Presentation is loading. Please wait.

Presentation is loading. Please wait.

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

Similar presentations


Presentation on theme: "G 22 Pythagoras’ Theorem Subject Content References: G2.1, G2.1h GCSE Maths Geometry & Measures."— Presentation transcript:

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

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 Let’s take a look at a right-angled triangle with sides 3, 4 & 5 cm: The hypotenuse is 5 cm long, so the area of the square on the hypotenuse is 25 cm 2 This side is 3 cm long, so the area of the square on this side is 9 cm 2.. and this side is 4 cm long, so the area of the square on this side is 16 cm 2 9 16 25 As we can see, 9 + 16 = 25 In general, for any right-angled triangle of sides a, b & c, where c is the hypotenuse a 2 +b 2 = c 2 Example

3 Clearly, this works for a ‘3, 4, 5’ triangle, but does it work for all right angled triangles? 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), 5.8 2 + 2.9 2 should be equal to 6.5 2 5.8 2 + 2.9 2 = 33.64 + 8.41 = 42.05.. which is pretty close 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 a 2 + b 2 = c 2 for each triangle drawn Since for any right-angled triangle, a 2 + b 2 = c 2 (Pythagoras) we can say a 2 = c 2 - b 2 (subtracting b 2 from both sides), and 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.. b 2 = c 2 - a 2 (subtracting a 2 from both sides) The actual value of 6.5 2 = 42.25.. and my measuring may not have been accurate enough.. Example Exercise 1 Example

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 a 2 + b 2 = c 2 (Pythagoras) then 8 2 + 4 2 = c 2 so 64 + 16 = c 2 c 2 = 80 c = √80  c = 8.9cm (1 dp) 2) A right-angled triangle ABC has sides a = 8cm and c = 11cm. Calculate the length of side b to 1 decimal place: If a 2 + b 2 = c 2 (Pythagoras) then b 2 = c 2 - a 2 (subtracting a 2 from both sides to make b the subject of the equation) so b 2 = 11 2 - 8 2 b 2 = 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: 3) c, when a = 7km, b = 6km 1) a, when b = 7cm, c = 10cm 2) b, when c = 9m, a = 6m 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: 6m 2.7m Step 2: h In any right-angled triangle with sides a, b and c, where c is the hypotenuse, a 2 + b 2 = c 2 (Pythagoras) h 2 + 2.7 2 = 6 2 h 2 = 6 2 - 2.7 2 h 2 = 36 - 7.29 = 28.71 h = √28.71 = 5.36m answer Draw the diagram We need to show that we recognise this as a problem involving Pythagoras’ theorem, so state it:  From the diagram, therefore, .. subtracting 2.7 2 from both sides

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 B C D a) b)

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

9 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: AB CD PQ RS 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: AB CD WX YZ


Download ppt "G 22 Pythagoras’ Theorem Subject Content References: G2.1, G2.1h GCSE Maths Geometry & Measures."

Similar presentations


Ads by Google