LEARNING MATH WITH TERRAPIN LOGO: LESSONS 9-15

By Kathryn ShaferUpdated fall 2017

www.terrapinlogo.com/logo/weblogo.html

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Table of ContentsLesson 9: The Sine Function................................................................................................ 3

Lesson 10: Obtuse Scalene Triangles...............................................................................6

Lesson 11: Inverse Sine Function....................................................................................11

Lesson 12 Acute Scalene Triangles.................................................................................17

Lesson 13: Parallelograms................................................................................................ 21

Lesson 14: Kites..................................................................................................................... 28

Lesson 15: Transformations............................................................................................33

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Lesson 9: The Sine Function

Purpose: This lesson introduces the sine function, which is used to find an unknown side length in a given right triangle. Essential Question: Given a right triangle, three known angle measurements and one known side length, how can the sine function and the Pythagorean Theorem be used to calculate the missing side lengths?

A calculator can be used to find the sine of different angle measurements. Scientific calculators have more than one mode setting; so make sure that the mode is set to degrees. Some calculators require you to enter the sine button and then the angle’s degrees while others need the degree measurement entered first. Practice with your calculator on the following problems before completing Example 1.

sin(20°) ≈ .342 sin(45°) ≈ .707 sin(60°) ≈ .866 sin(30°) = .500

It is important to round results to at least three decimal places when calculating trigonometric functions that need rounding, because the values are so close together. If the sines of 45°, 46°, 47°, and 48° are rounded to the tenths place they would all have the same value of 0.7. There are two additional trigonometric functions called cosine and tangent. They are commonly found in most high school geometry textbooks and are omitted in this series of lessons.

Example 1: Triangle PQR has angle measurements 30°, 60°, 90°. The short side measures 20 units. Find the measurements of the other two sides.

The sine function is a trigonometric function. The sine of angle P equals the ratio of RQ to PQ.

sin(angle) = opposite side length / hypotenuse length

Figure 9.1

Note that sin(30°) = .5

Computation to find h:sin(30°) = 20/h .5 = 20/h.5 * h = 20

h = 40 units

Since we calculated the hypotenuse of triangle PQR as 40 units, we can now use the Pythagorean Theorem to find the third side length.

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Computation to find m(PR):

Logo Code Completed graphic

402 = 202 + PR2

1600 = 400 + PR2

1200 = PR2

SQRT(1200)= PR34.6 units ≈ PR

fd 34.6rt (180-30)fd 40rt (180-60)fd 20

Figure 9.2

LESSON 9 – TASK 1: Find the missing side lengths for the given triangles. The solutions are on the next page.

LESSON 9 – TASK 2: Find the measures of the remaining side and angle for each of the triangles in TASK 1.

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Figure 9.3

Figure 9.4

Figure 9.5

Figure 9.6

LESSON 9 – TASK 3: Write the Logo code for each of the triangles in Task 1. Test the code on your computer.

A. Figure 9.3 - Start at Pont L with setheading 0 and left turnsB. Figure 9.4 - Start at point B with setheading 55 and right turnsC. Figure 9.5 - Start at point G with setheading 300 and right turnsD. Figure 9.6 - Start at point K with setheading -25 and left turns

Lesson 9 Project

Sketch a right triangle on your paper and measure one acute angle, the hypotenuse and one leg. Calculate the measure of the missing side lengths and angles. Write and test the Logo code.

Task 1 and Task 2 SolutionsFigure 9.3

sin(54°) = m/17sin (54°) * 17 = mm is approximately equal to 13.8 pixelsm ≈ 13.8 pixels

Segment DE is 9.9 pixels longSQRT (172 – 13.82) = 9.9 pixels

Figure 9.4

sin(35°) = x/20sin (35°) * 20 = xx is approximately equal to 11.5 pixelsx ≈ 11.5 pixels

Segment AB is 16.4 pixels longSQRT (202 – 11.52) = 16.4 pixels

Figure 9.5

sin(36°) = 10/t sin (36°) * t = 10t = 10/[sin(36)]t is approximately equal to 17 pixelst ≈ 17 pixels

Segment FG is 13.7 pixelsSQRT (172 – 102) = 13.7 pixels

Figure 9.6

sin(28°) = 50/wsin (28°) * w = 50w = 50/[sin(28)]w is approximately equal to 106.5 pixelsw ≈ 106.5 pixels

Segment IJ is 94 pixelsSQRT (106.52 – 502) = 94 pixels

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Lesson 10: Obtuse Scalene Triangles

Purpose: You will use the sine function and the Pythagorean Theorem to find missing side measurements in obtuse scalene triangles. Logo will be used to check your work.Essential Questions: How can an obtuse scalene triangle be decomposed into two right triangles? How can the sine function and the Pythagorean Theorem be used to create the Logo code for an obtuse scalene triangle?

Please review the material in Lesson 7 before completing this lesson.

Example 1: Obtuse Scalene Triangle XYZ

In this example, you will use both the sine function and the Pythagorean Theorem (PT) to create the Logo code for an obtuse scalene triangle. Triangle XYZ has angles that measure 116°, 27° and 37°. This is clearly an obtuse scalene triangle. In the diagram below, an auxiliary line was drawn perpendicular from X to segment ZY specifically to form two right triangles (∆XPZ and ∆XPY). The purpose of drawing the auxiliary line was to decompose the obtuse scalene triangle into two right triangles. A segment drawn from a vertex perpendicular to the opposite side of a triangle is called the altitude.

In this example, I selected the three angle measures and I made segment XP = 35 units. You might wonder why I had to pick a measure for only one of the sides. If you think about it, the angle measures will determine the general shape of the triangle, but here are an infinite number of similar triangles (of different sizes) that will have angles that measure 116°, 27° and 37° degrees. The length of 35 for the altitude will determine the remaining dimensions.

Figure 10.1

Notice that all of the angle measurements are given for triangle XYZ. Since segment XP is perpendicular to segment ZY, we know that ∠XPZ and ∠XPY are right angles. Thus m(∠PXZ) is 53° and m(∠PXY) is 63° (the sum of the measures of angles in a triangle is 180°). The calculations used to find the missing side lengths and the Logo code are on the next page (start at point Z).

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Computation for triangle XPZ Computation for triangle XPY

sin(37°) = 35/a.602 ≈ 35/a.602 * a ≈ 35a ≈ 35/.602a ≈ 58.1 units

b2 = 58.12 – 352

b2 = 2150.61b ≈ 46.4 units

sin (27°) = 35/c.454 ≈ 35/c.454 * c ≈ 35c ≈ 35/.454c ≈ 77.1 units

d2 = 77.12 – 352

d2 = 4719.41d ≈ 68.7 units

Completed Figure

Figure 10.2

Finish the problem by creating the logo code and graphic.

Logo Code Template Logo Code Completed Graphic

RT (90 – 37)FD 58.1RT (180 – 116)FD 77.1RT (180 – 27)FD (68.7 + 46.4)

RT 53FD 58.1RT 64FD 77.1RT 153FD 115.1

Figure 10.3

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Example 2: Obtuse Scalene Triangle QRS

In this example, I will walk through the steps used to complete calculations for triangle QRS. Let m(QT) = w. Let m(RT) = x. Let m(RS) = y

Figure 10.4

My decision-making process generally depends on the following guidelines:A. Use the theorem that says the sum of the measures of a triangle equals 180°.B. If I am given one side length of a triangle, I use the sine function with the measure of the angle opposite the given side length.C. If I have two side lengths of the triangle, I use the PT with the two given lengths.

Step One – I see there are two right triangles (because of the auxiliary line drawn from point P perpendicular to segment QS). Apply guideline A.

m(∠RQT) = 40° and m(∠SRT) = 65°

Step Two – I see the hypotenuse of triangle RTQ measures 120 pixels. Apply guideline B.

sin (50°) = w/120 sin (50°) * 120 = w 92 pixels ≈ w

Step Three – I see two of the three side lengths for triangle RTQ. Apply guideline C.SQRT (1202 – 922) = x 77 pixels ≈ x

Step Four – I see the leg of triangle RTS measures 77 pixels. Apply guideline B.sin (25°) = 77/y sin (25°) * y = 77 182 ≈ y

Step Five – I see two of the three side lengths for triangle RTS. Apply guideline C.SQRT (1822 – 772) = z 165 pixels ≈ z

Logo Code Logo Code Completed Graphic

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TemplateRT 50FD 120RT (180-115)FD 182RT (180-25)FD (165 + 92)

RT 50FD 120RT 65FD 182RT 155FD 257

LESSON 10 - TASK 1: Write and test the Logo code for the obtuse triangle show here.

Figure 10.6

LESSON 10 - TASK 2: Write and test the Logo Code for the obtuse triangle shown here.

Figure 10.7

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Lesson 10 Project

Write and test the Logo code for the shape shown in figure 10.8. Use the grid to approximate segment lengths.

Hint – there are matching obtuse isosceles triangles.

Figure 10.8

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Lesson 11: Inverse Sine Function

Purpose: This lesson introduces the inverse sine function which is used to find an angle measure in a given right triangle. Essential Question: Given a right triangle and three known side measurements, how can the inverse sine function be used to calculate the missing angle measures?

What would happen if all of the side measurements of a right triangle are given, but the two acute angle measures are unknown? How would you find the two missing angle measures?

Let’s revisit the mathematics in a right triangle. Using what you know about Pythagorean triples, let the side lengths of triangle PAQ be 30, 40 and 50 turtle steps. Since these lengths are the Pythagorean triple 3, 4, 5, times a factor of 10, these side measurements will form a right triangle.

Greek letters are often used in mathematics as a variable for an unknown angle measurement. In the figure below the letters alpha (α) and beta (β) are used to indicate measures of the exterior angles at points P and Q, respectively.

The goal is to write the Logo code (starting at point A in a clockwise direction).

Figure 11.1

Logo Code Template

FD 30RT ? (exterior angle at point P)FD 50RT ? (exterior angle at point Q)FD 40 Try to estimate the angle measures before reading the next section.

Note that you do not know either the interior or the exterior angles at points P and Q. But, since this is a right triangle you can use the sine function to find the triangle’s interior angles. Recall that the sine function uses three values, one of the non-right angles, the opposite side length and the hypotenuse. The set up looks like this:sin(m(∠P)) = 40/50

In words, this reads “the sine of the measure of angle P equals 40 divided by 50.”

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To “undo” the sine function I must take the inverse. The inverse symbol for the sine function is written as “sine to the negative one power” or sin-1.

The next step looks like this:sin-1 [sin(m(∠P))] = sin-1(40/50)

When I take the inverse of the sine function on the left side of this equation, I must do so on the other side of the equation!

Simplify the equation: sin-1 [sin(m(∠P))] = sin-1(40/50) m(∠P) = sin-1 (40/50)

m(∠P) ≈ 53.1° (tenths is sufficient)

Repeat these steps to find the measure of angle Q. sin(m(∠Q)) = 30/50

sin-1 (sin(m(∠Q))) = sin-1 (30/50) m(∠Q) = sin-1 (30/50) ≈ 36.9°

A good check of the calculations would be to add 36.9° and 53.1°, to confirm that the total is 90°. You may have noticed that subtracting 53.1° from 90° could have been done to find the measurement of angle Q, because the sum of the angle measures for any triangle has to be 180°.

Return to the diagram and write in the angle measurements. To complete the Logo code, find the angle of turn for the turtle as you trace the path from point A clockwise around the triangle.

Figure 11.2

Logo Code Template

Let the variables alpha and beta represent the missing angle measurements.

FD 30RT αFD 50RT βFD 40

Solution

Find α Completed Logo Code:

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(180 – 53.1) = 126.9°

Find β (180 – 36.9) = 143.1°

FD 30RT 126.9FD 50RT 143.1FD 40 Figure 11.3

LESSON 11 - TASK 1:Following the process described above, find the missing angle measures and write the Logo code for a right triangle with side measurements 30 cm, 59.9 cm, and 67 cm.

A sketch of this triangle is shown to the right.

I suggest that you start at point E with a starting heading of 350° (move clockwise).

Figure 11.4

Obtuse Scalene Triangle EFG [see figure 11.5 below]In your job as a learner or teacher, you might have to approximate measurements for a triangle that already exists on a worksheet or textbook page. Triangle EFG appears to be an obtuse scalene triangle. Drop a perpendicular segment from point E to segment GF and apply what you know about right triangles. The measure of ∠EPF is 90° because segment EP is perpendicular to segment GF.

Look at triangle EFP and estimate that segment EP is longer than segment PF. Let EP = 25 units, and let PF = 20 units. Be careful to only pick two sides in this triangle!

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Figure 11.5

Calculate the missing measurements in triangle EPF.

252 + 202 = EF2 625 + 400 = EF2

1025 = EF2

32.0 units ≈ EF

m(∠F) = sin-1 (25/32) m(∠F) ≈ 51.4°

m(∠FEP) = 90 – 51.4m(∠FEP) = 38.6°

Figure 11.6

Now find the missing measurements in triangle EPG. We may only pick one measurement. Let GP = 60 units since it appears to be about three times as long as segment PF. (Note that we could have picked one of the missing angle measurements.)

602 + 252 = EG2

3600 + 625 = EG2

4225 = EG2

65 units ≈ EG

m(∠G) = sin-1 (25/65)m(∠G) ≈ 22.6°

m(∠GEP) = 90 – 22.6m(∠GEP) = 67.4°

A quick check indicates that m(∠GEF) totals 106°, forming an obtuse angle. If m(∠GEF) was not greater than 90°, the side measurement of 60 units would have to be increased and the last three calculations repeated.

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Figure 11.7 (in this figure, angle G was rounded up from 22.6 degrees)

To finish this example, the Logo code has been written as follows. Since the turtle will have to turn before drawing side GE, the code begins with a right turn at point G. In figure 11.7, the measures of ∠G and ∠F were rounded to 23° and 51° respectively. The Logo code follows the calculations shown above.

Logo Code Template Logo Code Completed Graphic

RT (90 – 22.6)FD 65RT (180 – 106)FD 32RT (180 – 51.4)FD (20 + 60)

RT 67.4FD 65RT 74FD 32RT 128.6FD 80 Figure 11.8

Common Mistake Number One:Selecting conflicting measurements!

The most common error pattern noticed when working with these types of problems is the learner who picks too many measurements. This will cause a conflict within the shape!

CAUTION: While rounding is not an error, the cumulative effect of repeated rounding will cause the measurements to be off by enough that the turtle will not end up back at the starting point. Therefore, it is a good idea to round to the tenths place for both the side and angle measurements.

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LESSON 11 - TASK 2: Complete the Logo code for the obtuse triangle show here.

Figure 11.9

Lesson 11 Project

Choose the appropriate number of side lengths in the graphic shown below. Use the inverse sine function to find the missing angles. Write and test the logo code.

Figure 11.10

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Lesson 12 Acute Scalene Triangles

Purpose: This lesson focuses in decomposing acute scalene triangles into right triangles. You will use what you know about the Pythagorean Theorem. The 45-45-90 and the 30-60-90 triangles are introduced in this lesson. Knowledge of the sine function is necessary in the problems marked with an asterisk. Essential Question: Can you apply what you have learned about the sum of the measures of a triangle and the Pythagorean Theorem to solve complex problems?

*Example 1 – Acute Scalene Triangle JKMIn this example, you will use both the sine function and the Pythagorean Theorem to write the Logo code for an acute scalene triangle (example 2 does not require the sine function). Begin the Logo code at Point J. Triangle JKM has three acute angles and an altitude has been drawn from point K to segment JM. The calculations have been rounded to the hundredths place to avoid rounding error.

Figure 12.1

Work for Triangle KTJ, on Left Work for Triangle KTM, on Right y2 = 2002 + 2002

y2 = 40,000 + 40,000 y ≈ SQRT(80,000) ≈ 282.84 pixels

Note that y = 200 * SQRT (2) = 282.84The leg measure times the SQRT (2) equals the measure of the hypotenuse. This is true for any 45-45-90 triangle.

sin(60) = 200 / zz = 200 / sin(60°)z ≈ 230.94 pixels

230.942 = 2002 + w2

230.942 - 2002 = w2

115.47 units ≈ w

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There is a special relationship between the three side lengths of triangle KTM. See if you can find it. The solution is at the end of this lesson.

Logo Template Logo DrawingRT 45FD 282.84RT (180 - 75)FD 230.94RT (180 - 60)FD (115.47 + 200)

Figure 12.2

Example 2: Acute Scalene Triangle ABC

An altitude from point C to side AB in triangle ABC is given in figure 12.3. This segment decomposes triangle ABC into two right triangles. Compute the missing angle measures and side lengths, then write/test the Logo code. The work is on the next page.

Figure 12.3

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x2 + 4.32 = 8.22

x2 = 8.22 – 4.32

x2 = 67.24 – 18.49 x = SQRT(48.75) x ≈ 7.0 cm

w2 = 9.62 + 72

w = SQRT( 92.16 + 49) w ≈ 11.9 cm

α = 180° – (85.1° + 58.6°) α = 36.3°

Figure 12.4

To finish the problem, write the logo code starting at point B, moving up and turning counterclockwise or left. Starting at point B is easier than trying to find the initial turn at point C. A scale factor of 10 is used for each of the side lengths, because the side lengths are too small to use in the Logo code.

Calculation Template Simplified Logo Code Logo Drawing

FD 96 + 43LT (180 – 36.3)FD 119LT (180 – 85.1)FD 82

FD 139LT 143.7FD 119LT 94.9FD 82

Figure 12.5

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LESSON 12 - TASK 1: Calculate the three missing side lengths for acute scalene triangle ABC. Write and test the Logo code.

Figure 12.6

Lesson Twelve Project

Sketch an acute scalene triangle on your paper. (This can be tricky)Measure the side lengths with a ruler and the angle measures with a protractor. Use a scale factor of 10 to write/test the Logo code.You may NOT use the same measurements shown in the examples above.

Discussion of Example #1 – The 30 – 60 – 90 triangle

The measure of the longer leg (across from the 60 degree angle) is equal to the short leg times the square root of 3. In the example, we see that (115.47 * SQRT (3)) ≈ 199.99 units or 200 units when rounded.

The hypotenuse is equal to the short leg times 2. In the example, we see that 115.47 * 2 = 230.94 units.

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Lesson 13: Parallelograms

Purpose: This lesson introduces the definitions and properties of parallelograms. The Logo program provides an excellent environment to explore different types of parallelograms. This lesson uses a procedure with three variables to efficiently draw a given parallelogram.

Essential Question: Can you apply your knowledge of interior and exterior angle measures to code and investigate the properties of parallelograms?

Definitions:A quadrilateral is a closed polygon with exactly four sides.A parallelogram is a quadrilateral that has two pairs of parallel sides.A rectangle is a parallelogram with a right angle.A square is a rectangle with all sides of equal length (equilateral).A rhombus is a parallelogram with all sides of equal length (equilateral).Recall that a regular polygon is both equiangular and equilateral. The square (introduced in Lesson #5) is the only regular quadrilateral.

Note that the precise wording of a definition will vary depending on the book you are using. For example, a square could be defined as a rhombus with a right angle.

LESSON 13 – TASK 1:The graphic in Figure 13.1 shows parallelogram ABCD with given side lengths of 135 pixels by 195 pixels and an acute interior angle of 40 degrees. An auxiliary line was drawn from point A perpendicular to line BC. Write and test the Logo code for this shape starting at point A. (check your work at the end of this lesson).

Figure 13.1

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A parallelogram provides a good example of pairs of lines that are cut by a transversal. For example, lines AB and DC are cut by transversal AD (and transversal BC). Lines AD and BC are cut by transversal AB (and transversal DC). It helps to use colored pencils when locating a transversal line!

Definitions and TheoremsVertical Angles are defined as two non-adjacent angles formed by the intersection of two distinct lines. This definition was introduced in Lesson #4.

Vertical Angles Theorem Vertical angles have the same measure (are congruent).

Alternate Interior Angles are located between the pair of lines cut by the transversal but on opposite sides of the transversal.

Alternate Interior Angle Theorem (AIA)If a pair of alternate interior angles are congruent, then the lines are parallel.If a pair of parallel lines are cut by a transversal, then the alternate interior angles are congruent.

*The two conditional statements of the AIA Theorem can be written as one bi-conditional statement. See the end of this lesson for more information.

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LESSON 13 - TASK 2:In Figure 13.2, the measure of angle BAD is 40°. Mark each of the remaining 40-degree angles with a single arc tic mark. An example of the single arc tic mark is shown at angle BAD. Arc tic marks of the same type indicate angle measures that are equal. After marking the diagram, complete the chart below.

In the first row of the chart, write the vertical angle pairs for all of the 40 ° degree angles.In the second row, write the alternate interior angle pairs for all of the 40° degree angles.

Figure 13.2

Vertical Angle Pairs

∠BAD and ∠MAN

Alternate Interior Angle Pairs

∠BAD and ∠TBA

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Types of ParallelogramsFigure 13.3 illustrates six parallelograms drawn on a grid (each section is 100 x 100). The first two parallelograms are rectangles that are not squares. The side dimensions are 50 x 80 pixels. The third shape is an 80 x 80 square. The next two shapes are parallelograms that are not rectangles or rhombi. Their side dimensions are 50 x 80. The last shape is an 80 x 80 rhombus that is not a square. The acute angle in the last three shapes is 60 degrees. It is helpful to create shapes that have certain restrictions when you are studying parallelograms.

Figure 13.3

In this lesson, I will use one procedure, titled ‘PARA’, to efficiently draw each of the six shapes shown above. When you write a program with variables, you must make an assignment for each variable. In the PARA procedure, I made the assignments as follows:

X = length of the first side drawnY = length of the second side drawn (adjacent side)A = measure of the base angle at the lower left corner of the shape

LESSON 13 – TASK 3:Open the editor and name the new procedure PARA (as shown in figure 13.4).

Type X Y A for the input variablesEnter REPEAT 2[FD X RT (180-A) FD Y RT A) in the command windowDefine (save) the procedure

Figure 13.4

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Run the procedure from the listener window by typing para 80 50 90 without any brackets. This is the code for the first rectangle in figure 13.3. Note that when you run the procedure from the listener window you must type three values after the PARA command.

A. Use the Logo code in the third column of the chart below to explore results of the PARA procedure. Try to predict what shape will be drawn by examining the information given in the first two columns.

B. Complete the chart by indicating if your Logo shape matches the definition for each quadrilateral. An example is shown for the first row.

Label each row with the most descriptive category i-iv.i. Parallelogram that is not a rhombus or rectangleii. Rhombus that is not a squareiii. Rectangle that is not a squareiv. Square

Angle Constraint Sample Code Parallelogram Rectangle Square Rhombus Labela=90 x < y Para 50 80 90 Yes Yes No No iii

a=90 x > y Para 80 50 90

a=90 x = y Para 80 80 90

a< 90 x < y Para 50 80 60

a< 90 x > y Para 80 50 60

a< 90 x = y Para 80 80 60

C. Complete each sentence with the missing word Always, Sometimes or Never. You can provide a sketch to the right of the sentence.

i. Parallelograms are _______________ quadrilaterals.

ii. Squares are _______________ parallelograms.

iii. Rhombi are _______________ squares.

iv. Squares are _______________ rectangles.

v. Rectangles are _______________ squares.

vi. Squares are _______________ rhombi.

vii. Rhombi are _______________ rectangles.

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D. Complete the following charts by indicating which properties of diagonals are true or false for each given shape. Use Logo to help you justify your decision (no guessing!). If you do not know what a line of symmetry is, you may look it up on the Internet or in the glossary of a geometry book.

Parallelogram that is not a Rectangle or a Rhombus

Rectangle that is not a Square

Diagonals are congruent

false Diagonals are congruent

Diagonals bisect each other

Diagonals bisect each other

Diagonals are lines of symmetry

Diagonals are lines of symmetry

Rhombus that is not a square SquareDiagonals are congruent

Diagonals are congruent

true

Diagonals bisect each other

Diagonals bisect each other

Diagonals are lines of symmetry

Diagonals are lines of symmetry

E. A Venn diagram can be used to show the relationship between sets of items. Label the regions of the Venn diagram shown in Figure 13.5 with the types of parallelograms listed below. Let the universal set be Polygons.

SquaresRectanglesRhombiQuadrilateralsParallelograms

Figure 13.5

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Lesson Thirteen Project

Sketch three different quadrilaterals that are a:a) Parallelogram that is not a Rectangle or a

Rhombusb) Rectangle that is not a Squarec) Rhombus that is not a square

Measure one angle of each parallelogram and two adjacent sides. Then, write/test the logo code for that parallelogram. Do not use measurements that were used in this lesson.

Solution for Lesson 13 - Task 1rt 50 fd 135 rt 40 fd 195 rt 140 fd 135 rt 40 fd 195

Discussion of a bi-conditional statementAn if-then statement is called a conditional statement. If you have two inverse if-then statements (in the form of if hypothesis then conclusion and if conclusion then hypothesis) that are true, they can be combined into one bi-conditional statement. The format for the bi-conditional statement is to use “if and only if” between the two conditions. The AIA Theorem can be written in either of the following statements.

A pair of alternate interior angles are congruent if and only if the lines are parallel.

Or

A pair of parallel lines are cut by a transversal if and only if the alternate interior angles are congruent.

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Lesson 14: Kites

Purpose: This lesson introduces kites and should be a review of what you have learned about the Pythagorean Theorem. Knowledge of the sine function is necessary to complete Example 2.

Essential Question: Given a convex kite with a few measurements provided, can you use the Pythagorean Theorem and the sine function to calculate the missing angle measures and side lengths? Can you use Logo to test your calculations and code? Can you use these calculations to make generalizations about Kites?

A kite is a quadrilateral with two distinct pairs of congruent adjacent sides. A kite can be either convex or concave. The diagonals of a kite intersect at a right angle. Note that the diagonals decompose the convex kite into two pairs of congruent right triangles. In Figure 14.1, triangles QAP and QAR are congruent. Triangles SAP and SAR are congruent. Two triangles are congruent if the corresponding angles and corresponding segments are congruent (i.e. segment AP is congruent to segment AR; angles QAP and QAR are congruent). You will use this relationship as you write the Logo code for a kite.

Example 1: Kite PQRSFind all of the unknown side and angle measures for the kite in figure 14.1. Write and test the Logo code starting at point P. The solution can be found at the end of this lesson.

Figure 14.1

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*Example 2: Kite ABCD [Need Inverse Sine Function]Find the unknown side and angle measures. Write and test the Logo code starting at point D. Note that Greek letters are used in figure 14.3 to designate angle measures. The unit of measure is pixels.

Try to find the missing measures and Logo code BEFORE you look at the next page!

Figure 14.2 Figure 14.3

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Logo Code Template with Variables

Calculations for Kite ABCD

RT βFD xRT (180 - 2 β)FD xRT (180 - δ - ϒ)FD yRT (180 - 2θ)FD y

Find x:x2 = 402 + 202

x2 = 2000x ≈ 44.7 units

Find y:y2 = 602 + 202

y2 = 4000y ≈ 63.2 units

Find β:sin(β) = 20/44.7 β = sin-1 (20/44.7) β ≈ 26.6°

Find δ :δ ≈ 90° – 26.6°δ ≈ 63.4°

Find ϒ:sin(ϒ) = 60/63.2 ϒ = sin-1 (60/63.2) ϒ ≈ 71.7°

Find θ:θ ≈ 90° – 71.7°θ ≈ 18.3°

Logo Code Template Finished Logo Code Logo Diagram

RT 26.6FD 44.7RT (180 – 2(26.6))FD 44.7RT (180 – 63.4 – 71.7)FD 63.2RT (180 – 2(18.3))FD 63.2

RT 26.6FD 44.7RT 126.6FD 44.7RT 44.9FD 63.2RT 143.4FD 63.2

Figure 14.4

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LESSON 14 - TASK 1: Find the missing measurements for kite ABCD. Write and test the Logo code.

Figure 14.5

Lesson 14 ProjectA. List all of the properties of a kite.B. What is the definition of a rhombus? How is a rhombus similar to and different from a kite?C. Sketch a kite on paper, estimate a few of the measurements. Write and test the Logo code. Add color to your graphic.

You may NOT use the same measurements shown in the examples above.

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Solution to Kite PQRS:Find the unknown angles Find the unknown hypotenuse

90° – 36.9° = 53.1°90° – 61.8° = 28.2°

PS2 = 752 + 1402

PS2 = 5,625 + 19,600PS = SQRT (25225) ≈ 158.8 pixels

Logo Code Template Finished Logo Code Logo Diagram

RT 90 - ___FD ?RT (180 – 2(___))FD ?RT (180 –___ – ___)FD ?RT (180 – 2(___))FD ___

RT 36.9FD 125RT 106.2FD 125RT 65.1FD 158.8RT 123.6FD 158.8

Figure 14.6

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Lesson 15: Transformations

Purpose: This final lesson in Learning Geometry with Logo, introduces the transformations; rotation, reflection and translation. Polar coordinates are introduced in the discussion of a translation vector.Essential Questions: What information is necessary to turn, slide or flip an image and how can these movements (transformations) be coded in Logo?

A rotation requires three pieces of information. 1. A center of rotation2. The degrees of turn3. The direction of the turn

The design shown in figure 15.1 is a flag drawn with right turns. The direction of turn is clockwise.

The rotated flags in figures 15.2 and 15.3 use (0, 0) as the center of rotation. In each figure, the flag with the vertical post (seth 0) is called the preimage.

The rotated flag graphics were created with a main calling procedure titled Rotate. The Rotate procedure needs one input value that will indicate the degree of rotation symmetry. If the input value is 3, the rotation is 120 degrees, because 360/3 = 120 degrees [recall Lesson 1].

Logo Code for Flag Procedure Logo Code for Rotate Procedure

TO FLAG FD 100 RT 135 FD (50 * SQRT (2)) RT 135 FD 50 LT 90 FD 50 LT 180END

TO ROTATE X REPEAT X [FLAG RT 360 / X]END

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Pre-image Rotation 120 degrees Rotation 60 degrees

Figure 15.1 Figure 15.2 Figure 15.3

A reflection requires one piece of information, a line of reflection, often called a mirror. In the first example, the mirror is a vertical line through the point (0, 0) [the y-axis]. The second example uses a slanted line through (0, 0) as the mirror. Note that the line of reflection has a slope of 1.

The reflected flag graphic (figure 15.4) was created with a procedure called Flip. The turtled was slowed down so you can watch as the shape is drawn. Note that I was able to use the Flag routine for the pre-image on the right (first turn is right) but I had to rewrite the Logo code to draw the flag on the left (first turn is left so it moves in a counterclockwise direction). This indicates that a reflection will change the orientation of the points as you move around the shape.

Figure 15.4

Logo Code for Flip Procedure

TO FLIP HOME SLOWTURTLE FD 600 PU SETXY [50 0] PD FLAG PU SETXY [-50 0] PD FD 100 LT 135 FD (50 * SQRT (2)) LT 135 FD 50 RT 90 FD 50 PUEND

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Figure 15.5

TO FLIP2 PU SETXY [0 20] PD FLAG PU SETXY [-20 -20] PD SETH 45 FD (140 * SQRT (2)) PU SETXY [20 0] PD SETH 90 FD 100 LT 135 FD (50 * SQRT (2)) LT 135 FD 50 HTEND

Polar coordinates describe a position on the plane in terms of the distance from the turtle's home in the center of the Graphics window at the point (0, 0) and the angle from 0 degrees. Polar coordinates measure degrees counter-clockwise from a 0 degree heading that is equivalent to 3 o'clock. Retrieved from: Terrapin Logo Help | Commands | setp

New Commands:Think of combining the setheading and the forward commands into one Logo command.

The setp command draws a vector. A vector has two components, magnitude and direction (in polar coordinates). The syntax is setp magnitude direction. The example setp 100 90 will move the turtle 100 steps straight up.

In Logo, there are at least three different command sequences that will draw the same vector. Figures 15.6 and 15.7 illustrate two different vectors.

Turtle Moves up and to the Left Turtle moves up and to the Rightsetp 100 135

seth -45 fd 100

seth 315 fd 100

Setp 100 60

Seth 60 fd 100

Seth -300 fd 100

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Figure 15.6 Figure 15.7

A translation (slide) requires two pieces of information, the direction and length. The two different translations shown here begin at the origin. The translation in Figure 15.8 moves 100 steps at a 0 degree heading (using the polar direction). The translation shown in Figure 15.9, moves 100 steps at a 135 degree heading (using the polar direction).

Translation to the Right Logo Code for Slide Procedure

Figure 15.8

TO SLIDE SETXY [0 0] SETWIDTH 2 FLAG SETP 100 0 SETWIDTH 4 FLAG HTEND

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Translation Up and to the Left Logo Code for Slide 2 Procedure

Figure 15.9

TO SLIDE 2 SETXY [0 0] SETWIDTH 2 FLAG SETP 100 135 SETWIDTH 4 FLAG HTEND

A glide reflection involves the completion of a translation and a reflection. Figure 15.10 shows an isosceles right triangle that was translated 100 units to the right and then reflected over the line y = -25.

Logo code for the Glide ReflectionCS SLOWTURTLETO TRI_1

FD 50RT (180 - 45)FD (50 * SQRT (2))RT (180 - 45)FD 50

END

TO TRI_2RT 90SETPC "REDSETXY [100 0]SETPC "BLACKTRI_1

END

TO TRI_3SETPC "REDSETXY [100 -50]SETPC "BLACKSETH 90TRI_1

END

Figure 15.10

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Example 1 – Tessellated Triangle Figure 15.11 shows an equilateral triangle that has been rotated to form a tessellation. The Logo code was written with a triangle sub-procedure and a main calling procedure.

Tessellation of a Triangle Logo Code for Tri Procedure

Figure 15.11

TO TRIREPEAT 3 [FD 80 RT 120]

ENDLogo Code for Rot_Tri Procedure

TO ROT_TRISETWIDTH 2SETXY [0 0]REPEAT 6 [RT 60 TRI]

END

LESSON 15 - TASK 1: Sketch a rotation tessellation for a square on graph paper, then write and test the Logo code.Challenge – what other polygons could you use? Sketch, code and test your conjecture.

Example 2 - Hexagon TessellationFigure 15.12 shows the tessellation pattern for a regular hexagon. The main procedure draws four rows of nine hexagons (the Logo code is below). I used the sine function to calculate 86.6 steps across the hexagon. See if you can figure out my solution strategy.

LESSON 15 - TASK 2:

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Figure 15.12

Sketch a translation tessellation for a square on graph paper, then write and test the Logo code. What other polygons could you use? Sketch, code and test your conjecture.

Lesson 15 Project

Sketch a design that incorporates at least two of the transformations learned in this lesson (rotation, reflection, translation). Write and test your Logo code.

Logo Code for Example 2 - Hexagon Tessellation

TO HEXSETH 0REPEAT 6 [FD 50 RT 60]

END

TO MOVERTPU SETH 90 FD 86.6 PD

END

TO MOVEUPSETH 30 FD 86.6 SETH 0

END

TO SOLIDPUSETXY [-425 -170] PDREPEAT 9 [HEX MOVERT]PUSETXY [-425 -170]MOVEUP PDREPEAT 9 [HEX MOVERT]PUSETXY [-425 -170]MOVEUP MOVEUP PDREPEAT 9 [HEX MOVERT]PUSETXY [-425 -170]MOVEUP MOVEUP MOVEUP PDREPEAT 9 [HEX MOVERT]

END

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