1. Pythagoras’s Theorem One of the most important rules you will meet in mathematics – and it’s 2500 years old….. x 2 + y 2 = z 2

2. Aims for today’s lesson: Briefly revisit some angle problems and the idea of similar triangles Briefly revisit some angle problems and the idea of similar triangles Understand a special connection between the lengths of the sides of some triangles; Understand a special connection between the lengths of the sides of some triangles; Know that this connection is called Pythagoras’s Theorem Know that this connection is called Pythagoras’s Theorem Use the Theorem in some problems.. Use the Theorem in some problems..

3. Quick recap Quiz 1. What is the total (sum) of all the exterior angles of ANY polygon? 2. What name do we give to two angles between two parallel lines which form a ‘Z’ shape? 3. What is the sum of all the angles inside a quadrilateral?

4. Quick recap Quiz 4. The diagram shows part of a regular polygon that has 8 sides (an octagon). What are the sizes of the angles x and y? x y

5. Quick recap Quiz 5. Look at the diagrams below. Find the sizes of angles a, b, c and d. ba 28 73 c d 6. What is the name given to angles 73 and c in the diagram above?

6. 1 2 3 4 A B C D E F 5 6 7 8 9 10 11 12 7.Give the three-letter code for angle 3.

7. Quick recap Quiz - ANSWERS 1. What is the total (sum) of all the exterior angles of ANY polygon? (360) 2. What name do we give to two angles between two parallel lines which form a ‘Z’ shape? (Alternating) 3. What is the sum of all the angles inside a quadrilateral? (360)

8. Quick recap Quiz 4. The diagram shows part of a regular polygon that has 8 sides (an octagon). What are the sizes of the angles x and y? x y X = 360 ÷ 8 = 45° and y = 180 – 45 = 135°

9. Quick recap Quiz - ANSWERS 5. Look at the diagrams below. Find the sizes of angles a, b, c and d. ba 28 73 c d 6. What is the name given to angles 73 and c in the diagram above? a = b = 76° c=73 d=107 F-shape, so CORRESPONDING

10. 1 2 3 4 A B C D E F 5 6 7 8 9 10 11 12 6.Give the three-letter code for angle 3. Angle 3 = CFD or DFC

11. SIMILAR SHAPES – a reminder Shapes are similar when one is an ENLARGEMENT of the other Shapes are similar when one is an ENLARGEMENT of the other The enlargement must be achieved by multiplying all the sides by the same amount, called the scale factor The enlargement must be achieved by multiplying all the sides by the same amount, called the scale factor Two shapes are NOT similar if we just add the same amount onto all the sides. Two shapes are NOT similar if we just add the same amount onto all the sides. If the two shapes are identical size, they are called CONGRUENT instead. If the two shapes are identical size, they are called CONGRUENT instead.

12. Examples: 1. Below are two SIMILAR triangles. Work out the length of the sides marked x and y, and the angle a. 4cm 4.5cm18cm 12cm 52° y x a

13. Examples: 1. Below are two SIMILAR triangles. Work out the length of the sides marked x and y, and the angle a. 4cm 4.5cm18cm 12cm 52° y x a The 4cm is enlarged to 12cm, so the scale factor is 3. So x = 4.5 x 3 = 13.5 cm and y is 18 ÷ 3 = 6 cm. Angles never change, so a = 52°

15. Your task: For the following triangles, use a compass and a ruler to draw them accurately, putting the middle sized length as the base; For the following triangles, use a compass and a ruler to draw them accurately, putting the middle sized length as the base; Then, for each triangle, multiply each side length by itself (square it), writing the three answers you get inside the triangle you have drawn. Then, for each triangle, multiply each side length by itself (square it), writing the three answers you get inside the triangle you have drawn. Do you notice anything about four of the triangles and the values you work out? Do you notice anything about four of the triangles and the values you work out? TRIANGLE 1: 3cm, 4cm, 5cm TRIANGLE 2: 5cm, 12cm, 13cm TRIANGLE 3: 10cm, 8cm, 6cm TRIANGLE 4: 3.5cm, 12cm, 12.5cm TRIANGLE 5: 4cm, 7cm, 9.5cm WHICH OF THESE TRIANGLES IS THE ODD ONE OUT ?

16. What you should have found: For any right-angled triangle the longest side (called the HYPOTENUSE) squared equals the total of the other two sides squared!! For any right-angled triangle the longest side (called the HYPOTENUSE) squared equals the total of the other two sides squared!! This rule is called PYTHAGORAS’S THEOREM This rule is called PYTHAGORAS’S THEOREM It won’t work if the triangle has not got a right angle…..Like number 5! It won’t work if the triangle has not got a right angle…..Like number 5!

17. Using the method: EXAMPLE: Work out the length marked x in this triangle: x cm 9 cm 40 cm

18. Using the method: EXAMPLE 2: Work out the length marked x in this triangle (give your answer to 1 d.p) x cm 7.3 cm 12.8 cm

19. Here’s my special method: Step 1: Write the three sides in order of size – like this: 7.312.8x (miss out cm) Step 2: put ‘squares’ onto each number – like this: 7.3 2 12.8 2 x 2 Step 3: put a + and an = in the two gaps – like this: 7.3 2 +12.8 2 =x 2 x cm 7.3 cm 12.8 cm HYPOTENUSE (make sure it’s at the end)

20. Here’s my special method: Step 4: Work out the two parts you can – like this: 53.29 + 163.84 =x 2 Step 5: Now add the first two answers – like this: 217.13 =x 2 Step 6: Now we need to know what number squared actually gives 217.13. For this we need the square root key – it looks like this: √ x = √217.13 x = 14.7 cm to 1dp x cm 7.3 cm 12.8 cm

21. ….and finally (but crucial!): Always CHECK your answer looks right. It has got to be bigger (longer) than the other two sides…. WHY???? Because it’s supposed to be the HYPOTENUSE - which is the longest side!! So x = 14.7 cm is probably OK. x cm 7.3 cm 12.8 cm

22. Now YOU try this one: Question 3: Work out the length marked x in this triangle (give your answer to 1 d.p) x cm 6.6 cm 11.9 cm

23. Answer: 6.611.9x 6.6 2 11.9 2 x 2 6.6 2 +11.9 2 =x 2 43.56 + 141.61 =x 2 43.56 + 141.61 =x 2 185.17 =x 2 x = √185.17 x = 13.6 cm to 1dp : Does it look right? CHECK: Does it look right? x cm 6.6 cm 11.9 cm

24. Now a selection for YOU: Questions: Work out the length marked x in these triangles (give your answer to 1 d.p) x cm 6.6 cm 4.5 cm Q1 3.7 cm 15 cm x cm Q2 Q3 d cm 7.4 cm 6.1 cm

25. ANSWERS: Q1: x = √63.81 = 8.0 cm (to 1 dp) Q2: x = √238.69 = 15.4 cm (to 1 dp) Q3: d = √91.97 = 9.6 cm (to 1 dp)

26. BUT…what if x is NOT the longest side?? EXAMPLE Work out the length marked x in this triangle (give your answer to 1 d.p) x cm 17.3 cm 13.6 cm Well, we stick with the same method as before!!

27. So now for the special method: Step 1: Write the three sides in order of size – like this: x13.617.3 Step 2: put ‘squares’ onto each number – like this: x 2 13.6 2 17.3 2 Step 3: put a + and an = in the two gaps – like this: x 2 +13.6 2 =17.3 2 17.3 cm x cm 13.6 cm HYPOTENUSE (Again, it’s at the end)

28. Here’s my special method: Step 4: Work out the two parts you can – like this: x 2 + 184.96 =299.29 x 2 + 184.96 =299.29 Step 5: Now subtract these two answers – like this: x 2 = 299.29 – 184.96 x 2 = 299.29 – 184.96 x 2 = 114.33 x 2 = 114.33 Step 6: Now we need to know what number squared actually gives 114.33. For this we need the square root key – it looks like this: √ x = √114.33 x = 10.7 cm to 1dp 17.3 cm x cm 13.6 cm

29. ….and finally (but crucial!): Now again CHECK your answer looks right. It has got to be smaller than the hypotenuse… WHY???? Because the HYPOTENUSE is the longest side!! So x = 10.7 cm is probably OK. x cm 17.3 cm 13.6 cm

30. Now YOU try this one: Question 3: Work out the length marked x in this triangle (give your answer to 1 d.p) x cm 12.5 cm 8.9 cm

31. Answer: x8.912.5 x 2 8.9 2 12.5 2 x 2 +8.9 2 =12.5 2 x 2 + 79.21 =156.25 x 2 =156.25 – 79.21 x 2 =156.25 – 79.21 x 2 =77.04 x 2 =77.04 x = √77.04 x = 8.8 cm to 1dp : Does it look right? CHECK: Does it look right? x cm 12.5 cm 8.9 cm

32. Now a selection for YOU: Questions: Work out the length marked x in these triangles (give your answer to 1 d.p) x cm 9.6 cm 14.5 cm Q1 4.1 cm 13.7cm x cm Q2 Q3 t cm 17 cm 24 cm

33. ANSWERS: Q1: x = √118.09 = 10.9 cm (to 1 dp) Q2: x = √170.88 = 13.1 cm (to 1 dp) Q3: d = √287 = 16.9 cm (to 1 dp)

34. Now to recap……. Look at the problem on the next slide Look at the problem on the next slide It’s like what you could get at Foundation level……. It’s like what you could get at Foundation level……. And shows how you might be asked to apply Pythagoras’s theorem And shows how you might be asked to apply Pythagoras’s theorem

35. A GCSE-type question: A boat leaves a harbour and sails due North for 18Km, then turns East and sails for a distance of 25km. How far is the direct route back to the Harbour? x km 25km 18km

36. Solution: x km 25km 18km 1825= x 18 2 25 2 = x 2 18 2 + 25 2 = x 2 324 +625= x 2 949= x 2 x = 30.8km Does the answer LOOK right??

37. Thank-you to Pythagoras!!