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## 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)

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.

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.

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

Question 4.

Measure the angles of the given figures using a protractor and identify the type of angle as acute, obtuse, right or straight.

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°

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°

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°

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°

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°

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°

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°

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°

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°.

Question 6.

From the figures given below, classify the following pairs of angles into complementary and non-complementary.

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.

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

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°

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

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.

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)

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°

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°.