 Teacher resources and professional development across the curriculum

Teacher professional development and classroom resources across the curriculum            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.   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.   Problem B3 Since the circumference is 2 • • r, where r is the radius, the base is half of this. The base length is • r.   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.   Problem B5

Here is the completed table:  Circle  Area of Radius Square Area of Circle Number of Radius Squares Needed  1 6 36 36 •  A little more than 3 2 4 16 16 • A little more than 3 3 3 9 9 • A little more than 3    Problem B6

 a. 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. b. The best estimate is somewhere between 3.1 and 3.2, which we know is roughly the value of .   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.   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.   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(2 r) A = • (kr)2 = k2 r2 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.   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.     