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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:

Parallel lines:

\(\overline { YX }\) and \(\overline { DC }\)

\(\overline { YD }\) and \(\overline { ZE }\)

\(\overline { XC }\) and \(\overline { YD }\)

\(\overline { YZ }\) and \(\overline { DE }\)

\(\overline { XW }\) and \(\overline { BC }\)

\(\overline { XC }\) and \(\overline { WB }\)

\(\overline { WV }\) and \(\overline { BA }\)

\(\overline { ZV }\) and \(\overline { EA }\)

\(\overline { VA }\) and \(\overline { WB }\)

\(\overline { ZE }\) and \(\overline { VA }\)

Concurrent Lines:

\(\overline { AB }\), \(\overline { AE }\), \(\overline { AV }\)

\(\overline { BA }\), \(\overline { BC }\), \(\overline { BW }\)

\(\overline { CB }\), \(\overline { CX }\), \(\overline { CD }\)

\(\overline { DC }\), \(\overline { DE }\), \(\overline { DY }\)

\(\overline { EA }\), \(\overline { EZ }\), \(\overline { ED }\)

\(\overline { XC }\), \(\overline { XY }\), \(\overline { XW }\)

\(\overline { YX }\), \(\overline { YZ }\), \(\overline { YD }\)

\(\overline { ZY }\), \(\overline { ZE }\), \(\overline { ZV }\)

\(\overline { VA }\), \(\overline { VW }\), \(\overline { VZ }\)

\(\overline { WB }\), \(\overline { WV }\), \(\overline { WX }\)

Question 3.

Name the following angles as shown in the figure.

Solution:

(i) ∠1 = ∠CBD or ∠DBC

(ii) ∠2 = ∠DBE or ∠EBD

(iii) ∠3 = ∠ABE or ∠EBA

(iv) ∠1 + ∠2 = ∠CBE or ∠EBC

(v) ∠2 + ∠3 = ∠ABD or ∠DBA

(vi) ∠1 + ∠2 + ∠3 = ∠ABC 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:

(i) and (v) are complementary angles.

(ii), (iii) and (iv) non-complementary angles.

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

- 70°
- 35°
- 165°
- 90°
- 0°
- 180°
- 95°

Solution:

How far we should go in the same direction to reach the straight angle (180°) is called the supplementary angle.

- Supplementary angle of 70° = 180° – 70° = 110°
- Supplementary angle of 35° is 180° – 35° = 145°
- Supplementary angle of 165° is 180° – 165° = 15°
- Supplementary angle of 90° is 180° – 90° = 90°
- Supplementary angle of 0° is 180° – 0° = 180°
- Supplementary angle of 180° is 180° – 180° = 0°
- 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 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:

Total of complementary angles = 90°.

The angles are in the ratio 7 : 2

Dividing total angles to 7 + 2 = 9 equal parts

One angle \(=\frac{7}{9} \times 90=70^{\circ}\)

Another angle \(=\frac{2}{9} \times 90=20^{\circ}\)

Two angles are 70° and 20°.

Question 16.

Two supplementary angles are in ratio 5 : 4. Find the angles.

Solution:

Let the angles be 5x and 4x

According to the problem

5x + 4x = 180°

9x = 180°

x = \(\frac{180°}{9}\)

x = 20°

∴ Two angles are

(i) 5x = 5 × 20° = 100°

(ii) 4x = 4 × 20° = 80°