© T Madas

How do we find this distance? A cuboidal box measures 30 cm by 40 cm by 20 cm. What is the longest stick which fits in this box? It should be obvious that the stick must be placed in a “diagonal” fashion Is this the longest? 20 cm 40 cm 30 cm How do we find this distance? © T Madas

Note: strictly ±50, the minus rejected since it is a length A cuboidal box measures 30 cm by 40 cm by 20 cm. What is the longest stick which fits in this box? Using Pythagoras: 30 2 + 40 2 = x 2 x 2 = 900 + 1600 d 20 cm x 2 = 2500 x = 2500 x 50 cm x = 50 40 cm 30 cm Note: strictly ±50, the minus rejected since it is a length © T Madas

A cuboidal box measures 30 cm by 40 cm by 20 cm. What is the longest stick which fits in this box? Using Pythagoras: 50 2 + 20 2 = d 2 d 2 = 2500 + 400 d 20 cm d 2 = 2900 d = 2900 50 cm d ≈ 53.9 cm 40 cm 30 cm © T Madas

A cuboidal box measures 30 cm by 40 cm by 20 cm. What is the longest stick which fits in this box? 53.9 cm 20 cm 40 cm 30 cm © T Madas

( ) θ ≈ 21.8° A cuboidal box measures 30 cm by 40 cm by 20 cm. What is the longest stick which fits in this box? What is the angle between the stick and one of the 30 x 40 faces? Using Trigonometry: 20 50 = tanθ 53.9 cm 2 5 20 cm tanθ = ( ) 50 cm θ 2 5 θ = tan-1 40 cm 30 cm θ ≈ 21.8° © T Madas

Another Example © T Madas

Another Example © T Madas

1. Calculate the length of one of its sloping edges A pyramid has a height of 20 cm and a square base with side length of 8 cm. 1. Calculate the length of one of its sloping edges Using Pythagoras: x 4 cm 4 2 + 4 2 = x 2 x 2 = 16 + 16 4 cm x 2 = 32 20 cm x = 32 [x ≈ 5.66 cm] 32 8 cm 8 cm © T Madas

1. Calculate the length of one of its sloping edges A pyramid has a height of 20 cm and a square base with side length of 8 cm. 1. Calculate the length of one of its sloping edges Using Pythagoras: 2 20 2 + 32 = y 2 y 20 cm y 2 = 400 + 32 y 2 = 432 20 cm y = 432 32 y ≈ 20.8 cm 32 8 cm 8 cm © T Madas

A pyramid has a height of 20 cm and a square base with side length of 8 cm. 1. Calculate the length of one of its sloping edges 2. Find the angle between one of its sloping faces and its base. Using Trigonometry: 20 4 = tanθ tanθ = 5 20 cm θ = (5) tan-1 θ ≈ 78.7° θ 8 cm 4 8 cm © T Madas

1. Calculate the length of one of its sloping edges A pyramid has a height of 20 cm and a square base with side length of 8 cm. 1. Calculate the length of one of its sloping edges 2. Find the angle between one of its sloping faces and its base. 3. Calculate the total surface area of the pyramid Using Pythagoras: 20 cm d 416 4 2 + 20 2 = d 2 d 2 = 16 + 400 d 2 = 416 θ d = 416 8 cm 4 [x ≈ 20.4 cm] 8 cm © T Madas

A pyramid has a height of 20 cm and a square base with side length of 8 cm. 1. Calculate the length of one of its sloping edges 2. Find the angle between one of its sloping faces and its base. 3. Calculate the total surface area of the pyramid 1 2 S = x 8 x 416 x 4 + 8 x 8 20 cm x 416 + 64 416 = 16 = 16 x 16 x 26 + 64 = 16 x 4 26 + 64 8 cm = 64 26 + 64 4 = 64[ ] 26 + 1 ≈ 390 cm2 8 cm © T Madas

© T Madas

1. the length AC C F A B E C D B M A The picture below shows a triangular prism. RABC is a right angle, BC = 5 cm, ABED is a square with a side length of 12 cm and M is the midpoint of AD. Calculate: 1. the length AC C F 5 A 12 B E AC 2 = 52 + 122 c C AC 2 = 25 + 144 c AC 2 = 169 c 5 AC = 169 D 13 B AC = 13 cm 6 12 M 6 All lengths in cm A © T Madas

2. the length BD B F E D A C D B M A The picture below shows a triangular prism. RABC is a right angle, BC = 5 cm, ABED is a square with a side length of 12 cm and M is the midpoint of AD. Calculate: 2. the length BD B F 12 E D A 12 C BD 2 = 122 + 122 c 5 BD 2 = 144 + 144 c D 13 B 288 BD 2 = 288 c 6 BD = 288 12 BD ≈ 16.97 cm M 6 All lengths in cm A © T Madas

3. the length CD C F D B E C D B M A The picture below shows a triangular prism. RABC is a right angle, BC = 5 cm, ABED is a square with a side length of 12 cm and M is the midpoint of AD. Calculate: 3. the length CD C F 5 D B 288 E CD 2 = 52 + 288 2 c C CD 2 = 25 + 288 c 313 5 CD 2 = 313 c D 13 B CD = 313 288 CD ≈ 17.69 cm 6 12 M 6 All lengths in cm A © T Madas

The picture below shows a triangular prism. RABC is a right angle, BC = 5 cm, ABED is a square with a side length of 12 cm and M is the midpoint of AD. Calculate: 4. RCAB C F 5 θ A 12 B E 5 12 tanθ = c C 313 5 5 12 θ = tan-1 c D 13 B 288 θ ≈ 22.6° 6 12 M 6 All lengths in cm A © T Madas

The picture below shows a triangular prism. RABC is a right angle, BC = 5 cm, ABED is a square with a side length of 12 cm and M is the midpoint of AD. Calculate: 5. RCDB C 313 F 5 φ D B 288 E C 5 288 tanφ = c 313 5 D 5 288 13 B φ = tan-1 c 288 6 φ ≈ 16.4° 12 M 6 All lengths in cm A © T Madas

The picture below shows a triangular prism. RABC is a right angle, BC = 5 cm, ABED is a square with a side length of 12 cm and M is the midpoint of AD. Calculate: 6. RCMB B F 12 E C M 6 A 5 D BM 2 = 62 + 122 c B BM 2 = 36 + 144 c 6 180 BM 2 = 180 c 12 M BM = 180 6 BM ≈ 13.42 cm All lengths in cm A © T Madas

The picture below shows a triangular prism. RABC is a right angle, BC = 5 cm, ABED is a square with a side length of 12 cm and M is the midpoint of AD. Calculate: 6. RCMB C F 5 α M B 180 E C 5 180 tanα = c 5 D 5 180 B α = tan-1 c 6 180 α ≈ 20.4° 12 M 6 All lengths in cm A © T Madas

© T Madas

A cuboid OABCDEFG 6 units long by 3 units wide by 2 units high is drawn on a set of 3 dimensional axes as shown below. 1. Write down the coordinates of points B and F. 2. Calculate the length of OF. z y G F ( 6 ,3 ,2 ) D 2 E C B ( 6 ,3 ,0 ) 3 O x 6 A © T Madas

A cuboid OABCDEFG 6 units long by 3 units wide by 2 units high is drawn on a set of 3 dimensional axes as shown below. 1. Write down the coordinates of points B and F. 2. Calculate the length of OF. B z 3 y G F ( 6 ,3 ,2 ) O A 6 OB 2 = 62 + 32 c D 2 E OB 2 = 36 + 9 c C B OB 2 = 45 c ( 6 ,3 ,0 ) OB = 45 3 45 O x 6 A © T Madas

A cuboid OABCDEFG 6 units long by 3 units wide by 2 units high is drawn on a set of 3 dimensional axes as shown below. 1. Write down the coordinates of points B and F. 2. Calculate the length of OF. z F 2 y G F ( 6 ,3 ,2 ) O B 45 D 2 OF 2 = 22 + 45 2 c E OF 2 = 4 + 45 c C B ( 6 ,3 ,0 ) OF 2 = 49 c 3 OF = 7 45 O x 6 A © T Madas

© T Madas

The figure below shows a prism ABCDEF, whose cross sections are the right angled triangles ABC and DEF. RBAC = REDF = 90°, AB = 7 cm, AC = 24 cm, CF = 32 cm, M is the midpoint of BE and N is the midpoint of AD. Calculate to an appropriate degree of accuracy: BC AF RBFA RMFN lengths in cm E M D B N F 7 A 32 24 C © T Madas

The figure below shows a prism ABCDEF, whose cross sections are the right angled triangles ABC and DEF. RBAC = REDF = 90°, AB = 7 cm, AC = 24 cm, CF = 32 cm, M is the midpoint of BE and N is the midpoint of AD. Calculate to an appropriate degree of accuracy: BC AF RBFA RMFN B lengths in cm 7 E C A 24 M D BC 2 = 72 + 242 c BC 2 = 49 + 576 c B N BC 2 = 625 c F BC = 625 7 BC = 25 cm A 32 24 C © T Madas

The figure below shows a prism ABCDEF, whose cross sections are the right angled triangles ABC and DEF. RBAC = REDF = 90°, AB = 7 cm, AC = 24 cm, CF = 32 cm, M is the midpoint of BE and N is the midpoint of AD. Calculate to an appropriate degree of accuracy: BC AF RBFA RMFN A lengths in cm E 24 F M C 32 D B AF 2 = 242 + 322 c N AF 2 = 576 + 1024 c F 7 AF 2 = 1600 c AF = 1600 A AF = 40 cm 32 24 C © T Madas

The figure below shows a prism ABCDEF, whose cross sections are the right angled triangles ABC and DEF. RBAC = REDF = 90°, AB = 7 cm, AC = 24 cm, CF = 32 cm, M is the midpoint of BE and N is the midpoint of AD. Calculate to an appropriate degree of accuracy: BC AF RBFA RMFN B lengths in cm E 7 θ F A 40 M D 7 40 tanθ = c B N 7 40 F θ = tan-1 c 7 A θ ≈ 9.9° 32 24 C © T Madas

The figure below shows a prism ABCDEF, whose cross sections are the right angled triangles ABC and DEF. RBAC = REDF = 90°, AB = 7 cm, AC = 24 cm, CF = 32 cm, M is the midpoint of BE and N is the midpoint of AD. Calculate to an appropriate degree of accuracy: BC AF RBFA RMFN N lengths in cm E 24 M D F 16 7 B N 28.844 F NF 2 = 242 + 162 c 7 NF 2 = 576 + 256 c A NF 2 = 832 c 32 24 NF = 832 NF ≈ 28.844 cm C © T Madas

The figure below shows a prism ABCDEF, whose cross sections are the right angled triangles ABC and DEF. RBAC = REDF = 90°, AB = 7 cm, AC = 24 cm, CF = 32 cm, M is the midpoint of BE and N is the midpoint of AD. Calculate to an appropriate degree of accuracy: BC AF RBFA RMFN M lengths in cm E 7 α F N 28.844 M D 7 28.44 7 tanα = c B N 7 28.84 28.844 F α = tan-1 c 7 A α ≈ 13.6° 32 24 C © T Madas

© T Madas

ABCD is a square of side 40 cm The diagram below shows a wedge in the shape of a right angled triangular prism. FBC = 20°. ABCD is a square of side 40 cm Find the angle the line AF makes with the plane ABCD E x2 = 402 + 402 F D x2 = 1600 + 1600 x2 = 3200 x x = A C 20° 40 cm 40 cm B © T Madas

ABCD is a square of side 40 cm The diagram below shows a wedge in the shape of a right angled triangular prism. FBC = 20°. ABCD is a square of side 40 cm Find the angle the line AF makes with the plane ABCD E y 40 = tan20° F D y = 40 tan20° 14.56 cm y y ≈ 14.56 cm A C 20° 40 cm 40 cm B © T Madas

The diagram below shows a wedge in the shape of a right angled triangular prism. FBC = 20°. ABCD is a square of side 40 cm Find the angle the line AF makes with the plane ABCD E 14.56 56.57 = tanθ F tanθ ≈ 0.257 D θ ≈ tan-1 (0.257) 14.56 cm θ θ ≈ 14.4° A C 20° 40 cm 40 cm B © T Madas

3D Pythagorean Quads © T Madas

Experience on using the Pythagoras Theorem in 3 dimensions tells us that: There are a few integer lengths which satisfy the Pythagorean law a 2 + b 2 + c 2 = d 2. Some of us are aware of 1,1,2,3 Even fewer of us know the 2,3,6,7 What about 3,4,12,13? How many of us know the 4,5,20,21? Does 51, 52, 2652,2653 also work? What is the pattern of these quads? Is there an infinite number of 3D Pythagorean Quads? Is there a way to generate such numbers? © T Madas

Prove that the product of any 2 consecutive positive integers a and b and their product c, satisfy the 3D Pythagorean relationship a 2 + b 2 + c 2 = d 2 , with d a positive integer Let a = n b = n + 1 c = n (n + 1) = n 2 + n n 2 + (n + 1)2 + (n 2 + n)2 = n 2 + n 2 + 2n + 1 + n 4 + 2n 3 + n = n 4 + 2n 3 + 2n 2 + 3n + 1 = ( n 2 + n + 1 )2 © T Madas

© T Madas