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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 B

See solutions for Problems: B1 | B2 | B3 | B4 | B5 | B6 | B7 | B8 | B9 | B10

Problem B1

The area of the figure is exactly the area of the circle, since no area has been removed or added, only rearranged.

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<< back to Problem B1 (non-interactive version)


Problem B2

The length of the base is one-half the circle's circumference, since the entire circumference comprises the scalloped edges that run along the top and bottom of the figure, and exactly half of it appears on each side. The base length is C/2.

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Problem B3

Since the circumference is 2 • • r, where r is the radius, the base is half of this. The base length is • r.

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Problem B4

As the number of wedges increases, each wedge becomes a nearly vertical piece. The base length becomes closer and closer to a straight line of length • r (or half the circumference), while the height is equal to r. The area of such a rectangle is • r • r, or • r2.

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Problem B5

Here is the completed table:


Radius of Circle

Area of Radius Square

Area of Circle

Number of Radius Squares Needed




36 •

A little more than 3




16 •

A little more than 3




9 •

A little more than 3

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Problem B6


In each case, it takes a little more than three radius squares to form the circle. If using approximations, it should always take around 3.14 of the squares to cover the circle.


The best estimate is somewhere between 3.1 and 3.2, which we know is roughly the value of .

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Problem B7

The formula for the area of a circle is A = • r2. The activity helps one understand that a bit more than three times a radius square is needed to cover the circle. Namely, it illustrates why the formula is • r2.

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Problem B8

Think about a circle with a radius equal to 1 (r = 1). The circumference and the area of this circle are as follows:

C = 2 • 1 • = 2
A = 12 =

Now double the radius to 2 units (r = 2). The circumference and the area of the new circle are as follows:

C = 2 • 2 • = 4
A = 22 = 4

The circumference of the new circle doubled, but the area is multiplied by a factor of 4 (the square of the scale factor). You can replace the 1 with any other number, or with a variable r, to see that this relationship will always hold.

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Problem B9

Scale factor 3:
C = 2 • (3r) = 6 r
A = • (3r)2 = 9 r2

Scale factor 2/3:
C = 2 • (2/3r) = (4/3)r
A = • (2/3r)2 = (4/9)r2

Scale factor k:
C = 2 • (kr) = k(2r)
A = • (kr)2 = k2r2

As with other similar figures, the circumference (or perimeter) of the shape is multiplied by the scale factor, while the area is multiplied by the square of the scale factor. This is also evident in the formulas for each; the circumference formula involves r, while the area formula involves r2.

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Problem B10

A reasonable approximation is 25 cm, but the margin of error will be larger than 0.2 cm. The actual area in square centimeters may be anywhere from (4.8)2 (lower limit) to (5.2)2 (upper limit). Since 4.82, or 23.04, and 5.22, or 27.04, are each about 2 units away from 52, the margin of error for the area is approximately 2 cm2.

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