface area of the cylinder is \(60\pi\) square centimeters.

Step-by-step solution

Find the radius of the cylinder.

We are given that the volume of the cylinder is \(150\pi\). We can use the volume formula for cylinders to set up an equation and solve for the radius, \(r\). The volume formula for a right circular cylinder is \(V = \pi r^2 h\). In our case, the volume \(V\) is \(150\pi\) and the height \(h\) is 6 cm. Plugging in the values, the equation becomes: \(150\pi = \pi r^2 (6)\)

Solve for the radius, r.

Now, we will solve for \(r\) from the equation we set up in step 1. \(150\pi = \pi r^2(6)\) Divide both sides by \(\pi\): \(150 = 6r^2\) Divide both sides by 6: \(25 = r^2\) Take the square root of both sides to find the value of \(r\): \(r = 5\) cm

Calculate the lateral surface area of the cylinder.

We are given the lateral surface area formula for a right circular cylinder: \(A = 2\pi r h\). We have found the radius, \(r\) (5 cm) and we already know the height, \(h\) (6 cm). Plugging these values into the formula, we get: \(A = 2\pi (5)(6)\)

Simplify the expression for lateral surface area.

Now, let's simplify the expression we found in step 3 to find the lateral surface area: \(A = 2\pi (5)(6) = 10\pi (6) = 60\pi\) square centimeters The lateral surface area of the cylinder is \(60\pi\) square centimeters, which corresponds to the answer choice D.

Key concepts

Right Circular Cylinder

Understanding the right circular cylinder is essential in the world of geometry. It's a 3-dimensional solid that has a circular base and a congruent circular top, with straight sides connecting the two circles, known as the lateral surface. The term 'right' indicates that the axis of the cylinder is perpendicular to the base, forming right angles or a 'right' cylinder. This is opposed to an oblique cylinder, where the sides lean over, creating a slant.

To imagine a right circular cylinder, think of a can of soup or a roll of quarters; both have the key attributes: circular bases and tops with straight sides. In our exercise, the cylinder's height (the distance between the bases) plays a crucial role in determining other properties like volume and lateral surface area.

Cylinder Volume

The volume of a cylinder represents the amount of space inside it, which can be filled with a substance like water or air. The formula to calculate the volume of a right circular cylinder is given by \( V = \pi r^2 h \), where \(r\) is the radius of the circular base, and \(h\) is the height of the cylinder.

In our geometry problem, we used the given volume of \(150\pi\) cubic centimeters and the height of 6 cm to solve for the radius. Grasping this relationship allows us to determine another unknown measurement by rearranging the equation. In this case, discovering the radius was essential to move on to finding the lateral surface area, demonstrating an interdependence that is common in geometry problem-solving.

Geometry Problem Solving

Mastering geometry problem-solving fuses together understanding shapes, formulas, and algebraic manipulation. A methodical approach is best, as highlighted in our cylinder exercise. Start by identifying what you're given and what you need to find. Next, recall the relevant formulas. For lateral surface area and volume of a cylinder, those are \( A = 2\pi rh \) and \( V = \pi r^2 h \) respectively.

Once the appropriate formulas are identified, substitute the given values and solve for the unknowns step by step. As we solved for the radius first using the volume and then found the lateral surface area, it emphasized the sequential nature of problem-solving. Remembering and applying these equations appropriately is crucial. And, as shown, sometimes working backward is necessary to solve for intermediary steps that will ultimately give you the final answer.