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Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Ex 4.4

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.4

Miscellaneous Practice Problems

Question 1.
Find the type of lines marked in thick lines (Parallel, intersecting or perpendicular)
Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q1
Solution:
(i) Parallel lines
(ii) Parallel lines
(iii) Parallel lines and Perpendicular lines
(iv) Intersecting lines

Question 2.
Find the parallel and intersecting line segments in the picture given below.
Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q2
Solution:
(a) Parallel line segments

  • \(\overline{\mathrm{YZ}} \text { and } \overline{\mathrm{DE}}\)
  • \(\overline{\mathrm{EA}} \text { and } \overline{\mathrm{ZV}}\)
  • \(\overline{\mathrm{VW}} \text { and } \overline{\mathrm{AB}}\)
  • \(\overline{\mathrm{WX}} \text { and } \overline{\mathrm{BC}}\)
  • \(\overline{\mathrm{YX}} \text { and } \overline{\mathrm{DC}}\)
  • \(\overline{\mathrm{YD}} \text { and } \overline{\mathrm{XC}}\)
  • \(\overline{\mathrm{XC}} \text { and } \overline{\mathrm{WB}}\)
  • \(\overline{\mathrm{WB}} \text { and } \overline{\mathrm{VA}}\)
  • \(\overline{\mathrm{VA}} \text { and } \overline{\mathrm{ZE}}\)
  • \(\overline{\mathrm{ZE}} \text { and } \overline{\mathrm{YD}}\)

(b) Intersecting line segments

  • DE and ZV
  • WX and DC

Question 3.
Name the following angles as shown in the figure.
Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q3
Solution:
(i) ∠1 = ∠DBC or ∠CBD
(ii) ∠2 = ∠DBE or ∠EBD
(iii) ∠3 = ∠ABE or ∠EBA
(iv) ∠1 + ∠2 = ∠EBC or ∠CBE
(v) ∠2 + ∠3 = ∠ABD or ∠DHA
(vi) ∠1 + ∠2 + ∠3 = ∠ABC or ∠B or ∠CBA

Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Ex 4.4

Question 4.
Measure the angles of the given figures using a protractor and identify the type of angle as acute, obtuse, right or straight.
Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q4
Solution:
(i) 90° – Right Angle
(ii) 45° – Acute Angle
(iii) 180° – Straight Angle
(iv) 105° – Obtuse Angle

Question 5.
Draw the following angles using the protractor.
(i) 45°
(ii) 120°
(iii) 65°
(iv) 135°
(v) 0°
(vi) 180°
(vii) 38°
(viii) 90°
Solution:
(i) 45°
Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q5.2
Construction:
1. Drawn the base ray PQ.
2. Placed the centre of the protractor at the vertex P. Lined up the ray \(\overrightarrow{\mathrm{PQ}}\) with the 0° line. Then drawn and labelled a pointed (R) at the 45° mark on the inner scale (a) anticlockwise and (b) outer scale (clockwise)
3. Removed the protractor and drawn at \(\overrightarrow{\mathrm{PR}}\) to complete the angle
Now ∠P = ∠QPR – ∠RPQ = 45°.

(ii) 120°
Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q5.1
Construction:
1. Placed the centre of the protractor at the vertex X. Lined up the ray \(\overline{\mathrm{XY}}\) with the 0° Line. Then draw and label a point Z at 120° mark on the (a) inner scale (anti-clockwise) and (b) outer scale (clockwise).
2. Removed the protractor and draw \(\overline{\mathrm{XZ}}\) to complete the angle.
Now, ∠X = ∠ZXY = ∠YXZ = 120°.

(iii) 65°
Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q5.2
Construction:
1. Placed the centre of the protractor at the vertex A. Line up the ray \(\overrightarrow{\mathrm{AB}}\) with the 0° line. Then draw and label a point C at the 65° mark on the (a) inner scale (anti-clockwise) (b) outer scale (clockwise).
2. Removed the protractor and draw \(\overrightarrow{\mathrm{AB}}\) to complete the angle.
Now ∠A = ∠BAC = ∠CAB = 65°.

(iv) 135°
Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q5.3
Construction:
1. Placed the centre of the protractor at the Vertex A. Lined up the ray \(\overrightarrow{\mathrm{EG}}\) with the 0° line. Then draw and label a point F at the 135° mark on the (a) inner scale (anti-clockwise) and (b) outer scale (clockwise)
2. Removed the protractor and draw \(\overrightarrow{\mathrm{EF}}\) to complete the angle.
Now ∠E = ∠FEG = ∠GEF = 135°.

(v) 0°
Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q5.4
Construction:
1. Placed the centre of the protractor at the vertex G. Lined up the ray \(\overrightarrow{\mathrm{GH}}\) with the 0° line. Then draw and label a point I at the 0° mark on the
(a) inner scale (anti-clockwise)
(b) outer scale (clockwise)
2. Removed the protractor and seen \(\overrightarrow{\mathrm{GI}}\) lies exactly on \(\overrightarrow{\mathrm{GH}}\)
Now ∠G = ∠HGI = ∠IGH = 0°, which is a zero angle.

(vi) 180°
Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q5.5
Construction:
1. Placed the centre of the protractor at the vertex I. Lined up the ray \(\overrightarrow{\mathrm{IJ}}\) with the 0° line. Then draw and labelled a point K at the 180° mark on the (a) inner scale (anticlockwise) (b) outer scale (clockwise)
2. Removed the protractor and draw \(\overrightarrow{\mathrm{IK}}\) to complete the angle.
Now ∠I = ∠JHK = ∠KIJ = 180°, which is a straight Angle.

(vii) 38°
Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q5.6
Construction:
1. Placed the centre of the protractor at the vertex L. Lined up the ray \(\overrightarrow{\mathrm{LM}}\) with the 0° line. Then draw and label a point N at 38° mark on the (a) inner scale (anticlockwise) and (b)*huter scale (clockwise).
2. Removed the protractor and draw \(\overrightarrow{\mathrm{LN}}\) to complete the angle.
Now ∠L = ∠MLN = ∠NLM = 38°.

(viii) 90°
Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q5.7
Construction:
1. Placed the centre of the protractor at the vertex ‘O’. Lined up the ray \(\overrightarrow{\mathrm{OP}}\) with the 0° line. Then draw and label a point Q at 90° mark on the (a) inner scale (anticlockwise) and (b) outer scale (clockwise)
2. Removed the protractor and draw \(\overrightarrow{\mathrm{OQ}}\) to complete the angle.
Now ∠O = ∠POQ = ∠QOP = 90°.

Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Ex 4.4

Question 6.
From the figures given below, classify the following pairs of angles into complementary and non-complementary.
Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q6
Solution:
We know that the two angles are complementary if they add up to 90°.
Therefore (a) (i) is complementary.
In (v) ∠ABC and ∠CBD are complementary
(b) (ii), (iii), (iv) and (v) are non-complementary

Question 7.
From the figures given below, classify the following pairs of angles into supplementary and non-supplementary.
Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q7
Solution:
If two angles add up to 180°, then they are supplementary angles.
(a) In (ii) ∠AOB and ∠BOD are supplementary. In (iv) the pair is supplementary
(b) (i) and (iii) are not supplementary.

Question 8.
From the figure.
(i) name a pair of complementary angles
(ii) name a pair of supplementary angles
Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q8
Solution:
(i) ∠FAE and ∠DAE are complementary
(ii) ∠FAD and ∠DAC are supplementary

Question 9.
Find the complementary angle of
(i) 30°
(ii) 26°
(iii) 85°
(iv) 0°
Solution:
When we have an angle, how far we need to go to reach the right angle is called the complementary angle.
(i) Complementary angle of 30° is 90° – 30° = 60°
(ii) Complementary angle of 26° is 90° – 26° = 64°
(iii) Complementary angle of 85° is 90° – 85° = 5°
(iv) Complementary angle of 0° is 90° – 0° = 90°
(v) Complementary angle of 90° is 90° – 90° = 0°

Question 10.
Find the supplementary angle of
(i) 70°
(ii) 35°
(iii) 165°
(iv) 90°
(v) 0°
(vi) 180°
(vii) 95°
Solution:
How far we should go in the same direction to reach the straight angle (180°) is called supplementary angle.
(i) Supplementary angle of 70° = 180° – 70° = 110°
(ii) Supplementary angle of 35° is 180° – 35° = 145°
(iii) Supplementary angle of 165° is 180° – 165° = 15°
(iv) Supplementary angle of 90° is 180° – 90° = 90°
(v) Supplementary angle of 0° is 180° – 0° = 180°
(vi) Supplementary angle of 180° is 180° – 180° = 0°
(vii) Supplementary angle of 95° is 180° – 95° = 85°

Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Ex 4.4

Challenging Problems

Question 11.
Think and write and object having.
(i) Parallel Lines
1. _____________
2. _____________
3. _____________
(ii) Perpendicular lines
1. _____________
2. _____________
3. _____________
(iii) Intersecting lines
1. _____________
2. _____________
3. _____________
Solution:
(i) 1. Opposite edges of a Table.
2. Path traced by the wheels of a car on a straight road
3. Opposite edges of a black board
(ii) 1. Adjacent edges of a Table.
2. Hands of the block when it shows 3.30
3. Strokes of the letter ‘L’
(iii) 1. Sides of a triangle
2. Strokes of letter ‘V’
3. Hands of a scissors

Question 12.
Which angle is equal to twice of its complement?
Solution:
We know that the sum of complementary angles 90°
Given Angle = 2 × Complementary angle
Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q12
By trial and error, we find that Angle = 2 × Complement for 60°
The required angle = 60°
Another method:
Let the angle be x given
x = 2 (90 – x)
⇒ x = 180 – 2x
⇒ x + 2x = 180
⇒ 3x = 180
⇒ x = 60°

Question 13.
Which angle is equal to two-thirds of its supplement.
Solution:
Supplementary angles sum upto 180°
Given Angle = \(\frac{2}{3}\) × Supplement.
Forming the Table.
Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q13
By trial and error, we find that angle = \(\frac{2}{3}\) × supplement for 72°.
The required angle 72°.

Question 14.
Given two angles are supplementary and one angle is 20° more than other. Find the two angles.
Solution:
Given two angles are supplementary i.e. their sum = 180°.
Let the angle be x
Then another angle = x + 20 (given)
Samacheer Kalvi 6th Maths Term 1 Chapter 4 Geometry Ex 4.4 Q14
The two angles are 80° and 100°.

Question 15.
Two complementary angles are in the ratio 7 : 2. Find the angles.
Solution:
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°

Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Ex 4.4

Question 16.
Two supplementary angles are in ratio 5 : 4. Find the angles.
Solution:
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°.

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