Estimating the circumference of a circle

A state department of education, as a question on the public-release form of a statewide standardized test, released a question to the public in which students have to choose the value that most closely estimates the circumference of a circle, given its radius and a value to use for π.

The diagram looks something like this:

Diagram of a circle, with a point at the center and a line extending from that point to the circle, labeled 4.5 in., which looks like a radius of the circle

We are asked to choose the value from four options that is closest to the circumference of the circle, using 3.14 for π. This problem involves a main skill (applying a formula) and a secondary skill (estimating). Here, we take “estimating” to mean rounding, which is not really an estimation skill after all, leaving us with only a main skill of applying the formula for circumference of a circle given the radius and a value for π.

If you don’t know the formula, this problem was lifted from a statewide test, and formula sheets are provided by the state. Whether you know it by heart or get it from the formula sheet, the circumference of a circle, C, is twice the radius times π:

C = 2 π r

We have in our catalog at VoxLearn.org a nice list of Web pages that describe the radius of a circle. One of them at Math Is Fun defines the radius of a circle as “the distance from the center to the edge of a circle … half the diameter.” And what do we have in this case? Exactly that: 4.5 inches.

Given the formula for the circumference above, we simply plug in the value for the radius (r) and use our calculator:

2 × 3.14 × 4.5 = 28.26

Therefore, the circumference of the circle is about 28.26 inches (I say “about” because 3.14 is not the exact value for π). Our choices are 14, 20, 28, and 63 inches, with 28 being the closest estimate.

If you thought the right answer was 14, you forgot to multiply by 2, or you thought the line itself represented the diameter of the circle, or you got confused in some other way. The answers of 20 and 63 just don’t make any sense to me. I can’t figure out how you might have gotten those answers.

The Answers

(1) Area represents the amount of space inside a geometric figure. It isn’t really something we can draw directly, so before we can tackle this problem, we have to come up with something we can measure directly, such as the length of something. In order to measure the length of something, we would need to use a ruler, and if you are ever asked to draw something on a test, you should be provided with a few basic tools, such as a ruler.

We know the formula for the area of a circle, either from memory or because we looked on the formula sheet provided by the state. That formula is

A = π r²

Radius, r, is definitely something we can measure, but the problem tells us only the area. We can assume 3.14 for π and plug in the values to find the radius, but first, there’s a little bit of algebra to take care of with the formula.

A = π r²

First, we move π over to the left-hand side:

A / π = r²

Take the square root of both sides (we need r, not r²).

Thus we find that the radius of our circle will be the square root of the quantity, 15 (the area) divided by π. 15 divided by 3.14 is about 4.78, and taking the square root of that, we get 2.19, or somewhere in between 2 and one-eighth and 2 and one-quarter inches.

That means the diameter of the circle needs to be twice that. One way to do this is to draw the circle with a diameter of about 4.38 inches, as shown in red below. (We know that 4.375 is right at the 4 and three-eighths-inch mark, so if we draw it with a diameter right about to there, we’re close enough for a drawing.)

Drawing of a circle on top of a ruler, with the circle measuring somewhere between 4.25 and 4.375 inches in diameter

(2) In the second practice problem, we’re given a circumference of 20 and asked to find the diameter. We know that if the circumference, C, = 2 π r, then we know that

C = π d

where d is the diameter of the circle. Therefore, because we apply the inverse operation when solving the equation for d, we know that

d = C / π

That is, the diameter of a circle in which C = 20 cm is 20 / 3.14 = 6.37 (or thereabouts).

(3) This problem has many, many approaches you could take to solving it. It’s not easy, but if you feel challenged in any way, go ahead and try to find a few more than what I talk about below. I’m just going to show you two ways you might be able to solve this problem.

Method 1: Pythagorean Theorem

Before we do anything else, we need to draw a picture of the figure we’re talking about in the problem. This is always a good idea.

A circle with a radius diagrammed, plus a chord that is perpendicular to and bisects the radius segment, with a right-angle square showing that the chord is perpendicular to the radius

As you see, the chord, shown in red, is a perpendicular bisector of the radius segment (or as close as I could draw it freehand). The question asks us to find the length of the chord if the radius is 2 cm.

It’s not as easy as it looks. First of all, since the chord is a bisector, we know that the portion of the radius to the left of the chord has the same length as the portion of the radius to the right of the chord. Both are 1 cm, since the total radius length was given as 2 cm.

Thus, in our diagram, a = 1.

Next, we know that the length of the green line in our diagram is also 2 cm, since it is a line segment from the center of the circle to the edge of the circle. We know this is, by definition, a radius of the circle, and therefore, this line would have the length given in the problem for the radius, 2.

This means we have a right triangle that follows the Pythagorean Theorem:

a² + b² = c²

We know the hypotenuse of the right triangle is 2, one of the shorter sides is 1, and that means the other side must have a length equal to the square root of 3.

But that other side is only half the chord (the top half), so we double it to find the length of the entire chord. Thus the length of the chord is 2 × the square root of 3, or about 3.46 cm.

Method 2: Congruent triangles

Another possible way to solve this problem would be to notice that the triangle on top of the radius we drew in black is congruent to the triangle that would be made on the bottom. The hypotenuse of each triangle is a radius of the circle, so those sides are congruent. The half of the radius we drew in black forms a side of both triangles, so since those are the same side, they are congruent as well. Finally, the radius in black bisects the chord drawn in green, and therefore both parts of the chord (the upper part and the lower part) are congruent. By the side-side-side law, we have two congruent triangles.

Let x represent the length of one half of the red chord. We know the length of half the radius, a, is 1, and we know the length of the hypotenuse, a radius, is 2. Then, by the Pythagorean Theorem,

Algebraic work showing that length of chord = the square root of 3 times r in the general case

3 COMMENTS

Paul Katula Thursday, November 12, 2009 At 1:35 pm

Illinois Alignment

Grade Level: 8

Illinois Assessment Framework: 7.8.02. (8th grade) Solve problems involving perimeter/circumference and area of polygons, circles, and composite figures using diagrams, models, and grids or by measuring or using given formulas (may include sketching a figure from its description).

Illinois Learning Standard: 7.C.3b (middle school) Use concrete and graphic models and appropriate formulas to find perimeters, areas, surface areas and volumes of two- and three-dimensional regions.
Paul Katula Thursday, November 12, 2009 At 1:47 pm

Try a few on your own

(1) Draw a circle with an area of about 15 square inches, and justify why your circle is correct.

(2) What is the diameter of a circle that has a circumference of 20 cm? Explain how you found your answer.

(3) In a circle with a radius of 2 cm, how long is a chord that is also a perpendicular bisector of a radius? Explain how you found your answer.
Paul Katula Thursday, November 12, 2009 At 2:46 pm

The Answers

(1) Area represents the amount of space inside a geometric figure. It isn’t really something we can draw directly, so before we can tackle this problem, we have to come up with something we can measure directly, such as the length of something. In order to measure the length of something, we would need to use a ruler, and if you are ever asked to draw something on a test, you should be provided with a few basic tools, such as a ruler.

We know the formula for the area of a circle, either from memory or because we looked on the formula sheet provided by the state. That formula is

A = π r²

Radius, r, is definitely something we can measure, but the problem tells us only the area. We can assume 3.14 for π and plug in the values to find the radius, but first, there’s a little bit of algebra to take care of with the formula.

A = π r²

First, we move π over to the left-hand side:

A / π = r²

Take the square root of both sides (we need r, not r²).

Thus we find that the radius of our circle will be the square root of the quantity, 15 (the area) divided by π. 15 divided by 3.14 is about 4.78, and taking the square root of that, we get 2.19, or somewhere in between 2 and one-eighth and 2 and one-quarter inches.

That means the diameter of the circle needs to be twice that. One way to do this is to draw the circle with a diameter of about 4.38 inches, as shown in red below. (We know that 4.375 is right at the 4 and three-eighths-inch mark, so if we draw it with a diameter right about to there, we’re close enough for a drawing.)

(2) In the second practice problem, we’re given a circumference of 20 and asked to find the diameter. We know that if the circumference, C, = 2 π r, then we know that

C = π d

where d is the diameter of the circle. Therefore, because we apply the inverse operation when solving the equation for d, we know that

d = C / π

That is, the diameter of a circle in which C = 20 cm is 20 / 3.14 = 6.37 (or thereabouts).

(3) This problem has many, many approaches you could take to solving it. It’s not easy, but if you feel challenged in any way, go ahead and try to find a few more than what I talk about below. I’m just going to show you two ways you might be able to solve this problem.

Method 1: Pythagorean Theorem

Before we do anything else, we need to draw a picture of the figure we’re talking about in the problem. This is always a good idea.

As you see, the chord, shown in red, is a perpendicular bisector of the radius segment (or as close as I could draw it freehand). The question asks us to find the length of the chord if the radius is 2 cm.

It’s not as easy as it looks. First of all, since the chord is a bisector, we know that the portion of the radius to the left of the chord has the same length as the portion of the radius to the right of the chord. Both are 1 cm, since the total radius length was given as 2 cm.

Thus, in our diagram, a = 1.

Next, we know that the length of the green line in our diagram is also 2 cm, since it is a line segment from the center of the circle to the edge of the circle. We know this is, by definition, a radius of the circle, and therefore, this line would have the length given in the problem for the radius, 2.

This means we have a right triangle that follows the Pythagorean Theorem:

a² + b² = c²

We know the hypotenuse of the right triangle is 2, one of the shorter sides is 1, and that means the other side must have a length equal to the square root of 3.

But that other side is only half the chord (the top half), so we double it to find the length of the entire chord. Thus the length of the chord is 2 × the square root of 3, or about 3.46 cm.

Method 2: Congruent triangles

Another possible way to solve this problem would be to notice that the triangle on top of the radius we drew in black is congruent to the triangle that would be made on the bottom. The hypotenuse of each triangle is a radius of the circle, so those sides are congruent. The half of the radius we drew in black forms a side of both triangles, so since those are the same side, they are congruent as well. Finally, the radius in black bisects the chord drawn in green, and therefore both parts of the chord (the upper part and the lower part) are congruent. By the side-side-side law, we have two congruent triangles.

Let x represent the length of one half of the red chord. We know the length of half the radius, a, is 1, and we know the length of the hypotenuse, a radius, is 2. Then, by the Pythagorean Theorem,