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G 22 Pythagoras’ Theorem

GCSE Maths Geometry & Measures. G 22 Pythagoras’ Theorem. Subject Content References: G2.1, G2.1h. Pythagoras’ theorem states: For any right-angled triangle , the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides. Example.

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G 22 Pythagoras’ Theorem

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  1. GCSE MathsGeometry & Measures G 22 Pythagoras’ Theorem Subject Content References: G2.1, G2.1h

  2. Pythagoras’theorem states: For any right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides Example Let’s take a look at a right-angled triangle with sides 3, 4 & 5 cm: This side is 3 cm long, so the area of the square on this side is 9 cm2 The hypotenuse is 5 cm long, so the area of the square on the hypotenuse is 25 cm2 25 As we can see, 9 + 16 = 25 9 In general, for any right-angled triangle of sides a, b & c, where c is the hypotenuse 16 . . and this side is 4 cm long, so the area of the square on this side is 16 cm2 a2 +b2 = c2

  3. Clearly, this works for a ‘3, 4, 5’ triangle, but does it work for all right angled triangles? Example Let’s have a look . . here’s a right-angled triangle: I measured a, b and c and found: a = 5.8 cm, b = 2.9 cm and c = 6.5 cm Theoretically (using Pythagoras’ theorem), 5.82 + 2.92should be equal to 6.52 5.82 + 2.92 = 33.64 + 8.41 = 42.05 The actual value of 6.52 = 42.25 . . which is pretty close . . and my measuring may not have been accurate enough . . Exercise 1 In order to verify Pythagoras’ theorem for ourselves, draw two or three right angled triangles on squared paper and measure their sides as accurately as possible. To check, use the formula a2 + b2 = c2 for each triangle drawn Example Since for any right-angled triangle, a2 + b2 = c2 (Pythagoras) we can say a2 = c2 - b2(subtracting b2 from both sides), and b2 = c2 - a2(subtracting a2 from both sides) This is really important, because it means that if we know the lengths of any two sides of a right-angled triangle, we can calculate the length of the third side . .

  4. Examples 1) A right-angled triangle ABC has sides a = 8cm and b = 4cm. Calculate the length of side c to 1 decimal place: If a2 + b2 = c2 (Pythagoras) then 82 + 42 = c2 so 64 + 16 = c2 c2 = 80 c = √80 c = 8.9cm (1 dp) answer 2) A right-angled triangle ABC has sides a = 8cm and c = 11cm. Calculate the length of side b to 1 decimal place: If a2 + b2 = c2 (Pythagoras) then b2 = c2 - a2(subtracting a2 from both sides to make b the subject of the equation) so b2 = 112 - 82 b2 = 121 - 64 b = √57  b = 7.5cm (1 dp) answer

  5. Exercise 2 Using a right-angled triangle with sides a, b and c, where c is the hypotenuse, calculate (to 1 dp) the following: 1) a, when b = 7cm, c = 10cm 2) b, when c = 9m , a = 6m 3) c, when a = 7km, b = 6km 4) b, when c = 12m , a = 8m 5) c, when a = 3km, b = 2km 6) a, when b = 1.3cm, c = 5.1cm 7) c, when a = 3.8km, b = 4.5km 8) a, when b = 8.9cm, c = 13.4cm

  6. Example: Examination question A ladder of length 6m is leant up against the top of a vertical wall. It’s base is 2.7m away from the wall. Calculate the height of the wall to the nearest cm: Step 1: Draw the diagram Step 2: We need to show that we recognise this as a problem involving Pythagoras’ theorem, so state it: h 6m In any right-angled triangle with sides a, b and c, where c is the hypotenuse, a2 + b2 = c2 (Pythagoras) From the diagram, therefore, 2.7m h2 + 2.72 = 62 h2 = 62 - 2.72 . . subtracting 2.72 from both sides  h2 = 36 - 7.29 = 28.71  h = √28.71 = 5.36m answer

  7. Exercise 3 1) Point A is due north of point B at a distance of 12 km. Point C is due east of point B at a distance of 7 km. Calculate the distance between point C and point A to the nearest metre. 2) The hypotenuse of a right-angled triangle measures 22.3 cm. Its height measures 15.4 cm. Calculate the length of the base of the triangle to the nearest millimetre. 3) Look at the following grid (divided into equal squares). If each square is of side 1 metre, calculate the distance a) between A & B and b) between C & D to the nearest cm: a) A C B b) D

  8. Pythagoras’ theorem can also be used to solve 3-D problems that usually involve more than one triangle . . Example Below is a cuboid of width 5.8 cm, length 8.6 cm and height 3.5 cm. Calculate the length of the line BH to the nearest mm: Step 1: In order to calculate the length of line BH, we need to know the length of line CH . . . . because the right-angled triangle BCH will enable us to find BH using Pythagoras’ theorem B A Step 2: Identify the right-angled triangle CGH . . . . because the right-angled triangle CGH will enable us to find CH using Pythagoras’ theorem C D Step 3: Using Pythagoras’ theorem . . E F (CH)2 = 8.62 + 5.82 8.6 cm = 73.96 + 33.64 3.5 cm = 10.373 cm  CH = √107.6 Step 4: H Using Pythagoras’ theorem . . G 5.8 cm (BH)2 = 3.52 + 10.3732 = 12.25 + 107.6 = 10.9 cm answer  BH = √119.85

  9. A B D C P Q S R A B D C W X Z Y Exercise 4 1) Below is drawn a cuboid shaped room of height 2.2 m, width 5.6 m and length 9.7 m. Calculate the distance BS to the nearest cm: 2) This a cuboid shaped factory of height 14.5 m, width 30.6 m and length 72.7 m. Calculate the distance AY to the nearest cm:

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