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Question Video: Finding the Height of a Cylinder given Its Total Surface Area and Its Base Diameter Mathematics • 8th Grade

Find the diameter of the base of a cylinder if the surface area is 496πœ‹ square centimeters and its height is 23 centimeters.


Video Transcript

Find the diameter of the base of a cylinder if the surface area is 496πœ‹ square centimeters and its height is 23 centimeters.

A cylinder is usually described using two measurements, the height, β„Ž, of the cylinder, and the radius, π‘Ÿ, of its circular base. The surface area of a cylinder is equal to the sum of the areas of its three faces. Those are the circles on the top and base of the cylinder and then the curved surface area that wraps around it. The two circles each have an area of πœ‹π‘Ÿ squared. So, combined, they give a contribution of two πœ‹π‘Ÿ squared to the total surface area.

The curved surface area is actually a rectangle. The width of this rectangle is the height of the cylinder β„Ž. And the length of this rectangle is the circumference of the circular base of the cylinder. That’s two πœ‹π‘Ÿ. So, the contribution from the curved surface area is two πœ‹π‘Ÿβ„Ž.

In this question, we’ve been told that the surface area is 496πœ‹ square centimeters and the height of the cylinder is 23 centimeters. We want to calculate the diameter. But remember, the diameter is just twice the radius. So, if we can work out the radius, we’ll then be able to use this to calculate the diameter. So, we can make some substitutions into this formula. We substitute 496πœ‹ for the surface area and 23 for the height.

And now, we have an equation in terms of π‘Ÿ, which we’d like to solve. First, we note that we can cancel a factor of πœ‹ from each of the terms. We can also divide every term in our equation by two. 496 divided by two is 248. Two π‘Ÿ squared divided by two is π‘Ÿ squared, and two π‘Ÿ multiplied by 23 divided by two is just 23π‘Ÿ. So, we have 248 is equal to π‘Ÿ squared plus 23π‘Ÿ.

Now this is a quadratic equation in π‘Ÿ because the highest power of π‘Ÿ is two. We have the π‘Ÿ squared term. And in order to solve a quadratic equation, we need to first make that equation equal to zero, which we can do by subtracting 248 from each side. This gives zero is equal to π‘Ÿ squared plus 23π‘Ÿ minus 248. We can also swap the two sides of the equation round so that we have zero on the right-hand side if we prefer.

Now, we have three methods that we can use to solve a quadratic equation. We can see, first of all, whether the quadratic equation will factorize, or we can use the quadratic formula, or we can apply the completing the square method. If a quadratic equation does factorize, this will always be the quickest way. So, it’s worth doing a quick check, first of all, to see if it can be solved by factorizing.

As the coefficient of π‘Ÿ squared in our equation is just one, we’re looking for two numbers which sum to the coefficient of π‘Ÿ β€” that’s positive 23 β€” and multiply to the constant term β€” that’s negative 248. We won’t worry too much about the signs to begin with. We’ll begin by just listing some of the factors of 248. We have one and 248, then two and 124, then four and 62, and then eight and 31.

At this point, we know that as we want the two numbers to multiply to a negative number, we need one to be positive and one to be negative because a positive multiplied by a negative gives a negative answer. And if we make the eight negative in the final factor pair, then the sum of negative eight and 31, which is also 31 minus eight, is positive 23. So, this is the factor pair that we’re looking for. The two numbers that go in the brackets then are negative eight and 31. So, the factorized form of this quadratic is π‘Ÿ minus eight multiplied by π‘Ÿ plus 31 is equal to zero.

Next, we recall that if two expressions multiply to give zero, then at least one of them must itself be equal to zero. So, we take each bracket in turn and set it equal to zero, giving π‘Ÿ minus eight is equal to zero or π‘Ÿ plus 31 is equal to zero. These two linear equations can be solved in a relatively straightforward way to find the values of π‘Ÿ.

To solve the first equation, we add eight to each side, giving π‘Ÿ is equal to eight. To solve the second, we subtract 31 from each side, giving π‘Ÿ is equal to negative 31. But remember, π‘Ÿ represents the radius of the base of this cylinder, so it must take a positive value as it represents a length. Our value of π‘Ÿ equals negative 31 is a valid solution to the quadratic equation, but it isn’t a valid value of π‘Ÿ, so we can discard it. The radius of the cylinder then is eight.

But remember, we weren’t asked to give the radius of the cylinder. We were asked for the diameter, and the diameter is always twice the radius. So, we need to double this value of π‘Ÿ. Two multiplied by eight is 16, and the units for this diameter will be centimeters because the surface area was given in square centimeters and the height was given in centimeters. We found then that the diameter of the base of this cylinder is 16 centimeters.

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