Lesson Video: Areas and Circumferences of Circles | Nagwa Lesson Video: Areas and Circumferences of Circles | Nagwa

Lesson Video: Areas and Circumferences of Circles Mathematics • Second Year of Preparatory School

In this video, we will learn how to find the area and the circumference of a circle given its radius or diameter and use them to solve various problems.

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Video Transcript

In this video, we will learn how to find the area and the circumference of a circle given its radius or diameter and how to relate both the area and circumference to solve various problems.

Let’s begin by recapping some of the key language associated with circles. A circle is a set of points that have a constant distance from a point in the center, here denoted by the letter 𝑀. A radius of a circle is any line segment that connects the center of the circle to its edge, here labeled using the letter 𝑟. A diameter of a circle is a line segment whose endpoints both lie on the edge of the circle and that passes through the center. We denote the diameter using the letter 𝑑 and note that it is always twice as long as the radius of the same circle.

The circumference of a circle is the distance all the way around its edge. And we denote this using the letter 𝐶. Circles are very special shapes because their properties are related by a mathematical constant, which we call 𝜋 and denote using the Greek letter. 𝜋 is approximately equal to 3.142 to three decimal places. But in fact, 𝜋 is what is known as an irrational number, which means it has an infinitely long string of digits after the decimal point with no repeating pattern. If we tried to write all these digits down, we’d be trying forever.

The first key formula relating to circles that we need to know is this. The circumference of a circle is equal to 𝜋 multiplied by the diameter. 𝐶 is equal to 𝜋𝑑. As the diameter is twice the radius, this can also be written as 𝐶 is equal to two 𝜋𝑟.

Let’s begin by considering an example in which we calculate the circumference of a circle given the length of its diameter.

Work out the circumference of the circle, giving your answer accurate to two decimal places.

We recall first that the circumference of a circle is the distance all the way around its edge. The formula for calculating the circumference of a circle is 𝐶 equals 𝜋𝑑, where 𝐶 represents the circumference and 𝑑 represents the circle’s diameter. Looking at the diagram we’ve been given, we can see that this line segment is a diameter of the circle because its two endpoints each lie on the circle’s edge and it passes through the center of the circle. The length of this line is six, so the value of 𝑑 is six. The circumference of the circle is therefore 𝜋 multiplied by six, which we can write as six 𝜋.

Now if we needed to give an exact answer or if we didn’t have access to a calculator, we could leave our answer in this form. But we’re asked to give our answer accurate to two decimal places. So we need to evaluate this. Because 𝜋 is an irrational number, we get an answer that is also irrational, 18.84955 continuing. Rounding to two decimal places gives 18.85.

We weren’t given any units for the diameter in the question, so we don’t have any units for the circumference. But as the circumference is a measure of length, the units will be general length units. We found that the circumference of the circle to two decimal places is 18.85.

So we’ve now seen how to calculate the circumference of a circle by applying this formula. We can also apply these formulae to find the perimeter of other shapes related to circles. For example, suppose we had a semicircle. The perimeter of this figure is composed of two parts: a straight edge, which is the length of the circle’s diameter, and this curved portion, which we refer to as an arc.

As this is a semicircle, the length of this arc will be half the circumference of the full circle. So using the formula we’ve already written down, it will be 𝜋𝑑 over two. If we’re finding the total perimeter of this semicircle as opposed to just the curved length, we must remember to add on the straight edge. So we have that the perimeter of a semicircle, represented by 𝑃, is equal to 𝑑 plus 𝜋𝑑 over two.

We can apply the same reasoning to find a formula for the perimeter of a quarter circle. This time, we have two straight edges, each of which are radii of the original circle. We then have a curved edge which is quarter of the circle’s full circumference, so its length will be 𝜋𝑑 over four. The perimeter of the quarter circle is therefore equal to two 𝑟 plus 𝜋𝑑 over four. And as two 𝑟 is equal to 𝑑, we can also write this as 𝑑 plus 𝜋𝑑 over four.

We can also express each of these formulae in terms of 𝐶, the circumference of the original circle. For the semicircle, the perimeter is equal to 𝑑 plus 𝐶 over two. And for the quarter circle, the perimeter is 𝑑 plus 𝐶 over four. Let’s now consider an example involving portions of circles.

Using 3.14 to approximate 𝜋 and the fact that 𝐴𝐵𝐶𝐷 is a square, calculate the perimeter of the shaded part.

At first, it may seem that the perimeter of this shaded region will be tricky to work out as it’s an unusual shape. If we look carefully though, we see that each of these unshaded portions are quarter circles. The shaded region is enclosed by the curved portions of these four quarter circles. Each of these arc length is one-quarter of the circumference of a circle, but as there are four of them, together they form the full circumference of a circle.

We know that the formula for calculating the circumference of a circle is 𝐶 equals 𝜋𝑑, where 𝑑 represents the circle’s diameter. So the question is, what is the diameter of this circle?

Considering the quarter circle in the bottom left of the figure, we can see that the radius of this circle will be half of the side length of the square. That’s 68 over two. So the radius is 34 centimeters. The diameter of a circle is twice the radius, so the diameter is two times 34; it’s 68 centimeters. In fact, we could have deduced this from the figure without halving and then doubling again. Two of the radii of the quarter circles lie along the side length of the square, so twice the radius is 68 centimeters. And as the diameter is twice the radius, we find again that the diameter is 68 centimeters. We’ve already said that the perimeter of the shaded region is equal to the circumference of the full circle made up of these four quarter circles.

So using the formula 𝐶 equals 𝜋𝑑, we find that the perimeter of the shaded region is 𝜋 multiplied by 68. We’re told in the question to use 3.14 as an approximation for 𝜋. So multiplying gives 213.52. The units for this perimeter are the same as the length units used in the question. So we find that the perimeter of the shaded part using 3.14 as an approximation for 𝜋 is 213.52 centimeters.

Let’s now consider the area of circles. The area is also related to the circle’s dimensions by the constant 𝜋. In this case, the relationship is that the area is equal to 𝜋 multiplied by the radius squared. Area equals 𝜋𝑟 squared. As the radius is half the diameter, we could equivalently write this as 𝜋 multiplied by 𝑑 over two squared, or 𝜋𝑑 squared over four, but this is much less common. Usually, if we’re given the diameter of a circle, we would simply divide this value by two to calculate the circle’s radius and then use the first version of the formula. Let’s now consider an example in which we apply this formula to find the area of a circle.

Using 3.14 as an approximation for 𝜋, find the area of the circle.

We begin by recalling that the formula for finding the area of a circle is 𝜋𝑟 squared, where 𝑟 represents the radius of the circle. From the figure, we identify that the radius of this circle is 12 centimeters because we’ve been given the length of a line segment connecting the center of the circle to a point on the circle’s circumference.

So substituting 12 for 𝑟, we have that the area of this circle is equal to 𝜋 multiplied by 12 squared. We must remember that it is just the 12 we’re squaring. We’re not also squaring 𝜋. So this is equal to 𝜋 multiplied by 144. We’re told that we should use 3.14 as an approximation for 𝜋. So evaluating 3.14 multiplied by 144 gives 452.16. The units for area are square units.

So we found that the area of this circle, which has a radius of length 12 centimeters, is approximately 452.16 centimeter squared.

Just like we considered the perimeters of semi- and quarter circles, we can also find formulae for the areas of these shapes. And in fact, it’s a little more straightforward this time because we just need to divide the area of the full circle by two for a semicircle or by four for a quarter circle. So we have that the area of a semicircle is 𝜋𝑟 squared over two and the area of a quarter circle is 𝜋𝑟 squared over four. Let’s now consider an example in which we relate the area and the perimeter of a semicircle.

The area of the given semicircle is 51.04 centimeters squared. Find the perimeter of the semicircle to the nearest centimeter.

We recall that the formula for finding the area of a semicircle is 𝜋𝑟 squared over two, where 𝑟 is the radius of the circle. 𝜋𝑟 squared gives the area of the full circle, and then we divide it by two. The perimeter of a semicircle is made up of a straight edge, which is the circle’s diameter, and a curved portion, which is half of the circle’s circumference. So the perimeter of the semicircle is equal to 𝑑 plus 𝐶 over two. And the formula for finding the circumference of a circle is 𝜋𝑑, 𝜋 times the diameter. The perimeter of a semicircle can be written as 𝑑 plus 𝜋𝑑 over two.

In order to determine the perimeter of this semicircle then, we first need to calculate its diameter. Using the given area of the semicircle and the formula for calculating the area of a semicircle, we can form an equation: 𝜋𝑟 squared over two equals 51.04. If we can solve this equation to determine the value of 𝑟, we can then calculate the diameter of the circle by recalling that the diameter is twice the radius. We begin by multiplying both sides of the equation by two to give 𝜋𝑟 squared equals 102.08.

Next, we divide both sides of the equation by 𝜋, giving 𝑟 squared equals 102.08 over 𝜋, and then take the square root of each side of the equation. Evaluating gives 𝑟 equals 5.7002 continuing. So we found the radius of the circle, and we can double it to find the diameter, which gives 11.4005 continuing.

We’re now able to substitute this value of 𝑑 into our formula for the perimeter of a semicircle. And we need to make sure we keep this value as exact as possible. You may want to save this value in the memory of your calculator so that you can then recall it directly when evaluating the perimeter. Substituting this exact value of 𝑑 and then evaluating gives 29.3084. We’re asked for the perimeter to the nearest centimeter, so we round down to 29. By recalling the formulae for calculating both the area and perimeter of a semicircle then, we found that the perimeter of this semicircle to the nearest centimeter is 29 centimeters.

Let’s now consider one final example in which we’ll apply our knowledge of circles to a real-world problem.

A goat is tethered by a 10 meter long rope to the corner of a barn. What area of the field can the goat reach?

First, we need to think practically about how the goat can move around this field. If the goat was just tethered somewhere in the middle of the field, then this would be a much more straightforward problem. The goat could roam around anywhere within a circle of radius 10 meters. But because, instead, the goat is tethered to the corner of the barn, at some point the barn is going to get in the way.

Suppose first that the goat walks as far as they can along the top wall of the barn. As the rope is 10 meters long but this wall is 12 meters long, the goat won’t be able to get all the way to the end. So they’ll reach a point 10 meters away from the other corner before they run out of rope. If the rope stays fully stretched, the goat can then start walking in a circle centered at the corner of the barn with a radius of 10 meters.

In fact, we can trace out three-quarters of the circle until the goat arrives at this point here. But let’s think about what happens next. At this point, the rope is going to become caught on the bottom-right corner of the barn, and the goat is going to have to pivot about this point. Five meters of the rope will be used up because it would now be lying flat along the eastern edge of the barn. And that’s the length of this edge. So the goat only has five meters of rope left with which to pivot. The goat can therefore walk in a smaller circle with a radius of five meters until they reach the southern edge of the barn. Of course, all this is what happens if the goat stays right at the end of the rope, so this is tracing out the perimeter of the region the goat can cover.

The goat can also walk everywhere inside this region if the full length of rope isn’t used. So what we found then is that the total area the goat can cover is three-quarters of a circle with radius 10 meters and one-quarter of a circle with radius five meters.

We recall that the formula for calculating the area of a circle is 𝜋 multiplied by its radius squared. So for the three-quarter circle, the area is 𝜋 multiplied by 10 squared multiplied by three-quarters. And for the quarter circle, the area is 𝜋 multiplied by five squared multiplied by one-quarter. Evaluating gives 300 over four 𝜋 and 25 over four 𝜋. The total area of the field that the goat can reach then is the sum of these two values, which is 325 over four 𝜋 square meters.

Let’s now summarize the key points from this video. The circumference of a circle is equal to 𝜋 multiplied by the diameter, or two 𝜋 multiplied by the radius. The area of a circle is 𝜋𝑟 squared. The perimeter of a semicircle is equal to the diameter plus half the circumference, and the area is half the area of the full circle. It’s 𝜋𝑟 squared over two. The perimeter of a quarter circle is equal to the circle’s diameter plus one-quarter of the circle’s circumference. And the area is one-quarter of the area of the full circle. It’s 𝜋𝑟 squared over four. Given either the radius, diameter, circumference, or area of a circle, we can calculate any other of these measures using these formulae.

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