© T Madas. 20 cm 30 cm 40 cm A cuboidal box measures 30 cm by 40 cm by 20 cm. What is the longest...

© T Madas

20 c

m

30 cm 40 cm

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?

How do we find this distance?

© T Madas

30 cm 40 cm

20 c

m


x

Using Pythagoras:

30 2 + 40

2 = x 2

900 + 1600x 2 =

2500x 2 =

2500x =

50x =

Note: strictly ±50, the minus rejected since it is a length

50 cm

d


© T Madas

30 cm 40 cm

20 c

m



Using Pythagoras:

50 2 + 20

2 = d 2

2500 + 400d 2 =

2900d 2 =

2900d =

53.9 cmd ≈50 cm

d

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30 cm 40 cm

20 c

m



53.9 cm

© T Madas

( )30 cm 40 cm

20 c

m


53.9 cm

What is the angle between the stick and one of the 30 x 40 faces?

50 cm

Using Trigonometry:

2050

= tanθ

θ

tanθ = 25

θ = 25

tan-1

θ ≈ 21.8°


© T Madas

[x ≈ 5.66 cm]

8 cm

8 cm

A pyramid has a height of 20 cm and a square base with side length of 8 cm.

20 c

m

1. Calculate the length of one of its sloping edges

4 cm

4 c

m

Using Pythagoras:

4 2 + 4

2 = x 2

16 + 16x 2 =

32x 2 =

32x =

x

32

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+ 32

y ≈ 20.8 cm

8 cm

8 cm


20 c

m


20 c

m

Using Pythagoras:

20 2 = y

2

400 + 32y 2 =

432y 2 =

432y =

y

32

32

2

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8 cm

8 cm


20 c

m

1. Calculate the length of one of its sloping edges2. Find the angle between one of its sloping faces and

its base.

θ4

(5)

Using Trigonometry:

204

= tanθ

tanθ =

θ = tan-1

θ ≈ 78.7°

5

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2. Find the angle between one of its sloping faces and its base.

8 cm

8 cm


20 c

m


θ4

3. Calculate the total surface area of the pyramid

d

[x ≈ 20.4 cm]

Using Pythagoras:

4 2 + 20

2 = d 2

16+ 400d 2 =

416d 2 =

416d =

416

© T Madas

2. Find the angle between one of its sloping faces and its base.

8 cm

8 cm


20 c

m


4

3. Calculate the total surface area of the pyramid

S = x 8 x 4x 416

416

12

x 8+ 8

= 16 x 416 + 64

= 16 x 16 x 26 + 64

= 16 x 4 26 + 64

= 64 26 + 64

= 64[ ]26 + 1 ≈ 390 cm2

© T Madas

The picture below shows a triangular prism.

ABC 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

A

B

C

D

E

F

M

5

126

6All lengths in cm

A B

C

5

12

AC 2 =52+ 122

AC 2 =25+ 144

AC 2 =169

AC = 169AC = 13 cm

13

© T Madas


ABC 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

A

B

C

D

E

F

M

5

126

6All lengths in cm

D A

B

12

12

BD 2 =122+ 122

BD 2 =144+ 144

BD 2 =288

BD = 288BD ≈16.97 cm

13288

© T Madas


ABC 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

A

B

C

D

E

F

M

5

126

6All lengths in cm

D B

C

5

288

CD 2 =52+ 288 2

CD 2 =25+ 288

CD 2 =313

CD = 313CD ≈17.69 cm

13288

313

© T Madas


ABC 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. CAB

A

B

C

D

E

F

M

5

126

6All lengths in cm

13288

313

A B

C

5

12θ

tanθ = 512

θ = 512

tan-1

θ ≈22.6°

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ABC 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. CDB

A

B

C

D

E

F

M

5

126

6All lengths in cm

13288

313

D B

C

5φ

tanφ =5

288

φ =5

288

tan-1

φ ≈16.4°

288

313

© T Madas


ABC 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. CMB

A

B

C

D

E

F

M

5

126

6All lengths in cm

M A

B

12

6

BM 2 =62+ 122

BM 2 =36+ 144

BM 2 =180

BM =180BM ≈13.42 cm

180

© T Madas


ABC 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. CMB

A

C

D

E

F

M

5

126

6All lengths in cm

B

M B

C

5α

tanα =5

180

α =5

180

tan-1

α ≈20.4°

180

180

© T Madas

x

y

z

A

BC

DE

FG

O

(6,3,2)

(6,3,0)

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.

6

3

2

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x

y

z

A

DE

G

O




6

3

2

C

O A

B

3

6

OB 2 = 62+ 32

OB 2 =36+ 9

OB 2 =45

OB = 45

F (6,3,2)

B (6,3,0)

45

© T Madas

x

y

z

A

DE

G

O




6

3

2

C

O B

F

2

F (6,3,2)

B (6,3,0)

45

45

OF 2 =22+ 45 2

OF 2 = 4 + 45

OF 2 =49

OF = 7

© T Madas

7

24 32

The figure below shows a prism ABCDEF, whose cross sections are the right angled triangles ABC and DEF.

BAC = EDF = 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:1. BC

2. AF

3. BFA

4. MFN

A

B

C

D

E

F

M

N

lengths in cm

© T Madas

7

24 32




2. AF

3. BFA

4. MFN

A

B

C

D

E

F

M

N

lengths in cm

A

B

C7

24

BC 2 =72+ 242

BC 2 =49+ 576

BC 2 =625

BC = 625BC = 25 cm

© T Madas

7

24 32




2. AF

3. BFA

4. MFN

A

B

C

D

E

F

M

N

lengths in cm

C

A

F

24

32

AF 2 =242 + 322

AF 2 =576+ 1024

AF 2 =1600

AF = 1600AF = 40 cm

© T Madas

7

24 32




2. AF

3. BFA

4. MFN

A

B

C

D

E

F

M

N

lengths in cm

A

B

F7

40θ

tanθ = 740

θ = 740

tan-1

θ ≈ 9.9°

© T Madas

7

24 32




2. AF

3. BFA

4. MFN

A

B

C

D

E

F

M

N

lengths in cm

7

N

F

24

16

NF 2 =242 + 162

NF 2 =576+ 256

NF 2 =832

NF = 832NF ≈28.844 cm

28.844

© T Madas

7

24 32




2. AF

3. BFA

4. MFN

A

B

C

D

E

F

M

N

lengths in cm

7

28.844

N

M

F7

28.844α

tanα = 728.44

α = 728.84

tan-1

α ≈ 13.6°

© T Madas

.56 57 cmA

B

C

D

E

F

20°40 cm40 cm

The diagram below shows a wedge in the shape of a right angled triangular prism.

FBC = 20°.R ABCD is a square of side 40 cmFind the angle the line AF makes with the plane

ABCD

x

x2 = 402 + 402 Ûx2 = 1600 + 1600 Ûx2 = 3200 Ûx = 3200

.» 56 57 cm

© T Madas

14.5

6 cm

.56 57 cmA

B

C

D

E

F

20°40 cm40 cm



ABCD

y40 = tan20° Û

y

y = 40 tan20° Û

y ≈ 14.56 cm

© T Madas

14.5656.57

14.5

6 cm

.56 57 cmA

B

C

D

E

F

20°40 cm40 cm



ABCD

= tanθ Û

θ

tanθ ≈ Û0.257

θ ≈ Ûtan-1 (0.257)

θ ≈ 14.4°

© 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

n 2+ (n + 1)2+ (n

2 + n)2 = n 2 + n

2+ 2n + 1+ n 4+ 2n

3+ n

Let a = n b = n + 1 c = n (n + 1) = n

2 + n

= n 4 + 2n

3+ 2n 2+ 3n + 1

= n 2 + n + 1( )2

© T Madas. 20 cm 30 cm 40 cm A cuboidal box measures 30 cm by 40 cm by 20 cm. What is the longest...

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Transcript of © T Madas. 20 cm 30 cm 40 cm A cuboidal box measures 30 cm by 40 cm by 20 cm. What is the longest...