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Thales Theorem
Chapter 8-Constructions
Applications:
-Construction of the fourth
proportional
-Dividing a segment into equal
parts.
-Constructions of points
-Enlargement โ€“Reduction
Construction of the fourth proportional
โ€ข Question :construct the fourth proportional to 4,7 and 5.
Etape 1:Trace a straight line (d) and place 3 points O,A & B in the following order :OA = 4cm and OB =7cm.
O A B (d)
0 4 7
Etape 2:Trace a straight line (dโ€ฒ) that cuts (d) at O and place the point C such that OC = 5cm.
O A B (d)
0 4 7
C
5
D
x (dโ€ฒ)
Step 3: construct the parallel to (AC) passing through B, it cuts (dโ€™) at D.
According to Thalesโ€™ theorem, the length x of segment [OD] is: 5/x = 4/7 or x x 4 = 5 x 7
The length x of segment [OD] is the fourth proportional to 4,7 and 5.
Dividing a segment into equal parts
Dividing a segment into equal parts
Divide [AB] into 4 equal parts
Construction of points
Place point K on segment [AB] such that ๐ดM =
2
5
๐ด๐ต OR
AM
AB
=
2
5
Construction of points
Place points M on line (AB) such that
๐‘€๐ด =
2
5
๐‘€๐ต OR
๐‘€๐ด
๐‘€๐ต
=
2
5
Construction of points
Construction of points
Construction of points
Construction of points
Construction of points
Place points M on line (AB) such that ๐‘€๐ด =
2
3
๐‘€๐ต OR
๐‘€๐ด
๐‘€๐ต
=
2
3
Construction of points
Place points M on line (AB) such that ๐‘€๐ด =
2
3
๐‘€๐ต OR
๐‘€๐ด
๐‘€๐ต
=
2
3
Construction of points
Place points M on line (AB) such that
๐‘€๐ด
๐‘€๐ต
=2
Enlargement and Reduction
Enlargement and reduction can be seen in many
situations in real life.
Elements of enlargement and reduction
The elements of enlargement or reduction are:
1) The ratio or the coefficient k:
๐‘˜ =
๐‘™๐‘’๐‘›๐‘”๐‘กโ„Ž ๐‘œ๐‘“ ๐‘›๐‘’๐‘ค ๐‘ ๐‘–๐‘‘๐‘’ (๐‘–๐‘š๐‘Ž๐‘”๐‘’)
๐‘™๐‘’๐‘›๐‘”๐‘กโ„Ž ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘œ๐‘Ÿ๐‘–๐‘”๐‘–๐‘›๐‘Ž๐‘™
โ€ข If k>1,then the image is an enlargement
โ€ข If k<1, then the image is reduction
2) Center of enlargement or reduction
Application:
Given segment AB and Aโ€™Bโ€™ the image of AB by enlargement
A
B
Aโ€™
Bโ€™
1) Find the center of enlargement O
2)Suppose that OAโ€™=12 and OA=3
Find the ratio of enlargement
O
Enlarge triangle ABC of ratio 3 and center O
๐‘‚๐ตโ€ฒ
๐‘‚๐ต
= 3 ๐‘กโ„Ž๐‘’๐‘› ๐‘‚๐ตโ€ฒ
= 3๐‘‚๐ต
๐‘‚๐ถโ€ฒ
๐‘‚๐ถ
= 3 ๐‘กโ„Ž๐‘’๐‘› ๐‘‚๐ถโ€ฒ = 3๐‘‚๐ถ
๐‘‚๐ดโ€ฒ
๐‘‚๐ด
= 3 ๐‘กโ„Ž๐‘’๐‘› ๐‘‚๐ดโ€ฒ = 3๐‘‚๐ด
Enlarge triangle ABC of ratio 3 and center A
๐ด๐ตโ€ฒ
๐ด๐ต
= 3 ๐‘กโ„Ž๐‘’๐‘› ๐ด๐ตโ€ฒ
= 3๐ด๐ต
๐ด๐ถโ€ฒ
๐ด๐ถ
= 3 ๐‘กโ„Ž๐‘’๐‘› ๐ด๐ถโ€ฒ = 3๐ด๐ถ
๐ด๐ดโ€ฒ
๐ด๐ด
= 3 ๐‘กโ„Ž๐‘’๐‘› ๐ด๐ดโ€ฒ
= 3๐ด๐ด
But AA=0 then AAโ€™=0
This means that A and Aโ€™ are
confounded
Reduce triangle ABC of ratio 1/2 and center O
๐‘‚๐ตโ€ฒ
๐‘‚๐ต
=
1
2
๐‘กโ„Ž๐‘’๐‘› ๐‘‚๐ตโ€ฒ
=
1
2
๐‘‚๐ต
๐‘‚๐ถโ€ฒ
๐‘‚๐ถ
=
1
2
๐‘กโ„Ž๐‘’๐‘› ๐‘‚๐ถโ€ฒ =
1
2
๐‘‚๐ถ
๐‘‚๐ดโ€ฒ
๐‘‚๐ด
=
1
2
๐‘กโ„Ž๐‘’๐‘› ๐‘‚๐ดโ€ฒ =
1
2
๐‘‚๐ด
Reduce triangle ABC of ratio 1/2 and center A
๐ด๐ตโ€ฒ
๐ด๐ต
=
1
2
๐‘กโ„Ž๐‘’๐‘› ๐ด๐ตโ€ฒ
=
1
2
๐ด๐ต
๐ด๐ถโ€ฒ
๐ด๐ถ
=
1
2
๐‘กโ„Ž๐‘’๐‘› ๐ด๐ถโ€ฒ =
1
2
๐ด๐ถ
๐ด๐ดโ€ฒ
๐ด๐ด
=
1
2
๐‘กโ„Ž๐‘’๐‘› ๐ด๐ดโ€ฒ
=
1
2
๐ด๐ด
But AA=0 then AAโ€™=0
This means that A and Aโ€™ are
confounded

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Thales Theorem-construction for grade 9 .ppt

  • 2. Applications: -Construction of the fourth proportional -Dividing a segment into equal parts. -Constructions of points -Enlargement โ€“Reduction
  • 3. Construction of the fourth proportional โ€ข Question :construct the fourth proportional to 4,7 and 5. Etape 1:Trace a straight line (d) and place 3 points O,A & B in the following order :OA = 4cm and OB =7cm. O A B (d) 0 4 7 Etape 2:Trace a straight line (dโ€ฒ) that cuts (d) at O and place the point C such that OC = 5cm. O A B (d) 0 4 7 C 5 D x (dโ€ฒ)
  • 4. Step 3: construct the parallel to (AC) passing through B, it cuts (dโ€™) at D. According to Thalesโ€™ theorem, the length x of segment [OD] is: 5/x = 4/7 or x x 4 = 5 x 7 The length x of segment [OD] is the fourth proportional to 4,7 and 5.
  • 5. Dividing a segment into equal parts
  • 6. Dividing a segment into equal parts Divide [AB] into 4 equal parts
  • 7. Construction of points Place point K on segment [AB] such that ๐ดM = 2 5 ๐ด๐ต OR AM AB = 2 5
  • 8.
  • 9.
  • 10.
  • 11.
  • 12. Construction of points Place points M on line (AB) such that ๐‘€๐ด = 2 5 ๐‘€๐ต OR ๐‘€๐ด ๐‘€๐ต = 2 5
  • 17. Construction of points Place points M on line (AB) such that ๐‘€๐ด = 2 3 ๐‘€๐ต OR ๐‘€๐ด ๐‘€๐ต = 2 3
  • 18. Construction of points Place points M on line (AB) such that ๐‘€๐ด = 2 3 ๐‘€๐ต OR ๐‘€๐ด ๐‘€๐ต = 2 3
  • 19.
  • 20. Construction of points Place points M on line (AB) such that ๐‘€๐ด ๐‘€๐ต =2
  • 21.
  • 22. Enlargement and Reduction Enlargement and reduction can be seen in many situations in real life.
  • 23. Elements of enlargement and reduction The elements of enlargement or reduction are: 1) The ratio or the coefficient k: ๐‘˜ = ๐‘™๐‘’๐‘›๐‘”๐‘กโ„Ž ๐‘œ๐‘“ ๐‘›๐‘’๐‘ค ๐‘ ๐‘–๐‘‘๐‘’ (๐‘–๐‘š๐‘Ž๐‘”๐‘’) ๐‘™๐‘’๐‘›๐‘”๐‘กโ„Ž ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘œ๐‘Ÿ๐‘–๐‘”๐‘–๐‘›๐‘Ž๐‘™ โ€ข If k>1,then the image is an enlargement โ€ข If k<1, then the image is reduction 2) Center of enlargement or reduction
  • 24.
  • 25.
  • 26. Application: Given segment AB and Aโ€™Bโ€™ the image of AB by enlargement A B Aโ€™ Bโ€™ 1) Find the center of enlargement O 2)Suppose that OAโ€™=12 and OA=3 Find the ratio of enlargement O
  • 27. Enlarge triangle ABC of ratio 3 and center O ๐‘‚๐ตโ€ฒ ๐‘‚๐ต = 3 ๐‘กโ„Ž๐‘’๐‘› ๐‘‚๐ตโ€ฒ = 3๐‘‚๐ต ๐‘‚๐ถโ€ฒ ๐‘‚๐ถ = 3 ๐‘กโ„Ž๐‘’๐‘› ๐‘‚๐ถโ€ฒ = 3๐‘‚๐ถ ๐‘‚๐ดโ€ฒ ๐‘‚๐ด = 3 ๐‘กโ„Ž๐‘’๐‘› ๐‘‚๐ดโ€ฒ = 3๐‘‚๐ด
  • 28. Enlarge triangle ABC of ratio 3 and center A ๐ด๐ตโ€ฒ ๐ด๐ต = 3 ๐‘กโ„Ž๐‘’๐‘› ๐ด๐ตโ€ฒ = 3๐ด๐ต ๐ด๐ถโ€ฒ ๐ด๐ถ = 3 ๐‘กโ„Ž๐‘’๐‘› ๐ด๐ถโ€ฒ = 3๐ด๐ถ ๐ด๐ดโ€ฒ ๐ด๐ด = 3 ๐‘กโ„Ž๐‘’๐‘› ๐ด๐ดโ€ฒ = 3๐ด๐ด But AA=0 then AAโ€™=0 This means that A and Aโ€™ are confounded
  • 29. Reduce triangle ABC of ratio 1/2 and center O ๐‘‚๐ตโ€ฒ ๐‘‚๐ต = 1 2 ๐‘กโ„Ž๐‘’๐‘› ๐‘‚๐ตโ€ฒ = 1 2 ๐‘‚๐ต ๐‘‚๐ถโ€ฒ ๐‘‚๐ถ = 1 2 ๐‘กโ„Ž๐‘’๐‘› ๐‘‚๐ถโ€ฒ = 1 2 ๐‘‚๐ถ ๐‘‚๐ดโ€ฒ ๐‘‚๐ด = 1 2 ๐‘กโ„Ž๐‘’๐‘› ๐‘‚๐ดโ€ฒ = 1 2 ๐‘‚๐ด
  • 30. Reduce triangle ABC of ratio 1/2 and center A ๐ด๐ตโ€ฒ ๐ด๐ต = 1 2 ๐‘กโ„Ž๐‘’๐‘› ๐ด๐ตโ€ฒ = 1 2 ๐ด๐ต ๐ด๐ถโ€ฒ ๐ด๐ถ = 1 2 ๐‘กโ„Ž๐‘’๐‘› ๐ด๐ถโ€ฒ = 1 2 ๐ด๐ถ ๐ด๐ดโ€ฒ ๐ด๐ด = 1 2 ๐‘กโ„Ž๐‘’๐‘› ๐ด๐ดโ€ฒ = 1 2 ๐ด๐ด But AA=0 then AAโ€™=0 This means that A and Aโ€™ are confounded