1.2 Why all the fuss about pythagoras?  (Page 3/3)

2 A radical is an expression with a root sign.How to simplify radicals. Example: $\sqrt{2{\text{ab}}^{3}c\times 8{\text{abc}}^5}$ .

• The most important step is to write the expression under the root sign as simply as possible as products of powers: $\sqrt{2{\text{ab}}^{3}c\times 8{\text{abc}}^5}$ = $\sqrt{2^4{a}^2{b}^4{c}^6}$ .
• As we are working with a square root we group them into squares: $\sqrt{2^4{a}^2{b}^4{c}^6}$ = $\sqrt{{\left(2^2{\text{ab}}^2{c}^3\right)}^2}$
• and remove the root sign, so: $\sqrt{2{\text{ab}}^{3}c\times 8{\text{abc}}^5}$ = $\sqrt{{\left(2^2{\text{ab}}^2{c}^3\right)}^2}$ = $2^2{\text{ab}}^2{c}^3$ = $4{\text{ab}}^2{c}^3$

• Another example: $\sqrt[3]{\text{16}{x}^2{y}^5\times {\mathrm{2x}}^{2}y}$

• Write as products of powers: $\sqrt[3]{\text{16}{x}^2{y}^5\times {\mathrm{2x}}^{2}y}$ = $\sqrt[3]{2^4{x}^2{y}^5\times {\mathrm{2x}}^{2}y}$ = $\sqrt[3]{2^5{x}^4{y}^6}$
• This is a third root, so we group into third powers: $\sqrt[3]{2^5{x}^4{y}^6}$ = $\sqrt[3]{2^3{x}^3{y}^6\times 2^2{x}^1}$ = $\sqrt[3]{{\left(2{\text{xy}}^2\right)}^3\times \mathrm{4x}}$
• We can now remove the root sign over the part that can be simplified. $\sqrt[3]{{\left(2{\text{xy}}^2\right)}^3\times \mathrm{4x}}$ = $2{\text{xy}}^2\sqrt[3]{\mathrm{4x}}$
• 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 $\sqrt{\text{25}{a}^5{b}^3{c}^2}$

2.2 $\sqrt[3]{\text{81}{x}^9{y}^{\text{12}}}$

2.3 $\sqrt{\text{16}{\left(a+b\right)}^2}$

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

 good  average  not so good

Assessment

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 Δ A BC where  A is a right angle and b = 15 mm and c = 20 mm.

3. Δ PQR 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 Δ XYZ 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 . a ² = 15 ² + 20 ² = 225 + 400 = 625 a = 25 Hypotenuse is 25 mm

3. PR ² + QR ² = QP ² 2( PR ) ² = 15 ² 2( PR ) ² = 225 PR ² = 112,5 PR ≈ 10,6 cm

4. LK = 16 ² = 256

RK = 14 ² + 12 ² = 196 + 144 = 340

LK ≠ RK , so Δ DEF is not right–angled.

5. LK = 26 ² = 676

RK = 24 ² + 10 ² = 576 + 100 = 676

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

6. Write the following roots in the simplest form:

6.1 $\sqrt{\text{12}}$

6.2 $\sqrt{\text{50}{a}^3{b}^5}$

6.3 $\sqrt[4]{\text{64}{\left(a-1\right)}^{4}b}$

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

d ² = 12 ² + 5 ² = 144 + 25 = 169 = 13 ²

d = 13

2.1.2 XY = 4

3.1.1 hypotenuse ² = 81 + 81 = 162

hypotenuse ≈ 12,73 cm

3.1.2 PR ² + RQ ² = 2 ( PR ) ² – isosceles

2( PR ) ² = 13,5 ²

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 = c ² = 50 ² = 2500 mm ²

RK = a ² + b ² = 30 ² + 40 ² = 2500 mm ²

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

4.1.2 LK = 225 cm ² RK = 64 + 169 = 233 cm ²

LK ≠ RK so triangle is not right–angled.

4.1.3 LK = 242,11 cm ²

RK = 121 + 121 = 242 cm ²

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. $\sqrt{\text{64}}=8$ does not fit the table.

2.1 ${\mathrm{5a}}^2\text{bc}\sqrt{\text{ab}}$

2.2 ${\mathrm{3x}}^3{y}^4\sqrt[3]{3}$

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.