Term 1 · Class 6 Maths · Chapter 4

Samacheer Class 6 Maths - Geometry

24 textbook Q&AFree Content

Chapter-wise textbook exercise answers for Geometry with step-by-step solutions.

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1Book Back Questions24 questions
Q.1A line through two endpoints ‘A’ and ‘B’ is denoted by ______v
Answer:

\(\overleftrightarrow { AB }\)

Q.2Aline segment from point ‘B’ to point ‘A’ is denoted by ______v
Answer:

\(\bar { BA } \)

Q.3A ray has ______ endpoint(s).v
Answer:

one

Q.4Construct a line segment using a ruler and compass. (1) \(\overline { AB } \) = 7.5 cm (2) \(\overline { CD } \) = 3.6 cm (3) \(\overline { QR } \) = 10 cmv
Answer:

(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.

Q.5A line is denoted as __________v
  1. A. AB
  2. B. \(\overrightarrow{AB}\)
  3. C. \(\overleftrightarrow {AB} \)
  4. D. \(\overline { AB }\)
Answer:

(c) \(\overleftrightarrow {AB} \)

Q.620° and 70° are complementary.v
Answer:

True
Hint: 20°+ 70° = 90°

Q.788° and 12° are complementary.v
Answer:

False
Hint: 88° + 180° = 260° ≠ 1

Q.880° and 180° are supplementary.v
Answer:

False
Hint: 80° + 180° = 260° ≠ 1

Q.90° and 180° are supplementary.v
Answer:

True
Hint: 0° + 180° = 180°

Q.10Draw and label each of the angles. (i) ∠NAS = 90°n (ii) ∠BIG = 35° (iii) ∠SMC = 145°v
Answer:

Q.11In this figure, ∠AYZ = 45. If point ‘A’ is shifted to point ‘B’ along the ray, then the measure of ∠BYZ is v
  1. A. more than 45°
  2. B. 45°
  3. C. less than 45°
  4. D. 90°
Answer:

(b) 45°

Q.12Draw any line and mark any 3 points that are collinear.v
Answer:

Q.13Draw any line and mark any 4 points that are not collinear.v
Answer:

Q.14Draw any 3 lines to have a point of concurrency.v
Answer:

Q.15Draw any 3 lines that are not concurrent. Find the number of points of intersection.v
Answer:

Number of points of intersection = 3Objective Type Questions
Observe the Diagram and give answers

Q.16Find the type of lines marked in thick lines (Parallel, intersecting or perpendicular) v
Answer:

(i) Parallel lines
(ii) Parallel lines
(iii) Parallel and Perpendicular lines
(iv) Intersecting lines

Q.17Find the complementary angle of (i) 30° (ii) 26° (iii) 85° (iv) 0° (v) 90°v
Answer:

Q.18Find the supplementary angle of (i) 70° (ii) 35° (iii) 165° (iv) 90° (v) 0° (vi) 180° (vii) 95°v
Answer:

Challenging Problems

Q.19Think and write an object having (i) Parallel lines (1) ……….. (2) ………. (3) ……….. (ii) Perpendicular lines (1) ……… (2) ……… (3) ……….. (iii) Intersecting lines (1) ……….. (2) ………. (3) ……….v
Answer:

(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.

Q.20Which angle is equal to twice its complement.v
Answer:

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°

Q.21Which angle is equal to two-thirds of its supplement.v
Answer:

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°

Q.22Given two angles are supplementary and one angle is 20° more than the other. Find the two angles.v
Answer:

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°

Q.23Two complementary angles are in ratio 7 : 2. Find the angles.v
Answer:

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°

Q.24Two supplementary angles are in ratio 5 : 4. Find the angles.v
Answer:

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°.