1.  Only works in right angled triangles  Nothing to do with angles

2.  The hypotenuse is the longest side in a right angled triangle.  It is always the side opposite the right angle. h y p o t e n u s e

3. hypotenuse

4.  The area of the square drawn on the hypotenuse is equal to the sum of the area of the squares drawn on the other two sides  c 2 = a 2 + b 2 c b a

5. You can visualise the theorem b ac c2 =c2 =c2 =c2 = a2a2a2a2 + b 2

6. 18 10? c 2 = 10 2 + 18 2 Find the missing side. 100 324 100 + 324 = 424 c 2 = 424 c = √424 c = 20.6 (1.d.p.)

7. Give your answers to 1 d.p. 8cm 10cm 11m 7m 24km 5km c 2 = a 2 + b 2 c 2 = 10 2 + 8 2 c 2 = 100 + 64 c 2 = 164 c = √164 c = 12.8cm (1.d.p.) c 2 = a 2 + b 2 c 2 = 11 2 + 7 2 c 2 = 121 + 49 c 2 = 170 c = √170 c = 13.0m (1.d.p c = 13.0m (1.d.p.) c 2 = a 2 + b 2 c 2 = 24 2 + 5 2 c 2 = 576 + 25 c 2 = 601 c = √601 c = 24.5km (1.d.p.)

8. 18 10 ? a 2 = 18 2 - 10 2 Find the missing side. 100 324 – 100 = 224 324 a 2 = 224 a = √224 a = 15.0 (1.d.p.) 18 2 = a2a2a2a2 + 10 2

9. Give your answers to 1 d.p. 20 cm 12 cm 11m 17m 24km 5km a 2 = c 2 - b 2 a 2 = 20 2 - 12 2 a 2 = 400 - 144 a 2 = 256 a = √256 a = 16 cm a 2 = c 2 - b 2 a 2 = 17 2 - 11 2 a 2 = 289 - 121 a 2 = 168 a = √168 a = 13.0m (1.d.p a = 13.0m (1.d.p.) a 2 = c 2 - b 2 a 2 = 24 2 - 5 2 a 2 = 576 - 25 a 2 = 551 a = √551 a = 23.5km (1.d.p.)

10. Navigation problems are often solved using Pythagoras’ Theorem. N S EW

11. A plane leaves an airport and travels 32km west then it turns and travels 41km north. It develops a problem and has to return to the airport. How far is it? Step 1. Draw a diagram 32km Airport ? Step 2. Use Pythagoras 41km c 2 = a 2 + b 2 c 2 = 32 2 + 41 2 c 2 = 1024 + 1681 c 2 = 2705 c = √2705 c = 52.0km (1.d.p.)

12. Problems involving isosceles triangles are often solved using Pythagoras’ Theorem.

13. Draw perpendicular and mark lengths Use Pythagoras theorem bb a c c2 = a2 + b2

14. A roof on a house that is 6 m wide peaks at a height of 3 m above the top of the walls. Find the length of the sloping sides of the roof. 3 m 6 m c 2 = a 2 + b 2 c 2 = 3 2 + 3 2 c 2 = 9 + 9 c 2 =18 c = √18 c = 4.2 m (1.d.p.) ?? 3 m c Step 1. Draw a diagram Step 2. Use Pythagoras

15. draw a diagram for the problem that includes a right-angled triangle draw a diagram for the problem that includes a right-angled triangle label the triangle with the length of its sides from the question label the triangle with the length of its sides from the question label the unknown side ‘x’ label the unknown side ‘x’ if it’s the hypotenuse, then if it’s the hypotenuse, then “SQUARE, SQUARE, ADD, SQUARE ROOT” “SQUARE, SQUARE, ADD, SQUARE ROOT” if it’s one of the shorter sides, then if it’s one of the shorter sides, then “SQUARE, SQUARE, SUBTRACT, SQUARE ROOT” round your answer to a suitable degree of accuracy round your answer to a suitable degree of accuracy

16. Is this triangle possible?