Pythagoras. 1 of 60. (Pythagoras)
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1 Pythagoras 1 of 60 (Pythagoras)
2 2 of 60 The history of Pythagoras Theorem The theorem is named after the Greek mathematician and philosopher, Pythagoras. He lived around the year 400BC He also invented the music scale we now use. Although the Theorem is named after Pythagoras the result was known to many ancient civilizations including the Babylonians, Egyptians and Chinese, at least 1000 years before Pythagoras was born.
3 3 of 60 SLO Properties of right angled triangles
4 4 of 60 Copy into your notes Right-angled triangles A right-angled triangle contains a 90 degree angle. The longest side (opposite the right angle) is called the hypotenuse. 90 degree angle
5 5 of 60 Identify the hypotenuse
6 6 of 60 SLO Knowing what is Pythagoras Theorem
7 7 of 60 Copy into your notes Pythagoras Theorem Pythagoras Theorem states that the square formed on the hypotenuse of a right-angled triangle has the same area as the sum of the areas of the squares formed on the other two sides.
8 8 of 60
9 9 of 60 Pythagoras Theorem Label the length of the sides of a right-angled triangle a, b and h as follows, The area of the largest square is h h or h 2. a 2 a b h h 2 The areas of the smaller squares are a 2 and b 2. We can write Pythagoras Theorem as b 2 h 2 = a 2 + b 2 hypotenuse
10 10 of 60 Copy into your notes Pythagoras Theorem For a right-angled triangle with a hypotenuse of length h h 2 = a 2 + b 2 a h Pythagoras Theorem is used when: We have been given 2 sides of a triangle and have to find the third. No angles are involved. b
11 11 of 60 Find the correct equation
12 12 of 60 SLO Applying Pythagoras Theorem to triangles
13 13 of 60 Copy into your notes E.g. Finding the length of the hypotenuse Use Pythagoras Theorem to calculate the length of side a. 5 cm h 12 cm Using Pythagoras Theorem, h 2 = a 2 + b 2 h 2 = h 2 = h 2 = 169 h = 169 h = 13 cm
14 Your Turn: Use Pythagoras Theorem to calculate the length of side PR. 0.7 m P Q 2.4 m R Using Pythagoras Theorem PR 2 = PQ 2 + QR 2 Substituting the values we have been given, PR 2 = PR 2 = PR 2 = 6.25 PR = 6.25 PR = 2.5 m 14 of 60
15 15 of 60 Your Turn: Use Pythagoras Theorem to calculate the length of side x to 2 decimal places. 3 cm 6 cm x Using Pythagoras Theorem x 2 = x 2 = x 2 = 45 x = 45 x = 6.71 cm
16 16 of 60 E.g. Finding the length of the shorter sides Use Pythagoras Theorem to calculate the length of side a. 10 cm 26 cm a Using Pythagoras Theorem, a = 26 2 a 2 = a 2 = a 2 = 576 a = 576 a = 24 cm
17 Your Turn: Use Pythagoras Theorem to calculate the length of side AC to 2 decimal places. Using Pythagoras Theorem 5 cm B A 8 cm C AB 2 + AC 2 = BC 2 Substituting the values we have been given, AC 2 = 8 2 AC 2 = AC 2 = 39 AC = 39 AC = 6.24 cm 17 of 60
18 18 of 60 Your Turn: Use Pythagoras Theorem to calculate the length of side x to 2 decimal places. 15 cm x 7 cm Using Pythagoras Theorem, x = 15 2 x 2 = x 2 = x 2 = 176 x = 176 x = cm
19 Copy into your notes 19 of 60 In conclusion... Square Square Square-root
20 20 of 60 Questions to do from the books Achieve Merit Excellence Gamma P386 Ex27.01 Q4 15 IAS 1.7 P4 Q1-6
21 21 of 60 1 Your Turn: 53 m mm 101km 4 74 km
22 22 of 60 Your Turn: m 32.1 mm mm cm
23 23 of 60 SLO Merit Questions Finding the lengths of diagonals, heights of an isosceles triangle, height of an equilateral triangle and Word Problems
24 24 of 60 E.g. Calculate the length of the diagonal, d cm d 13.6 cm d 2 = d 2 = d 2 = 289 d = 289 d = 17 cm
25 25 of 60 Your Turn: Calculate the length d of the diagonal in a square of side length 7 cm. d 7 cm Using Pythagoras Theorem d 2 = d 2 = d 2 = 98 d = 98 d = 9.90 cm (to 2 d.p.)
26 26 of 60 Your Turn: Calculate the height h of this isosceles triangle. 5.8 cm h 8 cm Click for clue Using Pythagoras Theorem in half of the isosceles triangle (take care with h for height, not hypotenuse), we have 5.8 cm 4 cm h h = h 2 = h 2 = h 2 = h = h = 4.2 cm
27 27 of 60 Your Turn: Calculate the height h of an equilateral triangle with side length 4 cm. h 4 cm Click for clue Using Pythagoras Theorem in half of the equilateral triangle (take care h now equals height, not hypotenuse), we have h = 4 2 h 4 cm 2 cm h 2 = h 2 = 16 4 h 2 = 12 h = 12 h = 3.46 cm (to 2 d.p.)
28 Copy into your notes 28 of 60 Solving word problems using Pythagoras Theorem When solving a problem using Pythagoras Theorem always do the following: Draw a clearly labeled diagram. State Pythagoras Theorem for the given side lengths before substituting any values. Make sure that your answer is sensible given the lengths of the other two sides. Assume certain things such as all walls are built at 90 degrees to the ground.
29 E.g. The forwards in a rugby team have to run along the diagonal of a rugby field and back, while the backs run round the field. The field is 100 m by 70 m. How much further do the backs run each time they have run round the field once if the forwards have run the diagonal and back? No angles, so Pythagoras Theorem. Long side to find. d² = 100² + 70² = d = = 122 m (to 3 s.f.) Distance round field = 2( ) = 340 m Extra distance = 340 2(122) = 96 m (to 2 s.f.) Start and finish point d 100 m 70 m 29 of 60
30 Your Turn: A canvas length of 5 m is to be lifted and supported in the middle to form a tent. If the tent is to be 2.1 m high, how wide will the base of the tent be, if the canvas touches the ground? Click for clue total length 5 m No angles, so Pythagoras Theorem. Short side to find. d² = 2.5² 2.1² = 1.84 total length 5 m h = m d = m = m (to 5 s.f.) Width of tent = 2(1.3565) = 2.7 m (to 2 s.f.) d 30 of 60
31 The points A( 4, 5), B( 4, 2), and C (6, 2) form a right angled triangle. What is length of the line AC? Click for clue 1 31 of 60 Your Turn: A( 4, 5) 7 units y B( 4, 2) 3 10 units C (6, 2) 4 5 Click for clue 2 Using Pythagoras, AC 2 = AB 2 + AC 2 = = = 149 AC = 149 = units (to 2 d.p.) units (to 2 d.p.) x
32 32 of 60 Ladder problem
33 33 of 60 Flight path problem
34 34 of 60 Questions to do from the books Gamma Achieve Merit Excellence P386 Ex27.01 Q4 15 P387 Ex27.01 Q16 17 P389 Ex27.02 IAS 1.7 P4 Q1-6 P4 7 18
35 35 of 60 SLO Excellence Multi stage problems 3D problems Hard word problems
36 36 of 60 Other Triangle situations: Sometimes the right angled triangle will need to be found Remember to halve this length. We may have to subtract heights to find the height of the triangle. Sometimes we can make a triangle from a non triangle shape.
37 37 of 60 E.g. Tanya is making a party hat using a cone made out of paper. Determine the height of the cone. b 2 = c 2 a 2 h 2 = h 2 = cm h 2 = 144 h h cm h = 12 cm
38 38 of 60 Your Turn: A ladder is 2.8 m long and is leaning against the wall of a house from a point 1.2 m out from the base of the wall. The ladder can be extended to become 3.6 m long. How much further up the wall does the ladder reach when it is extended? Click for clue Normal height = d, extended = e. No angles, so Pythagoras Theorem, Short side to find. e² = 3.6² 1.2² = e = = 3.39 m (to 3 s.f.) d² = 2.8² 1.2² = 6.4 d = 6. 4 = 2.53 m (to 3 s.f.) Distance up wall = ( ) m = 0.86 m 2.8 m d m 3.6 m 1.2 m 1.2 m e m
39 39 of 60 Your Turn: A poster is to be made in the shape of an equilateral triangle (angles 60 ) with sides 48 cm long. If the poster is placed on a wall with the base 1.2 m from and parallel to the floor, how far from the floor will the highest point of the poster be? Click for clue No angles, so Pythagoras Theorem, short side to find. d² = 48² 24² = cm 48 cm 1.2m 48 cm d 48 cm 24 cm d = = 41.6 cm (to 3 s.f.) floor Distance from floor = ( ) m = 1.62 m
40 40 of 60 Your Turn: Show that the height h in an equilateral triangle of side length a can be found by the formula h = 3 a 2 Using Pythagoras Theorem in half of the equilateral triangle, we have h 2 + a 2 = a 2 We can think of 4 this as 1a 2 1 a 2 h a Click for clue a 2 h 2 = a 2 4 h a 4 a 2 h 2 = h = 3a a
41 Applying Pythagoras Theorem twice E.g. Find the length of side a. Click for clue To solve this problem we need to find the length of the other missing side which we can call b. b a 18 cm b 2 = 18 2 (4 + 9) 2 b 2 = b 2 = cm 9 cm Now, a 2 = b a 2 = a 2 = 171 a = cm (to 2 d.p.) 41 of 60
42 Your Turn: In the triangle, length y bisects the base of the triangle. Find the length y 7m x y 18m x Click for clue z 42 of a b h let2x z z (18) 2 49 z z 275 z 275 z 16.6m 16.6 x 2 x 8.3 m(1 dp) so we can use this to find y a b h ( y) y y y 10.9m y 2 42
43 43 of 60 E.g. The longest diagonal in a cuboid Pythagoras Theorem can also be applied to three-dimensional problems. For example, what is the length of the longest diagonal in a cuboid measuring 5 cm by 7 cm by 8 cm? E H F G Start by labeling the vertices of the cuboid. The longest diagonal is CE. 8 cm A D 7 cm B C 5 cm We could also use AG, BH or DF. CE is the hypotenuse of right-angled triangle ACE.
44 44 of 60 The longest diagonal in a cuboid Pythagoras Theorem can also be applied to three-dimensional problems. For example, what is the length of the longest diagonal in a cuboid measuring 5 cm by 7 cm by 8 cm? E 8 cm A H D 7 cm G F C 5 cm B Before we can find the length of CE, we need to find the length of AC. We can do this by applying Pythagoras Theorem to triangle ABC. AC 2 = AC 2 = AC 2 = 74
45 45 of 60 The longest diagonal in a cuboid Pythagoras Theorem can also be applied to three-dimensional problems. For example, what is the length of the longest diagonal in a cuboid measuring 5 cm by 7 cm by 8 cm? E H F G Now we know that AC 2 = 74, we can use Pythagoras Theorem to work out the length of CE. 8 cm A D 7 cm B C 5 cm CE 2 = AE 2 + AC 2 CE 2 = CE 2 = CE 2 = 138 CE = cm (to 2 d.p.)
46 46 of 60 Your Turn: 0 cm A chair has a square seat with 4 legs 65 cm long and 80 cm apart. A bar parallel to the floor joining two diagonally opposite legs is bolted 2 cm from the floor. Another bar is slanted to join diagonally opposite legs 2 cm from the floor and 2 cm from the seat on the other leg. What is the length of each of the bars between bolt holes? slanted bar floor parallel bar 80 cm b cm d cm 65 cm 61 cm Let Floor parallel bar = d. d²= 80² + 80² = d = = 113 cm (3 s.f.) Longer bar s length = b. With 2 cm above and below, leg part of triangle is 61 cm. b² = d² + 61² = = d = = 129 cm (3 s.f.) Click for clue d cm 80 cm 80 cm
47 What is the longest length that can fit in the box? Click for clue Your Turn: First we must calculate the hypotenuse of the base a b h (5) (11) x 2 x x 146 x x 12.1(1 dp) cm then we must calculate the diagonal y 2 y x y 2 y 2 y 210 x x 14.5(1 dp) cm 5cm 8cm x y 11cm Remember 12.1 came from 146 so when squared you must use of 60 47
48 48 of 60 Questions to do from the books Gamma Achieve Merit Excellence P386 Ex27.01 Q4 15 P387 Ex27.01 Q16 17 P389 Ex27.02 P394 Ex27.03 P413 Ex Q8 11 IAS 1.7 P4 Q1-6 P4 Q7 18
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