Solutions for Session 6, Part B

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

Problem B1

 a. The total area is 7.5 square units. b. The total area is 10 square units. c. The total area is 8 square units. d. The total area is 3.5 square units. e. The total area is 10 square units.

Problem B2

Possible solutions:

 a. A = 4 square units b. A = 4 square units c. A = 2 square units

Problem B3

 a. The area of the rectangle formed is 12 square units, so the triangle's area is 6 square units. b. The area of the rectangle formed is 2 square units, so the triangle's area is 1 square unit. c. This time, two rectangles must be formed. The one on top has an area of 6 square units, and the one on the bottom has an area of 4 square units, for a total area of 10 square units; therefore, the area of the figure is 5 square units. d. The area of the rectangle formed is 9 square units, so the triangle's area is 4.5 square units. e. Dividing the kite into four right triangles (formed by the kite's diagonals) and surrounding them by rectangles, gives an area of the rectangles of 12 square units. So the kite's area is 6 square units.

Problem B4

 a. The surrounding rectangle has an area of 6 square units, and there are two triangles with a total area of 3 square units to "subtract," so the figure's area is 3 square units. b. The surrounding rectangle has an area of 16 square units, and there are three triangles with a total area of 9.5 square units to "subtract," so the figure's area is 6.5 square units. c. The surrounding rectangle has an area of 12 square units, and there are two triangles with a total area of 8 square units to "subtract," so the figure's area is 4 square units. d. The surrounding rectangle has an area of 9 square units, and there are three triangles with a total area of 4.5 square units to "subtract," so the figure's area is 4.5 square units. e. The surrounding rectangle has an area of 9 square units, and there are three triangles with a total area of 5 square units to "subtract," so the figure's area is 4 square units.

Problem B5

a.

Here is one possible solution:

b.

Here is one possible solution:

In this example, the area of each is 8. Other solutions are also possible -- see, for example, Problem B2 (a) and (b). Regardless of what the area is, though, the square will always have the smaller perimeter.

c.

These can all be done using the geoboard:

 A = 5 square units A = 6 square units A = 7 square units

Problem B6

 a. On the geoboard, the area of a rectangle can be found by counting the unit squares or multiplying the length by the width, which is the same as the formula A = l • w. b. It's easy to visualize this in the case of the right triangle. Using the rectangle method to find the area of a right triangle with base b and height h, you enclosed the triangle in a rectangle with an area equal to b • h. You then divided the area in two, since one right triangle has half the area of a rectangle (in other words, two right triangles completely fill a rectangle). This is the same as the formula A = .

Problem B7

 a. The two figures have the same area since they are made up of the same shapes (take the parallelogram and transform it into a rectangle as shown below). The base of the rectangle is made up of the same shapes that form the base of the parallelogram, so it is still the same length (b). The height is the same as well. The area of the rectangle is base multiplied by height, or a • b, so the parallelogram's area must also be a • b. b. Any two identical triangles (isosceles, equilateral, scalene, etc.) can be put together to form a parallelogram. The base and height of the triangle and parallelogram will be equal. Therefore, to find the area of one of the triangles, divide the area of the parallelogram by half: A = Alternatively, you can arrive at the same result using what's known as the midline theorem. As you can see in the picture below, we've transformed a triangle into a parallelogram by cutting along the MP segment (M and P are the midpoints of their respective sides): MP, called the midline, is parallel to the base of the triangle and half as long; it divides the height of the triangle in half. The new parallelogram has the same base length as the triangle but half of its height, and, like the triangle, its area can be found with the formula (half the product of the base and the height). To learn more about the midline theorem, go to Learning Math: Geometry, Session 5.

Problem B8

 a. The area of the parallelogram is (b1 + b2) • h, since both the top and the bottom of the trapezoid comprise the base of the parallelogram. b. The areas of the two trapezoids add up to (b1 + b2) • h, so the area of each trapezoid is . Alternatively, you could find the area of one of the trapezoids by transforming it into a rectangle. The formula will be the same. To learn more about making such transformations, go to Learning Math: Geometry, Session 5, Part B.