(i) \(\overleftrightarrow { AB }\)
(ii) \(\bar { BA } \)
(iii) one
(i) \(\overleftrightarrow { AB }\)
(ii) \(\bar { BA } \)
(iii) one
(1)
(i) Draw line 1 and mark a point A on it.
(ii) Measure 7.5 cm using a compass, placing the pointer at ‘O’ and the pencil pointer at 7.5 cm.
(iii) Place the pointer of the compass at A then draw a small arc on the line 1 with the pencil pointer. It cuts line 1 at a point and name that point as B.
(iv) Now \(\overline { AB } \) is the required line segment of length 7.5 cm.
(2)
(i) Draw a line 1 and mark a point C on it.
(ii) Measure 3.6 cm using a compass, placing the pointer at O and the pencil pointer at 3.6 cm.
(iii) Place the pointer of the compass at C then draw a small arc on the line 1 with the pencil pointer. It cuts line 1 at a point and names the point as D.
(iv) Now \(\overline { CD } \) is the required line segment of length 3.6 cm.
(3)
(i) Draw a line 1 and mark a point Q on it.
(ii) Measure 10 cm using compass placing the pointer at O and the pencil pointer at 10 cm.
(iii) Place the pointer of the compass at Q then draw a small arc on the line 1 with the pencil pointer. It cuts the line 1 at a point and name that point as R.
(iv) Now \(\overline { QR }\) is the required line segment of length 10 cm.
(1)
(i) Draw line 1 and mark a point A on it.
(ii) Measure 7.5 cm using a compass, placing the pointer at ‘O’ and the pencil pointer at 7.5 cm.
(iii) Place the pointer of the compass at A then draw a small arc on the line 1 with the pencil pointer. It cuts line 1 at a point and name that point as B.
(iv) Now \(\overline { AB } \) is the required line segment of length 7.5 cm.
(2)
(i) Draw a line 1 and mark a point C on it.
(ii) Measure 3.6 cm using a compass, placing the pointer at O and the pencil pointer at 3.6 cm.
(iii) Place the pointer of the compass at C then draw a small arc on the line 1 with the pencil pointer. It cuts line 1 at a point and names the point as D.
(iv) Now \(\overline { CD } \) is the required line segment of length 3.6 cm.
(3)
(i) Draw a line 1 and mark a point Q on it.
(ii) Measure 10 cm using compass placing the pointer at O and the pencil pointer at 10 cm.
(iii) Place the pointer of the compass at Q then draw a small arc on the line 1 with the pencil pointer. It cuts the line 1 at a point and name that point as R.
(iv) Now \(\overline { QR }\) is the required line segment of length 10 cm.
- A. AB
- B. \(\overrightarrow{AB}\)
- C. \(\overleftrightarrow {AB} \)
- D. \(\overline { AB }\)
(c) \(\overleftrightarrow {AB} \)
Posted in Class 6 on January 22, 2025 January 23, 2025
Leave a Reply Cancel reply
You must be logged in to post a comment.
Facebook
Twitter
Instagram
Pinterest
Copyright © 2026 Samacheer Kalvi
(c) \(\overleftrightarrow {AB} \)
Posted in Class 6 on January 22, 2025 January 23, 2025
Leave a Reply Cancel reply
You must be logged in to post a comment.
Facebook
Twitter
Instagram
Pinterest
Copyright © 2026 Samacheer Kalvi
(i) True
Hint: 20°+ 70° = 90°
(ii) False
Hint: 88° + 180° = 260° ≠ 1
(iii) False
Hint: 80° + 180° = 260° ≠ 1
(iv) True
Hint: 0° + 180° = 180°
(i) True
Hint: 20°+ 70° = 90°
(ii) False
Hint: 88° + 180° = 260° ≠ 1
(iii) False
Hint: 80° + 180° = 260° ≠ 1
(iv) True
Hint: 0° + 180° = 180°
v- A. more than 45°
- B. 45°
- C. less than 45°
- D. 90°
(b) 45°
Posted in Class 6 on January 22, 2025 January 23, 2025
Leave a Reply Cancel reply
You must be logged in to post a comment.
Facebook
Twitter
Instagram
Pinterest
Copyright © 2026 Samacheer Kalvi
(b) 45°
Posted in Class 6 on January 22, 2025 January 23, 2025
Leave a Reply Cancel reply
You must be logged in to post a comment.
Facebook
Twitter
Instagram
Pinterest
Copyright © 2026 Samacheer Kalvi
Number of points of intersection = 3
Objective Type Questions
Observe the Diagram and give answers
Number of points of intersection = 3
Objective Type Questions
Observe the Diagram and give answers
- A. A, B, C
- B. A, F, C
- C. B, C, D
- D. A, C, D
(b) A, F, C
Hint: Collinear points are points on a line.
(b) A, F, C
Hint: Collinear points are points on a line.
- A. E
- B. F
- C. G
- D. H
(b) F
Posted in Class 6 on January 22, 2025 January 23, 2025
Leave a Reply Cancel reply
You must be logged in to post a comment.
Facebook
Twitter
Instagram
Pinterest
Copyright © 2026 Samacheer Kalvi
(b) F
Posted in Class 6 on January 22, 2025 January 23, 2025
Leave a Reply Cancel reply
You must be logged in to post a comment.
Facebook
Twitter
Instagram
Pinterest
Copyright © 2026 Samacheer Kalvi
v(i) Parallel lines
(ii) Parallel lines
(iii) Parallel and Perpendicular lines
(iv) Intersecting lines
(i) Parallel lines
(ii) Parallel lines
(iii) Parallel and Perpendicular lines
(iv) Intersecting lines
Challenging Problems
Challenging Problems
(i) Legs of the table, railway tracks, edges of the scale
(ii) Adjacent sides of a Board, Crossbars of windows, Adjacent sides of the textbook
(iii) Crossbars of windows, Ladder, blades of a scissor.
(i) Legs of the table, railway tracks, edges of the scale
(ii) Adjacent sides of a Board, Crossbars of windows, Adjacent sides of the textbook
(iii) Crossbars of windows, Ladder, blades of a scissor.
Let the angle be x
According to the problem, x = 2 × (90 – x)
x = 180 – 2x
x + 2x = 180
3x = 180
x = \(\frac{180}{3}\)
x = 60
∴ The angle is 60°
Let the angle be x
According to the problem, x = 2 × (90 – x)
x = 180 – 2x
x + 2x = 180
3x = 180
x = \(\frac{180}{3}\)
x = 60
∴ The angle is 60°
Let the angle be x
According to the problem,
x = \(\frac{2}{3}\) × (180° – x)
3x = 2(180 – x)
3x = 360 – 2x
3x + 2x = 360°
5x = 360°
x = \(\frac{360°}{5}\)
x = 72°
∴ The angle is 72°
Let the angle be x
According to the problem,
x = \(\frac{2}{3}\) × (180° – x)
3x = 2(180 – x)
3x = 360 – 2x
3x + 2x = 360°
5x = 360°
x = \(\frac{360°}{5}\)
x = 72°
∴ The angle is 72°
Let the angles be x and x + 20°
According to the problem,
x + x + 20 = 180°
2x + 20° = 180°
2x = 180° – 20°
2x = 160°
x = \(\frac{160°}{2}\)
x = 80°
x + 20 = 80° + 20°
= 100°
∴ The two angles are 80° and 100°
Let the angles be x and x + 20°
According to the problem,
x + x + 20 = 180°
2x + 20° = 180°
2x = 180° – 20°
2x = 160°
x = \(\frac{160°}{2}\)
x = 80°
x + 20 = 80° + 20°
= 100°
∴ The two angles are 80° and 100°
Let the angles be 7x, 2x
According to the problem,
7x + 2x = 90
9x = 90
x = \(\frac{90}{9}\)
x = 10
7x = 7 × 10
= 70
2x = 2 × 10
= 20
∴ Two angles are 70° and 20°
Let the angles be 7x, 2x
According to the problem,
7x + 2x = 90
9x = 90
x = \(\frac{90}{9}\)
x = 10
7x = 7 × 10
= 70
2x = 2 × 10
= 20
∴ Two angles are 70° and 20°
Total of two supplementary angles = 180°
Given they are in the ratio 5 : 4
Dividing total angles to 5 + 4 = 9 equal parts.
One angle \(=\frac{5}{9} \times 180=100^{\circ}\)
Another angle \(=\frac{4}{9} \times 180=80^{\circ}\)
Two angles are 100° and 80°.
Posted in Class 6 on January 22, 2025 January 23, 2025
Leave a Reply Cancel reply
You must be logged in to post a comment.
Facebook
Twitter
Instagram
Pinterest
Copyright © 2026 Samacheer Kalvi
Total of two supplementary angles = 180°
Given they are in the ratio 5 : 4
Dividing total angles to 5 + 4 = 9 equal parts.
One angle \(=\frac{5}{9} \times 180=100^{\circ}\)
Another angle \(=\frac{4}{9} \times 180=80^{\circ}\)
Two angles are 100° and 80°.
Posted in Class 6 on January 22, 2025 January 23, 2025
Leave a Reply Cancel reply
You must be logged in to post a comment.
Facebook
Twitter
Instagram
Pinterest
Copyright © 2026 Samacheer Kalvi