Problem Solving: Tips For TeachersAuthor(s): Phares G. O'Daffer, Gene Maier and Ted NelsonSource: The Arithmetic Teacher, Vol. 34, No. 2 (October 1986), pp. 34-36Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/41192975 .

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Problem totoing Tip> For Tczach<zr>

34 Arithmetic Teacher

^ | Strategy Spotlight ~"

Use Visual Reasoning

Visual reasoning is the process of forming conclu- sions based on what one sees or imagines. It can provide straightforward solutions that are insightful and convincing. Consider the following problem.

Problem: Fifty squares, each with edge 1 unit long, are placed side by side in a row. What is the perimeter of the resulting figure?

It is impractical to draw a sketch of 50 squares. However, one can draw a sketch of, say, 5 squares placed side by side and imagine what happens as more squares are added. (Good imagers may be able to see the squares in their mind's eye and find it unnecessary to sketch them.)

unnecessary to them.)

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• One person saw that the rectangles formed as squares are placed side by side have tops and bottoms whose lengths are equal to the number of squares. Their ends each have length 1 . Hence, if 50 squares are placed side by side, the resulting rectangle has a top and a bottom of length 50 and ends of length 1 . Hence its perimeter is (2 x 50) + 2, or 102.

Edited by Phares G. O'Daffer Illinois State University Normal, IL 61761

Prepared by Gene Maier and Ted Nelson Portland State University Portland, OR 97207

• Another person observed that each of the "in- side" squares contributes 2 sides to the perime- ter of the rectangle while the two "outside" squares contribute 3. If 50 squares are placed side by side, the resulting rectangle has 48 in- side squares and 2 outside squares. Hence its perimeter is (48 x 2) + (2 x 3), or 102.

• A third person noted that the perimeter of an in- dividual square is 4, and hence the perimeter of 50 individual squares is 200. However, when 50 squares are placed side by side, 49 pairs of edges come together and are not part of the perimeter of the resulting rectangle. Hence the perimeter of this rectangle is 200 - (2 x 49), or 102.

• A fourth person saw that each time a square is added, the length of the top and bottom of the rectangle each increase by 1 while the dimen- sions of the ends are unchanged. Hence adding a square increases the perimeter of the rectan- gle by 2. Since a single square has perimeter 4, adding 49 squares to it results in a rectangle whose perimeter is 4 + (2 x 49), or 102.

All these solutions are conclusions based on visual information and are thus examples of visual reasoning. They also illustrate the variety of ways in which the same situation can be viewed. The following problem extends the ideas considered here for squares, tri- angles, and hexagons.

Problem: Use visual reasoning to find the perimeter of the figures obtained when (a) 50 equilateral trian- gles are placed in a row and (b) 50 regular hexagons are placed in a row. Assume that the lengths of the sides of the triangles and hexagons are 1 . ^ coa:

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1 ' ^It^^^t Q Try This Method 1 : 'There are 20 cubes stacked up in

the middle and there are 19 in each Here is an activity to help students develop arm. So there are 20 + (4 x 19)." their visual reasoning abilities. Method 2 'There are (5 x 19) + 1 because • Use cubes to construct these three "build- each arm has 19 and there is 1 in

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ing would look like if it continued the pattern set # R thjs act¡v¡ w¡th Qther bu¡|d¡ by the 1 st three buildings. Then ask your stu- tems Here ̂ SQme s stions dents to imagine building the 20th building and ¿3» to determine the number of cubes needed to r-* |U construct it. Ask them to describe how they ar- ' ' I-

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rived at their answers. (For more advanced stu- ' ' I- ̂ 1 I I k

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Pan oí the Tip Board is reserved for techniques that you've found useful in teaching problem solving in your class. Send your ideas to the editor of the section. (continued On next page)

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36 Arithmetic Teacher

Q Classroom Climate

• Create an atmosphere in which students feel • Emphasize that many ways can be found to comfortable talking about their ideas for solv- approach a problem. Urge students to seek ing a problem. alternate solutions.

• Accept all ideas. Stress that every idea has • Encourage students to listen to others as merit, including those that don't lead to solu- they talk about their ideas for solving a prob- tions. Finding out why an idea doesn't work lem. Listening to others expands one's reper- provides useful information. tory of problem-solving approaches.

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