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Why all the fuss about Pythagoras?

Module by: Siyavula Uploaders. E-mail the author

MATHEMATICS

Grade 9

NUMBERS

Module 3

WHY ALL THIS FUSS ABOUT PYTHAGORAS?

INVESTIGATION

1.1 Work in a group but start on your own by drawing three right–angled triangles of different shapes and sizes. Work as accurately as possible. It will be a lot easier if you use squared paper. Now work even more accurately and measure the three sides of each triangle to the nearest millimetre. Complete the first three rows of the table. Now use your calculator to complete the rest of the table.

Table 1
  SYMBOL TRIANGLE A TRIANGLE B TRIANGLE C
Length of the shortest side a .................... .................... ....................
Length of the medium triangle b .................... .................... ....................
Length of the longest side c .................... .................... ....................
Square of the length of the shortest side a2 .................... .................... ....................
Square of the length of the medium side b2 .................... .................... ....................
Sum of the two squares above a2 + b2 .................... .................... ....................
Square of the length of the longest side c2 .................... .................... ....................

1.2 There should be something interesting about the shaded cells of the table. In your group write down carefully what you notice and (if you can) why it happens.

2. Take three lines:

Figure 1
Figure 1 (Picture 79.png)
  • The three given lines can be used to form a right–angled triangle.

Cutting–out:

2.1 Can the three given squares be used to form a triangle?

3 Write a neat summary of the results of the investigation.

end of INVESTIGATION

The Theorem of Pythagoras goes:

  • In a right–angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

The importance of the Theorem of Pythagoras is that we use it in two ways: Firstly, if we know that a triangle is right angled, then we can say something very important about its sides. Secondly, if we know that the three sides of a triangle have a certain relationship with each other, then we also know that the triangle must be right–angled.

CLASS WORK

1 We label triangles as follows:

  • Refer to the sketch alongside.
  • The three vertices (corners) get capital letters. (A, B and C).
  • The sides can be named as the two vertices betweenwhich the side lies (AB, BC and AC), or we can uselowercase letters, each corresponding to the vertexopposite the side (a, b and c).
    Figure 2
    Figure 2 (Picture 81.png)
  • At the moment we are dealing with right–angled triangles, but all triangles are labelled the same way.
    Figure 3
    Figure 3 (Picture 82.png)
  • We also use the same letter to refer both to the name of a side and to its length.
  • E.g. PR = 3,5cm or r = 5cm.ΔPRS means: triangle PRS

REMEMBER always to use a ruler for good sketches!

  • In the following exercise the first problem is always an example.

2 Problem: AEH has a right angle at H.AH = 6cm and EH = 8cm.Draw a sketch (not accurate) of the triangle and use the Theorem of Pythagoras to calculate the length of side AE.

Figure 4
Figure 4 (Picture 83.png)

Solution: Because we know that the triangle has a rightangle, we are allowed to say that AE2 = AH2 + EH2(or: h2 = e2 + a2 )

Substitution:AH2 + EH2 = (6)2 + (8)2 = 36 + 64 = 100 cm2 If E2 = 100 cm2 , then AE must be 10 cm.

2.1 Calculate the length of the third side of these triangles:

2.1.1 ΔDEF with D a right angle and e = 5 mm and f = 12 mm

2.1.2 ΔXYZ with Y a right angle and x = 3 cm and y = 5 cm.

3 Problem: What is the length of the shortest side (b) of the right–angled ΔABC if the other two sides are 6 cm and 9 cm? C is the right angle.

Solution: In a right–angled triangle the longest side is always the hypotenuse: the side opposite the right angle. Now we use the Theorem of Pythagoras in the other form.

  • If b is the shortest side, and C is a right angle, then c must be the longest side. So, you use:

b2 = c2 – a2 (note carefully where b2 is, and that we subtract)

b2 = (9)2 – (6)2 = 81 – 36 = 45 cm2 Calculator time!

b2 = 45. Use the size 12{ sqrt {} } {} – button on the calculator to find the value of b.

  • Your calculator supplies the answer: b = 6,7082039 . . . et cetera. But is this a sensible answer? Discuss whether the approximated (rounded) answer of 6,7cm is good enough for our purposes.

3.1.1 Calculate the length of the hypotenuse of a triangle with both of the other sides equal to 9 cm. (Label the triangle yourself.)

3.1.2 ΔPQR is right–angled and isosceles. Calculate the length of PR, if the hypotenuse is 13,5 cm.

4 Problem: Is ΔGHK right–angled if GK = 24 cm, GH = 26 cm and HK = 10 cm?

Solution: In this problem we know all three the sides’ lengths. If we want to find out whether it is right–angled, we have to confirm whether (hypotenuse)2 = (one side)2 + (other side)2 .

The hypotenuse is always the longest side. We have a very specific method whenever we have to confirm a result. We calculate the left–hand side and the right–hand side of the equation separately. Thus:

  • Left–hand side = (hypotenuse)2 = 262 = 676 cm2
  • Right–hand side = (one side)2 + (other side)2 = 242 + 102 = 576 + 100 = 676 cm2
  • Because the left–hand side and right–hand side come out the same, we can conclude that the triangle is right–angled.
  • Is it possible to know which angle is the right angle? You give the answer!

4.1 Are the triangles with the given side lengths right–angled? Which angle is the right angle?

4.1.1 a = 30 mm, b = 40 mm and c = 50 mm.

4.1.2 p = 8 cm, q = 13 cm and r = 15 cm.

4.1.3 MN = 15,56 cm, and NP = MP = 11 cm.

end of CLASS WORK

HOMEWORK ASSIGNMENT

1 Find the third side of the following triangles:

1.1 ΔABC with C = 90° and b = 5 mm and c = 13 mm

1.2 ΔMNO with O the right angle and m = 6 cm and n = 8 cm.

2 Determine whether the following triangles are right–angled, and which angle is 90° .

2.1 a = 9 mm, b = 11 mm and c 13 cm

2.2 XZ = 85 mm, XY = 13 mm and YX = 86 mm.

end of HOMEWORK ASSIGNMENT

The connection between roots and exponents

CLASS WORK

1 Eight of the equations in this list must be filled in the second row of the table under the equation in the top row where each fits the best.

25=525=5 size 12{ sqrt {"25"} =5} {} ; b=b2b=b2 size 12{b= sqrt {b rSup { size 8{2} } } } {} ; 9=39=3 size 12{ sqrt {9} =3} {} ; 646=2646=2 size 12{ nroot { size 8{6} } {"64"} =2} {} ; a33=aa33=a size 12{ nroot { size 8{3} } {a rSup { size 8{3} } } =a} {} ; 83=283=2 size 12{ nroot { size 8{3} } {8} =2} {} ;

814=3814=3 size 12{ nroot { size 8{4} } {"81"} =3} {} ; 64=864=8 size 12{ sqrt {"64"} =8} {} ; 49=749=7 size 12{ sqrt {"49"} =7} {}

Table 2
Exponential form 2 3 = 8 2 3 = 8 size 12{2 rSup { size 8{3} } =8} {} 9 = 3 2 9 = 3 2 size 12{9=3 rSup { size 8{2} } } {} 25 = 5 2 25 = 5 2 size 12{"25"=5 rSup { size 8{2} } } {} 7 2 = 49 7 2 = 49 size 12{7 rSup { size 8{2} } ="49"} {} 3 4 = 81 3 4 = 81 size 12{3 rSup { size 8{4} } ="81"} {} b × b = b 2 b × b = b 2 size 12{b times b=b rSup { size 8{2} } } {} 64 = 2 6 64 = 2 6 size 12{"64"=2 rSup { size 8{6} } } {} a × a × a = a 3 a × a × a = a 3 size 12{a times a times a=a rSup { size 8{3} } } {}
Root form ........... .......... ............ ............. ............ ............... ............. ....................

2 A radical is an expression with a root sign.How to simplify radicals. Example: 2ab3c×8abc52ab3c×8abc5 size 12{ sqrt {2 ital "ab" rSup { size 8{3} } c times 8 ital "abc" rSup { size 8{5} } } } {}.

  • The most important step is to write the expression under the root sign as simply as possible as products of powers: 2ab3c×8abc52ab3c×8abc5 size 12{ sqrt {2 ital "ab" rSup { size 8{3} } c times 8 ital "abc" rSup { size 8{5} } } } {}= 24a2b4c624a2b4c6 size 12{ sqrt {2 rSup { size 8{4} } a rSup { size 8{2} } b rSup { size 8{4} } c rSup { size 8{6} } } } {}.
  • As we are working with a square root we group them into squares: 24a2b4c624a2b4c6 size 12{ sqrt {2 rSup { size 8{4} } a rSup { size 8{2} } b rSup { size 8{4} } c rSup { size 8{6} } } } {} = 22ab2c3222ab2c32 size 12{ sqrt { left (2 rSup { size 8{2} } ital "ab" rSup { size 8{2} } c rSup { size 8{3} } right ) rSup { size 8{2} } } } {}
  • and remove the root sign, so: 2ab3c×8abc52ab3c×8abc5 size 12{ sqrt {2 ital "ab" rSup { size 8{3} } c times 8 ital "abc" rSup { size 8{5} } } } {} = 22ab2c3222ab2c32 size 12{ sqrt { left (2 rSup { size 8{2} } ital "ab" rSup { size 8{2} } c rSup { size 8{3} } right ) rSup { size 8{2} } } } {} = 22ab2c322ab2c3 size 12{2 rSup { size 8{2} } ital "ab" rSup { size 8{2} } c rSup { size 8{3} } } {} = 4ab2c34ab2c3 size 12{4 ital "ab" rSup { size 8{2} } c rSup { size 8{3} } } {}
  • Another example: 16x2y5×2x2y316x2y5×2x2y3 size 12{ nroot { size 8{3} } {"16"x rSup { size 8{2} } y rSup { size 8{5} } times 2x rSup { size 8{2} } y} } {}
  • Write as products of powers: 16x2y5×2x2y316x2y5×2x2y3 size 12{ nroot { size 8{3} } {"16"x rSup { size 8{2} } y rSup { size 8{5} } times 2x rSup { size 8{2} } y} } {} = 24x2y5×2x2y324x2y5×2x2y3 size 12{ nroot { size 8{3} } {2 rSup { size 8{4} } x rSup { size 8{2} } y rSup { size 8{5} } times 2x rSup { size 8{2} } y} } {} = 25x4y6325x4y63 size 12{ nroot { size 8{3} } {2 rSup { size 8{5} } x rSup { size 8{4} } y rSup { size 8{6} } } } {}
  • This is a third root, so we group into third powers: 25x4y6325x4y63 size 12{ nroot { size 8{3} } {2 rSup { size 8{5} } x rSup { size 8{4} } y rSup { size 8{6} } } } {} = 23x3y6×22x1323x3y6×22x13 size 12{ nroot { size 8{3} } {2 rSup { size 8{3} } x rSup { size 8{3} } y rSup { size 8{6} } times 2 rSup { size 8{2} } x rSup { size 8{1} } } } {} = 2xy23×4x32xy23×4x3 size 12{ nroot { size 8{3} } { left (2 ital "xy" rSup { size 8{2} } right ) rSup { size 8{3} } times 4x} } {}
  • We can now remove the root sign over the part that can be simplified. 2xy23×4x32xy23×4x3 size 12{ nroot { size 8{3} } { left (2 ital "xy" rSup { size 8{2} } right ) rSup { size 8{3} } times 4x} } {} = 2xy24x32xy24x3 size 12{2 ital "xy" rSup { size 8{2} } ` nroot { size 8{3} } {4x} } {}
  • The simplified part is a coefficient; the rest remains as a radical.

Please note that this can be done only if the root contains factors. In other words, it cannot be done with a sum expression.

  • Simplify these radicals as far as possible:

2.1 25a5b3c225a5b3c2 size 12{ sqrt {"25"a rSup { size 8{5} } b rSup { size 8{3} } c rSup { size 8{2} } } } {}

2.2 81x9y12381x9y123 size 12{ nroot { size 8{3} } {"81"x rSup { size 8{9} } y rSup { size 8{"12"} } } } {}

2.3 16a+b216a+b2 size 12{ sqrt {"16" left (a+b right ) rSup { size 8{2} } } } {}

end of CLASS WORK

ENRICHMENT ASSIGNMENT

1 As you may have noticed, most right–angled triangles do not have natural numbers as side lengths. But those that do have whole–number side lengths are very interesting. The well–known (3 ; 4 ; 5)-triangle is one example. These groups of three numbers are called Pythagorean triples.

1.1 Take groups of three numbers from these numbers, trying to find all the Pythagorean triples you can.

3 ; 4 ; 5 ; 12 ; 13 ; 35 ; 36 ; 37 ; 77 ; 84 ; 85

end of ENRICHMENT ASSIGNMENT

There are many different ways to prove the Theorem of Pythagoras.

  • An American mathematician had a hobby of collecting as many different proofs as he could. He eventually published a book of these proofs – over four hundred.

Assessment

Pythagoras ω

Table 3
I can . . . ASs Now I have to . . .
Name triangles correctly 4.4       <
Use Pythagoras to calculate sides 4.4        
Identify right–angled triangles 4.4        
Calculate square roots 4.4       >

good average not so good

Table 4
For this learning unit . . .      
I worked very hard yes no  
I neglected my work yes no  
Did very little yes no Date:

Table 5
Learner can . . . ASs 1 2 3 4 Remarks
Name triangles correctly 4.4          
Use Pythagoras to calculate sides 4.4          
Identify right–angled triangles 4.4          
Calculate square roots 4.4          
Table 6
Critical outcomes 1 2 3 4
Identification and creative solution of problems        
Diagrammatic communication        
Accuracy        
Cooperation in groups        
Table 7
Educator:
Signature: Date:
Table 8
Feedback from parents :
 
 
 
Signature: Date:

Assessment

Table 9
LO 4
MeasurementThe learner will be able to use appropriate measuring units, instruments and formulae in a variety of contexts.
We know this when the learner :
4.1 solves ratio and rate problems involving time, distance and speed;
4.2 solves problems (including problems in contexts that may be used to develop awareness of human rights, social, economic, cultural and environmental issues) involving known geometric figures and solids in a range of measurement contexts by:
4.2.1 measuring precisely and selecting measuring instruments appropriate to the problem;
4.2.2 estimating and calculating with precision;
4.2.3 selecting and using appropriate formulae and measurements;
4.3 describes and illustrates the development of measuring instruments and conventions in different cultures throughout history;
4.4 uses the Theorem of Pythagoras to solve problems involving missing lengths in known geometric figures and solids.

Memorandum

TEST

Where appropriate, give answers accurate to one decimal place.

1. Write the complete Theorem of Pythagoras down in words.

2. Calculate the hypotenuse of ΔABC where A is a right angle and b = 15 mm and c = 20 mm.

3. ΔPQR has a right angle at R. PR = QR. Calculate the lengths of sides PR and QR if QP = 15 cm.

4. Is ΔDEF right–angled if DF = 16cm, DE = 14 cm and EF = 12 cm?

5. What kind of triangle is ΔXYZ if YZ = 24 cm, XY = 10cm and XZ = 26 cm? Give complete reasons.

TEST 3

1. In a right–angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

2. Hypotenuse = a. a2 = 152 + 202 = 225 + 400 = 625 a = 25 Hypotenuse is 25 mm

3. PR2 + QR2 = QP2 2(PR)2 = 152 2(PR)2 = 225 PR2 = 112,5 PR ≈ 10,6 cm

4. LK = 162 = 256

RK = 142 + 122 = 196 + 144 = 340

LK ≠ RK, so ΔDEF is not right–angled.

5. LK = 262 = 676

RK = 242 + 102 = 576 + 100 = 676

LK = RK, so ΔXYZ is right–angled with Y the right angle.

6. Write the following roots in the simplest form:

6.1 1212 size 12{ sqrt {"12"} } {}

6.2 50a3b550a3b5 size 12{ sqrt {"50"a rSup { size 8{3} } b rSup { size 8{5} } } } {}

6.3 64a−14b464a−14b4 size 12{ nroot { size 8{4} } {"64" left (a - 1 right ) rSup { size 8{4} } b} } {}

INVESTIGATION

  • If there is confusion about the a, b, c symbols, do draw a triangle as guidance while learners complete the table Learners with poor measuring skills might need individual support, if they cannot get reasonable answers.
  • Photocopy the squares so that they can be cut out and fitted.

2.1 This is the well-known “proof” of the Theorem of Pythagoras. This work is addressed again when working with similarity.

CLASS WORK

Encourage learners to get into the habit of making realistic sketches.

2.1.1

EF = d

d2 = 122 + 52 = 144 + 25 = 169 = 132

d = 13

2.1.2 XY = 4

3.1.1 hypotenuse 2 = 81 + 81 = 162

hypotenuse ≈ 12,73 cm

3.1.2 PR2 + RQ2 = 2 (PR)2 – isosceles

2(PR)2 = 13,52

PR ≈ 9,55 cm

4. Because GH is the longest side, it has to be the hypotenuse – so K is a right angle.

4.1.1 LK = c2 = 502 = 2500 mm2

RK = a2 + b2 = 302 + 402 = 2500 mm2

LK = RK, triangle is right–angled; C is the right angle.

4.1.2 LK = 225 cm2RK = 64 + 169 = 233 cm2

LK ≠ RK so triangle is not right–angled.

4.1.3 LK = 242,11 cm2

RK = 121 + 121 = 242 cm2

LK ≠ RK but almost!

P is very close to 90°.

HOMEWORK ASSIGNMENT

1.1 a = 12 mm

1.2 o = 10 cm

2.1 No

2.2 Very close –Z ≈ 90°

CLASS WORK

1. 64=864=8 size 12{ sqrt {"64"} =8} {} does not fit the table.

Table 10
c 9 = 3 2 9 = 3 2 size 12{9=3 rSup { size 8{2} } } {} 25 = 5 2 25 = 5 2 size 12{"25"=5 rSup { size 8{2} } } {} 7 2 = 49 7 2 = 49 size 12{7 rSup { size 8{2} } ="49"} {} 3 4 = 81 3 4 = 81 size 12{3 rSup { size 8{4} } ="81"} {} b × b = b 2 b × b = b 2 size 12{b times b=b rSup { size 8{2} } } {} 64 = 2 6 64 = 2 6 size 12{"64"=2 rSup { size 8{6} } } {} a × a × a = a 3 a × a × a = a 3 size 12{a times a times a=a rSup { size 8{3} } } {}
8 3 = 2 8 3 = 2 size 12{ nroot { size 8{3} } {8} =2} {} 9 = 3 9 = 3 size 12{ sqrt {9} =3} {} 25 = 5 25 = 5 size 12{ sqrt {"25"} =5} {} 49 = 7 49 = 7 size 12{ sqrt {"49"} =7} {} 81 4 = 3 81 4 = 3 size 12{ nroot { size 8{4} } {"81"} =3} {} b = b 2 b = b 2 size 12{b= sqrt {b rSup { size 8{2} } } } {} 64 6 = 2 64 6 = 2 size 12{ nroot { size 8{6} } {"64"} =2} {} a 5 5 = a a 5 5 = a size 12{ nroot { size 8{5} } {a rSup { size 8{5} } } =a} {}

2.1 5a2bcab5a2bcab size 12{5a rSup { size 8{2} } ital "bc" sqrt { ital "ab"} } {}

2.2 3x3y4333x3y433 size 12{3x rSup { size 8{3} } y rSup { size 8{4} } ` nroot { size 8{3} } {3} } {}

2.3 4(a+b)

ENRICHMENT ASSIGNMENT

  • Group learners to check one another’s work so that the whole class can decide on the answer.

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