Problem Solving: Tips For Teachers

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  • 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 .Accessed: 15/06/2014 22:48

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  • Problem totoing Tip> For Tczach

    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.)

    iqj=

    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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  • Ottober I9H6 35

    202 S| 9jn6i aqi jo jaiewued ag 'aouen ' I ifluei -ioq pue o' ag jo situai am |o tuns aqi 'sniji a|

    sbu. pua ipeg o(H M)6ua| dAeg qoea wouoq pue -ueui goee ujojj apis auo jo pasodwoo aje ajn6;j 6u; do; at|j 'snuj. wouoq aq; sj os iuoBxaq ipee aiojj -unsaj ay' jo uuouoq au.1 pue o' aifl jaiuaoi (fe)

    sapa o/vu jo pasoduuoo sj ajnjj agi jo doj am (q) uaAi s; uo!jn|os aiqjs 29 S! ajnu agj jo jajeiuuad -sod auo luaiqojd sjmj o' uoijnios b jb oaujb o' sbm

    aij; 'aoudH ' I si pua qoBd jo mdua' am 09 s; luoi |BJdAds ui pasn aq ubo u^uiqj |BnsjA :suon/os

    Tip Board - 771^05 unequal? (Add) 1

    IO Tips trOm """are? ? 'Agence? (SuW-ac

    ' p a. l^fSnsi (Subtracl)

    9 Louisiana Te* Uraverenv

    1

    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

    ings" yT7j the middle."

    >q7i LZ SEA Method 3: "You start with 1 cube and then add ZZ7I /jjtA / / J~ 'r A 5 nineteen times."

    (a) '') ' f') ') I 'ry(

    'r I ') Method 4: "There are 20 cubes in each of the

    - ^ ''s 5 arms coming out from the middle, s. . u * i ^ u i-* but that counts the center one 5 Discuss with your students s. . what u * a i fourth ^ u build- i-* timeSj SQ , have tQ subtract 4

    ing would look like if it continued the pattern set # R thjs actv wth 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-

    ZA ^ (''ft I CT'' fHI rived at their answers. (For more advanced stu-

    ' ' I- 1 I I k "7

    dents one might ask for the number of cubes in / / a /ryirX] S^^^TQ the 50th, 100th, or nth building.) (c) LL^ I ' I 'X f f ( f Qr

    Solutions to (a): The 20th building will require yqpi -&J) fl^J^j^^ 96 cubes. Here are some ways students used (d) D' (5^3^ fuI'M') '-lr'-jr'-y visual reasoning to arrive at that number:

    '-lr'-jr'-y

    Pan o 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.

    5S^ fo,,ow,ng m* enioy > y " ~ -1

    fo,,ow,ng s/,uation m* enioy > y

    Solution comments I Instigation e diagonal

    comments 0' a 4 /

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    Article Contentsp. 34p. 35p. 36

    Issue Table of ContentsThe Arithmetic Teacher, Vol. 34, No. 2 (October 1986), pp. 1-61Front MatterEditorial: Manuscripts Needed [pp. 2-2]One Point of View: How Many Ways Can You "Understand"? [pp. 3-3]Readers' Dialogue [pp. 4, 53]Building Understanding with Blocks [pp. 5-11]A Problem-solving Component for Junior High School Mathematics [pp. 14-17]Sex-Role Assignments in Elementary School Mathematics Textbooks [pp. 19-21]Estimation and mental Computation [pp. 24-25]Ideas [pp. 26-32]Problem Solving: Tips For Teachers [pp. 34-36]AT Classic: The Revolution in Arithmetic [pp. 38-42]Computer Corner [pp. 44-45]Reviewing and ViewingComputer MaterialsReview: untitled [pp. 46-46]Review: untitled [pp. 46-47]Review: untitled [pp. 47-47]Review: untitled [pp. 47-47]Review: untitled [pp. 47-48]Review: untitled [pp. 48-48]Review: untitled [pp. 48-48]Review: untitled [pp. 48-49]Review: untitled [pp. 49-49]Review: untitled [pp. 49-50]

    From the File [pp. 49-49]Reviewing and ViewingNew BooksFor TeachersFrom NCTMReview: untitled [pp. 50-50]Review: untitled [pp. 50-50]

    From other PublishersReview: untitled [pp. 51-51]Review: untitled [pp. 51-51]Review: untitled [pp. 51-52]Review: untitled [pp. 52-52]

    EtceteraReview: untitled [pp. 52-52]Review: untitled [pp. 52, 28]

    President's Report: Better Teaching, Better Mathematics: Are They Enough? [pp. 54-59]Research ReportThe Cognitive Content of Elementary School Mathematics Textbooks [pp. 60-61]

    Back Matter