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Learning Math Home
Measurement Session 7: Solutions
Session 7 Part A Part B Homework
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Session 7 Materials:

A B 


Solutions for Session 7, Part A

See solutions for Problems: A1 | A2 | A3 | A4 | A5 | A6 | A7 | A8 | A9 | A10

Problem A1

Here is the completed table:


Design 1

Design 2

Design 3

Diameter of the Circle

2 cm

4 cm

6 cm

Perimeter of the Hexagon

6 cm

12 cm

18 cm

Perimeter of the Square

8 cm

16 cm

24 cm

Approximate Circumference of the Circle

6.3 cm

12.6 cm

18.9 cm

<< back to Problem A1


Problem A2

The measurements stay in scale. In all three, the diagonal of the hexagon is twice the length of the hexagon's side. Also, as we move from one design to the next, the length of each side of the inscribed hexagon increases by 1; the length of each side of the inscribed hexagon is equal to the radius of the circle (as shown by the inscribed equilateral triangles):

The length of each side of the square is the same as the diameter of the circle inscribed within; the ratio of the length of the diameter of the circle to the length of one side of the hexagon is 2/1 for all three designs.

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Problem A3


The perimeter of the hexagon is three times the diameter of the circle, and the perimeter of the square is four times the diameter of the circle.


The circumference of the circle is between these two values, and closer to the hexagon's perimeter.


The circumference appears to be between 3.1 and 3.2 times larger than the diameter. If a circle had a diameter of 7 cm, you might predict its circumference to be somewhere near 22 cm. As we explore this further, we will see that the relationship between the circumference and diameter is a constant value.

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Problem A4


Answers will vary.


The measured ratio of C/d should be approximately the same for all circular objects. It seems that the relationship between diameter and circumference is linear, and there is some number k so that C = k • d for every circle's circumference and diameter. For now, we can say that this number is just slightly larger than 3.

<< back to Problem A4


Problem A5


Answers will vary. The points should roughly form a straight line. If you were to place a line of best fit onto your scatter plot, the line would be y = 3.14 • x.

Notice that the ratio C/d is about 3.14 no matter what the size of the circle. This is called a constant ratio since the value is constant, regardless of the circle. Constant change is represented by a straight-line graph and is sometimes referred to as a linear relationship. If the ratio between circumference and diameter differed for every circle, the graph would not be a straight line.


This suggests that the diameter and circumference are in direct variation; that is, the circumference is a direct multiple of the diameter. Note that since you measured the circumference and diameter, there are likely to be measurement errors which will affect the graphed data.

<< back to Problem A5


Problem A6

Answers will vary, but should be close to 3.14 (an approximation for ). Finding the mean minimizes any measurement errors in the calculations of Problem A5.

<< back to Problem A6


Problem A7

We've seen that = C/d, where C is the circumference and d is the diameter of a circle. We can multiply both sides of the equation by d to get a new equation as C = d. Because the diameter of a circle is always twice its radius, we can write the new equation as C = • 2r, which is what we wanted.

<< back to Problem A7


Problem A8

These forms make calculations involving easier by using an approximation. In cases where there may already be measurement error, it doesn't make sense to use an overly accurate version of . Fractions like 22/7 or decimals like 3.14 do the job nicely in different situations.

In practical terms, it is impossible to buy, for example, a length of fencing that measures 4. In applications, we often want an approximation that we can measure and work with. In mathematics problems, however, it is almost always preferable to use the symbol .

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Problem A9

One or the other may be rational, but not both. If they were both rational, their ratio (which is ) would also have to be rational, which it is not. A circle may have a diameter of exactly 12 cm with an irrational circumference, or a circumference of exactly 100 m with an irrational diameter.

They can, however, both be irrational.

<< back to Problem A9


Problem A10

This answer provides the only way to write the answer exactly. Additionally, it is easier to perform arithmetic with 4 than with a decimal approximation of it. An approximation may be substituted later if needed.

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