Algebra through visual patterns, volume 2

4 Ask the students to construct squares, and find their areas and side lengths, using the following Ask the students to examine their squares for relationships between square roots like

The Math Learning Center is a nonprofiit organization serving the

education community Our mission is to inspire and enable individuals todiscover and develop their mathematical confidence and ability We offerinnovative and standards-based professional development, curriculum,materials, and resources to support learning and teaching To find out morevisit us at www.mathlearningcenter.org

The Math Learning Center grants permission to classroom teachers toreproduce blackline masters in appropriate quantities for their classroom use

This project was supported, in part, by the National ScienceFoundation Opinions expressed are those of the authors and notnecessarily those of the Foundation

Prepared for publication on Macintosh Desktop Publishing system

Printed in the United States of America

ISBN 1-886131-60-0

LESSON 1 Tile Patterns & Graphing 1

VOLUME 2

START-UP FOCUS FOLLOW-UP

integral area, and thus the square root of any positive

integer, are developed One of these constructions

leads to the Pythagorean Theorem.

Overview

Students construct squares of

integral areas and establish the

relationship between squares

and square roots

Materials

Centimeter grid paper (see

Appendix), 2-3 sheets per

a single square In the process,they arrive at the PythagoreanTheorem They dissect rect-angles and reassemble thepieces to form squares and, in

so doing, construct squareroots Students examine therelationship between products(quotients, sums) of squareroots and square roots ofproducts (quotients, sums)

Materials

Centimeter grid paper (seeAppendix), 2-3 sheets perstudent

Scissors, 1 pair per student

Start-Up Master 9.1, 1 copyper student and 1 transpar-ency

Focus Masters 9.1-9.2,

1 copy of each per student

Focus Master 9.3, 1 copy perstudent and 1 transparency

Overview

Students solve problemsinvolving squares and squareroots, using the PythagoreanTheorem as necessary Theyrelate the arithmetic mean andthe geometric mean of twopositive numbers to the con-struction of squares

Materials

Follow-Up 9, 1 copy perstudent

roots transparency.

1 Distribute centimeter grid paper

to the students Tell them that

1 square represents 1 unit of area.

For each of the integers 1 through

25 ask them to construct, if possible,

a square whose vertices are grid

intersection points and whose area

is the given integer For each square

they draw, ask the students to

indi-cate its area and the length of its

con-Of the integers 1 through 25, there are 13 for which a squareexists that satisfies the conditions of Action 1 Start-Up Master9.1 attached at the end of this activity shows a square of eacharea A square of area 25 can also be obtained by carrying out a3,4 pattern as described below

One way to obtain a square that fits the conditions is to pick twointersection points as successive vertices In the instance shown

below, one can get from point P to point Q by going 3 units in

one direction and 1 in the other Repeating this 3,1 pattern, asshown, results in a square

Square generated by a 3,1 pattern.

1

1

1 1

continued next page

SQUARES AND SQUARE ROOTS LESSON 9

START-UP BLACKLINE MASTER 9.1

1 1

2 2

20

25 18

2 Discuss with the students why

they can be certain that the 13

integers mentioned in Comment 1

are the only ones in the range 1

through 25 for which squares exist

that satisfy the conditions of Action 1.

1 continuedThe area of this square can be found by subtracting the area of theshaded regions from the area of the circumscribed square (see the

figure) Note that regions A and C combine to form a rectangle of area 3 as do rectangles B and D Thus, the area of the inscribed

square is 16 – 6, or 10 Since the area of the square is 10, thelength of its side is 10 An approximation of 10 can be obtained

by measuring the side of the square with a centimeter ruler

If n is non-negative, the positive square root of n, written n, is the length of the side of a square of area n.

2 One way to see there are only 13 different areas is to notethat if a square is to have area no greater than 25, then thedistance between successive vertices must be less than or equal

to 5 Thus, if P and Q are successive vertices, Q must lie on or within a circle of radius 5 whose center is at P In the sketch, the

13 intersections marked with an x are possibilities for Q that lead

to 13 differently sized squares Any other choice for Q leads to a

square the same size as one of these 13

A

B

x x xx

x

xx

xx

xx

x

x

P

3 Point out to the students that

20 = 2 5 Ask the students to

examine the squares they have

constructed for other relationships

of this type.

3 The square of area 20 is composed of 4 squares of area 5, asshown in the sketch Hence, the side of a square of area 20 is twicethe length of the side of a square of area 5 Thus, 20 = 2 5

The square of area 8 is composed of 4 squares of area 2 and thesquare of area 18 is composed of 9 squares of area 2 Thus,

as the lesson continues

4 Ask the students to construct

squares, and find their areas and

side lengths, using the following

Ask the students to examine their

squares for relationships between

square roots like those discussed in

Action 3.

squares and reassemble the pieces

to form a single square In the

pro-cess, they arrive at the Pythagorean

Theorem They dissect rectangles

and reassemble the pieces to form

squares and, in so doing, construct

square roots Students examine the

relationship between products

(quotients, sums) of square roots

and square roots of products

1 Give each student a sheet of

centimeter grid paper and a pair of

scissors Ask them to construct a

square using a 4, 2 pattern Have

them divide their square into 3

parts as shown below Then ask the

students to cut out these 3 parts

and reassemble them to form 2

adjacent squares, and determine the

length of the sides of these squares.

1 Rotating the right triangles around a vertex, as shown, forms the original square into 2 squares The lengths of the sides

trans-of the squares are 2 and 4 Notice that these are the lengths trans-ofthe legs of the rotated right triangles

Focus Master 9.3, 1 copy per student and 1 trans- parency.

2 Ask the students to construct a

square using a 6,3 pattern and then

use the method of Action 1 to

transform it into a 3 x 3 and a 6 x 6

square.

3 Distribute a copy of Focus

Mas-ter 9.1 to each student Have the

students cut out the square and

triangle Then have them dissect the

square and reassemble it into 2

squares whose sides have length a

and b Discuss.

2 You can ask for volunteers to describe how they proceeded

3 The students can cut out the triangle and use it to divide thesquare into the following regions, which can be cut out andreassembled into the desired squares

Notice that the area of the large square is c2 and the areas of the

2 smaller squares are a2 and b2 Thus, c2 = a2 + b2 Since c is the hypotenuse of the given right triangle and a and b are the legs of

the triangle, we have shown that the square of the hypotenuse of

a right triangle is the sum of the square of its legs This result is

known as the Pythagorean Theorem.

b

b

SQUARES AND SQUARE ROOTS LESSON 9

FOCUS BLACKLINE MASTER 9.1

c 2

b c

4 a) The distance d is the hypotenuse of a right triangle whose legs are 3 and 7 (See the figure.) Thus, d2 = 32 + 72 = 58 Hence

d = 58 ≈ 7.62

a)

c)

5 This is the converse of the dissection done in Action 1 It can

be accomplished by locating the point P on the long side of the

figure, and then cutting and reassembling as shown

4 Ask the students to use the

Pythagorean Theorem to find the

following distances on a coordinate

5 Distribute a copy of Focus

Mas-ter 9.2 to each student Ask them to

dissect the two squares shown

below so they can be reassembled

into a single square Ask for

volun-teers to show how they dissected

d (3,7)

11 10 9 8 7 6 5 4 3 2 1

d (2,10)

–2 1

–1 –2 –3 –4 –5 –6 –7

–3 –1 1 2 3 4 –4 5 6 7 8

cut

cut

a

a b

b P

d 2 = 10 2 + 4 2

= 116; d = 116 ≈ 10.77

6 Place a transparency of Start-Up

Master 9.1 on the overhead Discuss

with the students how squares not

appearing on the transparency can

be constructed using the method of

Action 5 Then distribute a copy of

Start-Up Master 9.1 to each student.

Ask them to dissect and reassemble

squares from Start-Up Master 9.1 to

form squares of area 11 and 21.

7 Tell the students that squares can

also be constructed by dissecting

and reassembling rectangles, for

example, a square of area 21 can be

constructed from a 3 x 7 rectangle.

To begin an investigation of how

square can be dissected and

reas-sembled into squares, distribute

centimeter grid paper to the

stu-dents, ask them to dissect a 9 x 16

rectangle so that it can be

reas-sembled into a square Ask them to

do this by making as few cuts as

possible.

6 Figure 1 shows a square of area 11 constructed from squares

of areas 2 and 9 The students may find it helpful to tape thesquares together in the position shown before dissecting them.Figure 2 shows a square of area 21 constructed from squares ofarea 5 and 16 A square of area 21 can also be constructed fromsquares of areas 1 and 20 or areas 4 and 17 or areas 8 and 13.Note that all the squares of area less than 25 that do not appear

on Start-Up Master 9.1 can be constructed by this method

7 Transforming a 9 x 16 rectangle into a square is simpler thantransforming a 3 x 7 rectangle into a square since, in the formercase, the side of the square is integral

The students will recognize that they need to form a 12 x 12square and they most likely will proceed by cutting the 9 x 16rectangle along grid lines By doing this, a square can be obtained

by making 3 cuts as shown below

a

b 2

9 b

However, only 2 cuts are required:

The students are not likely to find this dissection One way toproceed is to tell the students that you can get a square bymaking only 2 cuts and show them the first cut in silhouette onthe overhead (See the figure below.) Since the students know thelength of the side of the square is 12, they have a clue to thelocation of this diagonal cut

cut 1

cut 2

8 Distribute a copy of Focus

Mas-ter 9.3 to each student Tell the

students to cut off the bottom

portion of the page and set it aside

for use in the next Action Then ask

the students to cut out one of the

rectangles in the top portion of

Focus Master 9.3 and dissect it

using the “diagonal cut” method of

Action 7 and reassemble the

result-ing pieces to form a square Have

them measure their resulting figure

to determine whether or not it is a

square If not, ask them to dissect a

second rectangle so that the

result-ing figure is more “squarelike.”

9 Using a transparency of the

bottom half of Focus Master 9.3 on

the overhead, discuss with the

students how cut lines can be

deter-mined so that the resulting pieces

can be arranged to form a square.

Then ask the students to dissect the

rectangle on the bottom half of

Focus Master 9.3 and reassemble it

ing distance AE somewhat.

9 Note that the dissection in Action 8 would have resulted in a

square if, in the last figure in Comment 8, vertices A and C coincided:

C B

E

A

D

C E

A E

B D

C

D

C E

SQUARES AND SQUARE ROOTS LESSON 9

FOCUS BLACKLINE MASTER 9.3

CUT

If dotted lines are added to the previous figure to show theoriginal location of the pieces, the figure becomes:

Notice that the sum of the two angles at the top of the

figure is a right angle and segments AB and CD have the

same length, since they were originally opposite sides of

a rectangle These observations point to the followingprocedure, illustrated below for determining cut lines

Beginning with rectangle ABCD, the cut lines can be

determined as follows

1 Locate point P on an extension of BC at a distance h from C (See Figure 1.)

2 Extend side CD (See Figure 1.)

3 Place a sheet of paper with a square corner so that

the corner is on the extension of CD and the edges of the paper go through points B and P (See Figure 2.)

4 Let E be the intersection of the edge of the paper and segment AD The first cut line is the segment BE The distance d from A to E is the side of the desired square.

(See Figure 2.)

5 Using a square corner of a sheet of paper as a guide,

starting from a point F on BC at a distance d from C, draw the perpendicular from F to the first cut line This

is the second cut line (See Figure 3.)

A

B

D

C A

B

C B

h

P h

A

sheet of paper

E d

C B

F

sheet of paper

10 Using the procedure described in Comment 6, a 3 x 7rectangle can be dissected and reassembled as shown below toobtain a square of area 21 Note that the dimension of the square

is the distance between vertex C of the rectangle and square corner S Hence, the distance CS is 21

11 If the students start with a 1 by 7 rectangle, an additionalcut is required (Additional cuts are required whenever the length

of a rectangle is more than 4 times its height.)

10 Distribute centimeter grid

paper to the students Tell them that

each square is 1 unit of area Then

ask them to dissect a rectangle of

area 21 and reassemble it to form a

12 If the procedure of Comment 9 is carried out on a

3 x 5 rectangle, the result is a square of area 15 The dimension of

this square is distance CS in the following figure Hence, segment

CS has length 15.

Segment CS has length 15.

13 The students may arrive at their conclusions by considering

specific values for S and T If so, you can ask them if they have reason to believe their conclusions hold for all values of S and T.

The equality of the expressions in a) and b) are useful in ing radicals

simplify-a) A general conclusion can be reached by noting that S T is

the length of the side of a square whose area is ( S T )( S T ) and ST is the length of the side of a square whose area is ST Since ( S T )( S T ) = ( S S )( T T ) = ST, these two squares

have the same area Hence the lengths of their sides are equal

b) Since ( S⁄ T)( S⁄ T) = S S⁄ T T = S⁄T, both S⁄ T and S⁄T arethe lengths of the sides of squares whose area is S⁄T Hence thetwo expressions are equal

line segment of length 15.

expressions on the overhead or

chalkboard:

a) S T ; ST

c) S + T; S + T

For each pair of expressions, ask

the students to determine, for

posi-tive numbers S and T, whether the

first expression is less than, equal to,

or greater than the second

expres-sion Discuss their conclusions.

13 continued

c) As shown in the sketch below, S + T is the combined length

of adjacent squares of areas S and T respectively If the area S where distributed around the square of area T to obtain a square

of area S + T, the length of the side of this square would be less

then the combined lengths of the sides of the original squares

Hence, S + T > S + T , that is the sum of the square roots of

two positive numbers is greater than the square root of theirsum

Some students may be curious about the relationship between

S – T and S – T Assuming S is greater than T so that S – T is

positive, from the last statement in c) above, the sum of the

square roots of S – T and T is greater than the square root of their sum, which is S Thus S – T + T > S and so, subtracting

T from both expressions, S – T > S – T

As determined in a) and b) above, ST = S T and S⁄T = S⁄ T.These results can be used to “simplify radicals.” For example,

b c

A D

CUT

2 A 14 foot ladder is leaning against a wall The foot of the ladder is 6 feet from the wall How far is the top of the ladder above the ground?

(A diagonal of a cube is a line segment that connects two vertices of the cube and goes through its center.)

b) What is the length of a diagonal if an edge is s inches long?

1 square centimeter.

6 5 4 3 2 1

Square S has the same area as rectangle R.

Square T has the same perimeter as rectangle R.

Find the lengths of the sides of squares S and T.

b) (Challenge) The geometric mean of a and b is the length of the side of square S The

arithmetic mean of a and b is the length of the side of square T Determine which of these

means is the greater Explain how you arrived at your conclusion.

4 a) A diagonal d of a cube is the hypotenuse of a

triangle whose legs are an edge of the cube and a

diagonal of the base The edge of a cube is 1 and the

diagonal of a base is 2 Hence, d2 = 12 + ( 2)2 =

b) Square T is larger than square S, hence the arithmetic

mean is greater than the geometric mean One can see

that square T is larger than square S by comparing areas.

Since R and S have the same area, T can be compared

with R.

If R is superimposed on T as shown in the following

sketch, the area of region A is ( a⁄2 + b⁄2)(b⁄2 – a⁄2) while

that of region B is a( b⁄2 – a⁄2) Since a < b, a⁄2 + a⁄2 < a⁄2 + b⁄2 .

Thus, the area of A is greater than the area of B Hence T,

which consists of regions A and C, has a greater area

than R, which consists of regions B and C.

START-UP FOCUS FOLLOW-UP

equations Coordinate graphs of the values of the

arrangements establish a relationship between algebra

and geometry and illustrate solutions to systems of

linear and quadratic equations.

Overview

Students relate graphs of

points that lie along a linear

path to sequences of counting

piece arrangements and

formu-las for the nth arrangement of

such sequences

Materials

Algebra pieces including

frames), 1 set per student

Start-Up Masters 10.1 and

10.3, 1 copy of each per

student and 1 transparency

tions of the nth arrangements

of extended sequences ofcounting piece arrangements

They use Algebra Pieces andgraphs to represent and solvelinear and quadratic equations

of each

Coordinate grid paper (seeAppendix), 2 sheets pergroup and 1 transparency

Algebra Pieces for theoverhead

1⁄4″ grid paper, 2-4 sheetsper student

Overview

Students create sequencesthat satisfy specific conditions.They write formulas for, graph,and solve linear and quadraticequations They complete thesquare to solve quadraticequations

formulas for the nth arrangement of

such sequences.

Start-Up Masters 10.1 and 10.3, 1 copy of each per student and 1 trans- parency of each.

1 Give Algebra Pieces to each

student Write the following chart

on the overhead Have the students

form the –2nd through 2nd and the

nth arrangements of an extended

sequence of counting piece

arrange-ments which fits the data on the

chart (n) indicates the arrangement

number and v(n) is the value of

arrangement n) Ask the students to

write a formula for v(n) Discuss.

–1 (–1) + 0

0 2(0) + 1

1 2(1) + 1

2 2(2) + 1

.

n 2(n ) + 1

.

2 Shown below is one extended sequence that fits the data.

For this sequence, v(–3) = –10, v(3) = 8 and v(4) = 11 Coordinate

points for these 3 cases are circled on the copy of Start-UpMaster 10.1 shown on the left The ordered pair (–4,–13) lies offthe graph

It may be instructive here to review the use of terms such as

horizontal axis, ver tical axis, and origin (The origin is the

coordi-nate (0,0) which is the point of intersection of the horizontal andvertical axes.)

2 Give each student a copy of

Start-Up Master 10.1 Have the

students form the –3rd through 3rd

and the nth arrangements of an

ex-tended sequence of counting piece

arrangements which fits the data

displayed in graphical form on

Start-Up Master 10.1 Ask them to

deter-mine v(–4), v(–3), v(3), and v(4) for

the sequence and, if possible, add

this information to their graph.

LINEAR & QUADRATIC EQUATIONS LESSON 10

START-UP BLACKLINE MASTER 10.1

v (n )

n

10 9 8 7 6 5 4 3 2 1

–1 –2 –3 –4 –5 –6 –7 –8 –9 –10 –11

–2 –7

–1 –4

0 –1

1 2

.

2 5

3 8

3 Ask the students to:

a) record in the space provided on

Start-Up Master 10.1 a formula for

v(n) for the sequence they

con-structed in Action 2;

b) label the coordinates of each

point on the graph;

c) record 4 or 5 observations about

the graph.

Discuss, encouraging observations

about relationships between the

numbers in the students’ formulas

for v(n) and their graphs.

3 a) The arrangements shown in Comment 2 suggest the formula

v(n) = 3n – 1 The students may have other equivalent formulas.

b) The coordinates are labeled on the copy of Start-Up Master10.1 shown on the left below

c) Students’ observations will vary Following are examples ofobservations that students have made about the graph

The points of the graph lie along the path of a straight line.

The points are equally spaced.

Moving left to right, to get from one point to the next, go 1 unit to the right and 3 units up.

The increase in height from point to point is always the same.

There are only points on the graph where n is an integer.

Plotting points for v(n) = 3n – 1 is just like plotting points for v(n) = 3n after shifting the coordinate axes down 1 unit.

In the formula, v(n) = 3n – 1, 3 is the coefficient of n and is the amount the value, v(n), increases as n increases by 1 The constant

term, –1, is the value of the 0th arrangement It indicates wherethe graph intersects the vertical axis When the points on a graph

lie along a straight line, the graph is called linear.

Some students may draw a line connecting the points of the graph,

implying there are arrangements for non-integral values of n The

students may even suggest ways of constructing such ments (see Lesson 11); however, for this extended sequence,

arrange-there are only points on the graph for integral values of n.

The intent throughout this lesson is to promote intuitions aboutrelationships between graphs, formulas, and the sequences ofarrangements the graphs and formulas represent Terminology

such as slope and x- or y-intercept are introduced in Lesson 11,

after extended sequences of arrangements are augmented sotheir graphs are continuous

LINEAR & QUADRATIC EQUATIONS LESSON 10

START-UP BLACKLINE MASTER 10.1

v(n) = 3n – 1

v (n )

n

10 9 8 7 6 5 4 3 2 1

–1 –2 –3 –4 –5 –6 –7 –8 –9 –10 –11

4 Place a transparency of Start-Up

Master 10.2 on the overhead,

re-vealing the top half only (see next

page) Tell the students the

arrange-ment shown is the nth arrangearrange-ment

of an extended sequence of

count-ing piece arrangements Ask the

students to form the –3rd to 3rd

arrangements of this sequence.

5 Distribute a copy of Start-Up

Master 10.3 to each student For

the sequence of Action 4, ask the

students to record a formula for v(n),

construct its graph (see completed

activity below), and record their

observations about the graph

Intro-duce the concept of function, and

the domain and range of a function.

4 Arrangements numbered –3 through 3 are shown on the

bottom half of Start-Up Master 10.2 Recall that a –n-frame contains red tile if n is positive and black tile if n is negative It contains no tile if n is 0.

5 The formula for v(n) can be written in various forms One possibility is v(n) = 4 – n Another is v(n) = 4 + (–n).

In general, a function is a rule that relates 2 sets by assigning each element in the 1st set (called the domain) to exactly one element

in the 2nd set (called the range) Hence, the relationship v(n) = 4 + (–n) is a function that relates the variable n to v(n) so that, for any arrangement number, n, there is exactly one value of the arrangement, v(n) The set of all values for n—in this case, the integers—is the domain of the function v(n) = 4 + (–n) The set of all possible values for v(n)—in this case, also the integers—is the

range of the function (In Lesson 11, students explore functionswhose domains and ranges include all real numbers.)

LINEAR & QUADRATIC EQUATIONS LESSON 10

START-UP BLACKLINE MASTER 10.3

–1 –2 –3 –4

v(n) =

n

10 9 8 7 6 5 4 3 2 1

–1 –2 –3 –4 –5 –6 –7 –8 –9 –10 –11

1

–3 –4

11

…

v(n) =

Observations about the graph:

n

10 9 8 7 6 5 4 3 2 1

–1 –2 –3 –4

2

–6 –8

11

–3 –1 –5

–7

12

nth arrangements of extended

sequences of counting piece

arrange-ments They use Algebra Pieces and

graphs to represent and solve linear

and quadratic equations.

Focus Masters 10.1, 10.2, and 10.4, 1 transparency

of each.

Focus Masters 10.3, 10.5, and 10.6, 1 copy of each per student and 1 trans- parency of each.

1 Arrange the students in groups

and distribute Algebra Pieces to

each student Ask the groups to form

the –3rd through 3rd and nth

ar-rangements of an extended

se-quence of counting piece

arrange-ments for which v(n) = n2 + 2n + 1.

1 Shown below is one possible set of arrangements

One possible nth arrangement is shown below Notice the use of the two frames to represent 2n These frames are needed because they may represent a black n-strip or a red n-strip, depending on the value of n.

2 Tell the students there exists a

sequence of square arrangements

which fits the criterion of Action 1.

Ask the groups to show how their

arrangements from Action 1 can be

formed into a sequence of squares,

using edge pieces to show the values

of the edges of the squares Discuss.

2 Some students may have formed square arrangements inAction 1 If so, you may call the other students’ attention to thesearrangements

Below is a set of square arrangements with edge pieces

Other edges are possible For example, here is another possibilityfor the –3rd arrangement:

The nth arrangement formed in Action 1 can be rearranged to

form a square with 2 possibilities for edges, as shown below

Notice that the figures show that (n + 1)2 and (–n – 1)2 are

equivalent expressions for v(n).

If edge frames were not discussed earlier, you will need to do sonow Edge frames are edge pieces whose color, like that of frames,

differs for positive and negative n Edge frames are obtained by

cutting frames into thirds lengthwise