Area Part B: Exploring Area With a Geoboard (50 minutes)

One standard approach to finding the area of a shape is to divide the shape into subshapes, determine the area of each subshape, and then add the areas together. You have used this approach to answer Problems B1 and B2.

A second approach for finding area is to surround the shape in question with another shape, such as a rectangle. For this approach, you first determine the areas of both the rectangle and the pieces of the rectangle that are outside the original shape, and then you subtract those areas to determine the area of the original shape.

Here are three examples of how to surround a right triangle with a rectangle:

You can also divide a triangle into right triangles, form rectangles around each triangle, and then calculate the areas of the rectangles:

In each case, the area of the triangle is half the area of the rectangle that surrounds it.

Problem B3

Use the rectangle method to find the area of each figure:

Does this method work for non-right triangles? For example, how might you find the area of a triangle like Δ BDE below?

Here’s how to do it: First, form rectangle ABCD around Δ BDE. Determine the area of rectangle ABCD and then subtract the areas of Δ ABE and Δ BCD. (Use the rectangle method to determine the areas of these two triangles.) This will give you the area of Δ BDE:

Area of ABCD = 9 square units
Area of Δ ABE = 3 square units
Area of Δ BCD = 4.5 square units
Area of Δ BDE = ABCD – ABE – BCD = 9 – 3 – 4.5 = 1.5 square units

Use the Interactive Activity to work on the geoboard problems in Part B. For a non-interactive version, use an actual geoboard and rubber bands, or print the dot paper worksheet (PDF).

Problem B4

Use this method to find the area of each of the following:

Completely surround the figure with a rectangle; the vertices of the figure should touch the sides of the rectangle. Then find the areas of the outside spaces — the parts of the rectangle that are not inside the figure in question. Some people like to cover all of the rectangle except the section they are working on so as not to be distracted by overlapping lines and shapes.

Take It Further

Problem B5

Construct the following shapes:

1. A triangle with an area of 3 square units
2. A triangle and a square with equal areas (which one has the smaller perimeter?)
3. Triangles with areas of 5, 6, and 7 square units, respectively

Use a guess-and-check strategy: First make a triangle on the geoboard. Next determine its area, using one of the methods mentioned earlier. Adjust the shape of your triangle as needed (i.e., make it larger or smaller), and repeat the process, refining the size of your triangle as you get closer to the desired area.

Most of us use formulas to determine the area of common polygons, such as triangles and rectangles. The formula for the area of a rectangle is A = l • w, where l represents the length of the rectangle and w represents the width.

The formula, for the area of a triangle is b•h where b represents the length of the base of the triangle and h represents the height of the triangle. (Height is the length of the segment from a vertex perpendicular to the opposite side.)

Problem B6

Explain how the formulas below relate to using the geoboard to find an area:

1. Rectangle: A = l • w
2. Right triangle: A =

Think about the rectangle method you used on the geoboard to find the area of a right triangle.

Why does the formula A =   work for triangles other than right triangles? To answer this question, let’s look at parallelograms, since we’ll derive the triangle formula from the formula for the area of a parallelogram.

Problem B7

1. Explain how the area of a parallelogram with height a and base b (as shown below) is found using the formula A = a • b. How does it compare to the area of a rectangle with the same height and base?
   
2. Examine the figure below, in which two congruent triangles are placed together to form a parallelogram. Using this figure, explain how the formula for the area of a triangle relates to the formula for the area of a parallelogram.
3. 

Note the height of the triangle, h, is the length of the line segment perpendicular to the base and adjoining it to the opposite vertex. This is equal to the height of the parallelogram.

Take It Further

Problem B8

• Make a parallelogram out of two trapezoids as follows: Fold a piece of paper in half.
• Draw any trapezoid on the piece of paper.
• Label the top base b1, the bottom base b2, and the height h.
• Cut out two identical copies of your trapezoid, and arrange them to form a single parallelogram.

1. What is the area of this parallelogram?
2. How does the area compare to the area of a single trapezoid? What is the area of one trapezoid?

Note 2

If you are working in a group, take this opportunity to learn from one another by sharing your approaches and solution strategies to Problems B1-B4.

Problem B1

1. The total area is 7.5 square units.
2. The total area is 10 square units.
3. The total area is 8 square units.
4. The total area is 3.5 square units.
5. The total area is 10 square units.

Problem B2

Possible solutions:

1. A = 4 square units
   
2. A = 4 square units
   
3. A = 2 square units
   

Problem B3

1. The area of the rectangle formed is 12 square units, so the triangle’s area is 6 square units.
2. The area of the rectangle formed is 2 square units, so the triangle’s area is 1 square unit.
3. 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.
4. The area of the rectangle formed is 9 square units, so the triangle’s area is 4.5 square units.
5. 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

1. 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.
   
2. 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.
   
3. 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.
   
4. 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.
   
5. 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

1. Here is one possible solution:
   
2. 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.
3. These can all be done using the geoboard:
    A = 5 square units
    A = 6 square units
    A = 7 square units

Problem B6

1. 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.
2. 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

1. 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.
2. 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 =

1. 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):
2. 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

1. 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.
2. 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.