Class 6

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 9 2D-3D Shapes InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 9th Lesson 2D-3D Shapes InText Questions

(Page No. 125)

Question 1.
Draw six different types of rough sketches of polygons in your notebook. In which case, it is not possible to form a polygon ?
Hence, what is the least number of sides needed to form a polygon ? Obviously three.
Solution:

From the above figure we conclude that with one side and two sides cannot form a polygon. So, atleast three sides needed to form a polygon.

Check Your Progress (Page No. 127)

Look at the adjacent figure.

Question 1.
Which points are marked in the interior of A GHI ?
Solution:
The points marked in the interior of AGHI are A, B and 0.

Question 2.
Which points sure marked on the triangle?
Solution:
The points marked on the triangle are G, P, H, I and Y.

Question 3.
Which points are’marked in the exterior of A GHI ?
Solution:
The points marked in the exterior of AGHI are M, R, S, X and Z.

(Page No. 130)

Question 1.
Take a rectangular sheet (like a post-card). Fold it along its length so that one half fits exactly over the other half. Is this fold a line of symmetry ? Why ?
Solution:
Yes. Because the two parts of the rectangular sheet (post-card) coincide exactly with each other.
So, the folded line is the line of symmetry.

Question 2.
Open it up now and again fold along its width in the same way. Is this second fold also a line of symmetry ? Why ?
Solution:
Yes. Because the two parts of the rectangular sheet (post-card) coincide exactly with each other.
So, the second folded line also a line of symmetry.

Question 3.
Do you find that these two lines are the lines of symmetry ?
Solution:
Yes. These two lines are the lines of symmetry. One is line of symmetry along length and the other is line of symmetry along width.

Project (Page No. 130)

Question 1.
Collect symmetrical figures from your surroundings and prepare a scrap book.
Solution:

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 9 2D-3D Shapes Ex 9.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 9th Lesson 2D-3D Shapes Ex 9.2

Question 1.
Look at the given triangle and answer the following questions.
i) Which points are marked in the exterior of the triangle?
ii) Which points are marked on the triangle?
iii) Which points are marked in the interior of the triangle?

Solution:
i) The points marked in the exterior of the triangle are X, Y and Z.
ii) The points marked on the triangle are A, I, B, J and C.
iii) The points marked in the interior of the triangle are K, L and O.

Question 2.
Look at the adjacent figure. Answer the following questions.
i) How many sides are there in the triangle? What are they?
ii) How many vertices lie there on the triangle? What are they?
iii) What is the side opposite to the vertex P?
iv) What is the vertex opposite to \(\overline{\mathrm{PR}}\) ?

Solution:
In the given ∆PQR,
i) It has three sides. They are \(\overline{\mathrm{PQ}}, \overline{\mathrm{QR}}\) and \(\overline{\mathrm{PR}}\)
ii) It has three vertices. They are P, Q and R.
iii) The side opposite to the vertex P is \(\overline{\mathrm{QR}}
iv) The vertex opposite to the side [latex]\overline{\mathrm{PR}}\) is Q.

Question 3.
Look at the given triangle and answer the following questions.
i) How many angles are there in the triangle? What are they?
ii) What is the angle opposite to \(\overline{\mathrm{MN}}\) ?
iii) Where is the right angle in the given triangle ?

Solution:
In the given ∆MON,
i) It has three angles, they are ∠MNO, ∠NOM and ∠OMN.
ii) Angle opposite to the side \(\overline{\mathrm{MN}}\) is ∠MON or ∠O.
iii) In the triangle ∠MON is the right angle.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 9 2D-3D Shapes Ex 9.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 9th Lesson 2D-3D Shapes Ex 9.1

Question 1.
What is the name of four sided polygon? Draw the rough sketch.
Solution:
A four sided polygon is called a quadrilateral.

ABCD is a quadrilateral.

Question 2.
Draw a rough sketch of pentagon.
Solution:
A five sided polygon is called a pentagon.
So, pentagon has five sided figure.

ABCDE is a pentagon.

Question 3.
Write all the sides of the given polygon ABCDEF.
Solution:
Given polygon ABCDEF is a Hexagon.
Its six sides are \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CD}}, \overline{\mathrm{DE}}, \overline{\mathrm{EF}} \text { and } \overline{\mathrm{FA}}\)
So, it is called a Hexagon.

Question 4.
Write the interior angles of the polygon PQRST.
Solution:
Given the polygon PQRST has five sides.
So, it is called a Pentagon.
It has five interior angles.
They are ∠TPQ, ∠PQR, ∠QRS, ∠RST and ∠STP.
They can also be written as ∠P, ∠Q, ∠R, ∠S, ∠T.

Question 5.
Measure the length of the sides of the polygon PQRST.
Solution:
The given polygon has five sides.
They are \(\overline{\mathrm{PQ}}\) = 2cm; \(\overline{\mathrm{QR}}\) = 2.5 cm;
\(\overline{\mathrm{RS}}\) .= 2.4 cm; \(\overline{\mathrm{ST}}\) = 2.2 cm, \(\overline{\mathrm{PT}}\) = 2.5 cm
It’s a pentagon.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 8th Lesson Basic Geometric Concepts InText Questions

(Page No. 111)

i) How many rays are there ?
Solution:
Four

ii) How many coins are close to player Q ?
Solution:
B, C and D coins are close to player Q.

iii) While striking with striker there is a possibility of a coin touches with any other. Draw all such possibilities in the given picture by means of the line segments.
Solution:
\(\overline{\mathrm{CB}}\) and \(\overline{\mathrm{DE}}\) .

iv) How many such line segments can be drawn in the picture ?
Solution:
\(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{ED}}, \overline{\mathrm{AC}}, \overline{\mathrm{AE}}, \overline{\mathrm{CD}}, \overline{\mathrm{CE}}, \overline{\mathrm{AD}}, \overline{\mathrm{BD}} \text { and } \overline{\mathrm{BE}}\)

Let’s Do (Page No. 112)

Question 1.
Observe the table and their notations and fill the gaps.

Solution:

Lets Explore (Page No. 114)

Question 1.
Measure the lengths of all line segments in the given figures by using divider and scale. Then compare the sides of the given figures.

Solution:
i) In ΔABC; \(\overline{\mathrm{AB}}\) = 2.2 cm; \(\overline{\mathrm{BC}}\) = 2 cm and \(\overline{\mathrm{AC}}\) = 2.2 cm ‘
2.2 cm = 2.2 cm > 2 cm i.e., \(\overline{\mathrm{AB}}=\overline{\mathrm{AC}}>\overline{\mathrm{BC}}\)
Two sides are equal and one side is different in length.

ii) In PQRS rectangle \(\overline{\mathrm{PS}}=\overline{\mathrm{QR}}\) = 2.7 cm; \(\overline{\mathrm{P Q}}=\overline{\mathrm{RS}}\) = 1.8 cm and \(\overline{\mathrm{PR}}=\overline{\mathrm{QS}}\) = 3.2 cm.
Opposite sides are equal and diagonals are equal in length,

iii) In KLMN square \(\overline{\mathrm{KL}}=\overline{\mathrm{LM}}=\overline{\mathrm{MN}}=\overline{\mathrm{KN}}\) = 1.8 cm
All sides are equal in length.

Let’s Do (Page No. 115)

Question 1.
(i) Find the parallel lines in the below figure. Name, write and read them.

Solution:
a) l, m are parallel lines.
We denote this by writing l || m and can be read as l is parallel to m.
b) n, o are parallel lines.
We denote this by writing n||o and can be read as n is parallel to o.
c) p, q are parallel lines.
We denote this by writing p||q and can be read as p is parallel to q.

ii) Find the intersecting lines in the above figure. Name, write and read them.
Solution:
Intersecting lines are (l, q); (m,q); (m, r); (n,q); (p, r); (o, r); (o, q); (q, r);

iii) Find the concurrent lines in the above figure. Name, write and read them.
Solution:
Three or more lines passing through the same point are called concurrent lines. Concurrent lines are (l, o, p) & (m, n, p).

iv) Find the perpendicular lines in the above figure. Name, write and read them.
Solution:
a) p, o are perpendicular lines.
We denote this by writing p ⊥ o and can be read as p is perpendicular to o.
b) p, n are perpendicular lines.
We denote this by writing p ⊥ n and can be read as p is perpendicular to n.
c) n, q are perpendicular lines.
We denote this by writing n ⊥ q and can be read as n is perpendicular to q.
d) q, o are perpendicular lines.
We denote this by writing q⊥ o and can be read as q is perpendicular to o.

(Page No. 119)

Question 1.
Measure the angles at the vertices.

Solution:

In triangle ABC,
m∠BAC = 60°
m∠ABC = 60°
m∠ACB = 60°

In triangle XYZ,
m∠YXZ = 40°
m∠XYZ = 70°
m∠XZY = 70°

In triangle PQR,
m∠QPR = 35°
m∠PQR = 38°
m∠PRQ = 107°

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 7th Lesson Introduction to Algebra InText Questions

Let’s Explore (Page No. 102)

Question 1.
Arrange 2 matchsticks to form the shape Continue the same shape for 2 times, 3 times and 4 times. Frame the rule for repeating the pattern.
Solution:
To make the given shape 2 matchsticks are needed.
To make the given 2 shapes 4 matchsticks are needed.
To make the given 3 shapes 6 matchsticks are needed.
To make the given 4 shapes 8 matchsticks are needed.
Continue and arrange the information in the following table.

Number of matchsticks required = 2 × Number of shapes to be formed
= 2 × x = 2x

Question 2.
Rita took matchsticks to form the shape
She repeated the pattern and gave a rule.
Number of matchsticks needed = 6.y, where y is the number of shapes to be formed. Is it correct ? Explain.
What is the number of sticks needed to form 5 such shapes ?
Solution:

To make the given shape 6 matchsticks are needed.
To make the given 2 shapes 12 matchsticks are needed.
To make the given 3 shapes 18 matchsticks are needed.
Continue and arrange the information in the following table.

Yes, it is correct.
Number of matchsticks required = 2 × Number of shapes to be formed
= 2 × y = 2y
Number of matchsticks needed to form 5 such shapes = 6 × 5 = 30

Let’s Explore (Page No. 103)

Question 1.
A line of shapes is constructed using matchsticks.

Shape-1 Shape-2 Shape-3 Shape-4
i) Find the rule that shows how many sticks are needed to make a line of such shapes ?
ii) How many matchsticks are needed to form shape -12 ?
Solution:

Number of matchsticks 3 5 7 9
i) Let us know the pattern
S₁ = 3 = 2 + 1 = (1 × 2) + 1
S₂ = 5 = 4 + 1 = (2 × 2) + 1
S₃ = 7 = 6 + 1 = (3 × 2) + 1
S₄ – 9 = 8 + 1 = (4 × 2) + 1
Now the rule for this pattern is number of matchsticks.

ii) Used to make ‘n’ number of shapes is Sn = (n × 2) + 1 = 2n + 1
Number of matchsticks needed to form shape – 12 is
S₁₂ = 2(12) + 1 = 24 + 1 = 25 sticks.

Check Your Progress (Page No. 105)

Question 1.
Fill the following table as instructed. One is shown for you.

Solution:

Let’s Explore ? (Page No. 106)

Question 1.
Find the general rule for the perimeter of a rectangle. Use variables T and ‘b’ for length and breadth of the rectangle respectively.
Solution:
Given length of rectangle = l
breadth of rectangle = b
We know that the perimeter of rectangle is twice the sum of its length and breadth.
Sum of length and breadth = l + b
Twice the sum of length and breadth = 2 × (l + b)
Rule for the perimeter of a rectangle = 2(l + b)

Question 2.
Find the general rule for the area of a square by using the variable ‘s’ for the side of a square.
Answer:
Given side of a square = s
We know that the area of a square is the product of side and side.
Area of a square = side × side
Rule for the area of a square = s.s

(Page No. 107)

Question 1.
Find the nth term in the following sequences.
0 3, 6, 9, 12, ii) 2, 5, 8, 11, iii) 1, 4, 9, 16,
Solution:
i) Given number pattern is 3, 6, 9, 12,……………..
To find the nth term in the given pattern, we put the sequence in a table.

First number = 3 × 1
Second number = 3 × 2
nth number = 3 × n = 3n
So, the nth term of the pattern 3, 6, 9, 12, is 3n.

ii) Given number pattern is 2, 5, 8, 11,
To find the nth term in the given pattern, we put the sequence in a table.

First number = 2 = 3 × 1 – 1
Second number = 5 = 3 × 2 – 1
Third number = 8 = 3 × 3 – 1
nth number = 3 × n – 1 = 3n – 1
So, the nth term of the pattern 2, 5, 8, 11 is 3n – 1.

iii) Given number pattern is 1, 4, 9, 16,
To find the nth term in the given pattern, we put the sequence in a table.

First number =1 = 1 × 1
Second number = 4 = 2 × 2
Third number =9 = 3 × 3
nth number = n × n = n²
So, the nth term of the pattern 1, 4, 9, 16 is n².

Check Your Progress (Page No. 108)

Question 1.
Complete the table and find the value of ‘p’ for the equation \(\frac{\mathbf{p}}{\mathbf{3}}\) = 4

Solution:

Question 2.
Write LHS and RHS of following simple equations.
i) 2x + 1 = 10
ii) 9 = y – 2
iii) 3p + 5 = 2p + 10
Solution;
i) 2x+ 1 = 10
Given equation is 2x + 1 = 10
L.H.S = 2x + 1
R.H.S = 10

ii) 9 = y – 2
Given equation is 9 = y – 2
LHS = 9
RHS = y – 2

iii) 3p + 5 = 2p + 10
Given equation is 3p + 5 = 2p + 10
LHS = 3p + 5
RHS = 2p + 10

Question 3.
Write any two simple equations and write their LHS and RHS.
Solution:
i) Consider 8x + 3 = 4 is a simple equation.
L.H.S = 8x + 3
RHS = 4

ii) Consider 5a + 6 = 8a – 3 is a simple equation.
LHS = 5a + 6
RHS = 8a – 3

Let’s Explore (Page No. 109)

Observe for what value of m, the equation 3m = 15 has both LHS and RHS become equal.
Solution:
Given equation is 3m = 15
If m = 1, then the value of 3m = 3(1) = 3≠15 ∴ LHS ≠RHS
If m = 2, then the value of 3m = 3(2) = 6 ≠ 15 ∴ LHS ≠ RHS
If m = 3, then the value of 3m = 3(3) = 9≠15 ∴ LHS ≠ RHS
If m = 4, then the value of 3m = 3(4) = 12 ≠ 15 ∴ LHS ≠ RHS
If m = 5, then the value of 3m = 3(5) = 15 = 15 ∴ LHS = RHS
From the above when m = 5 the both LHS and RHS are equal

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Unit Exercise Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 8th Lesson Basic Geometric Concepts Unit Exercise

Question 1.
In the given figure, measure the length of AC. Check whether
i) AB + AC > AC
ii) AC > AD – DC
Solution:
In the given figure, AB = 4.2 cm; BC = 5.5 cm
AC = 5.4 cm; CD = 3 cm; AD = 4 cm.
i) AB + AC = 4.2 + 5.4 = 9.6 cm > 5.4 cm
AB + AC > AC
ii) AD – DC = 4 – 3 = 1 cm < 5.4 cm
AD – DC < AC (or) AC > AD – DC

Question 2.
Draw a line segment \(\overline{\mathbf{A B}}\). Put a point C in between \(\overline{\mathbf{A B}}\). Extend \(\overline{\mathbf{C B}}\) upio D such that CD > AB. Now check whether AC and BD are equal length.
Solution:

Let draw a line \(\overline{\mathbf{A B}}\) = 5 cm and mark a point C on \(\overline{\mathbf{A B}}\) such that \(\overline{\mathbf{A C}}\) = 3 cm (or) \(\overline{\mathbf{BC}}\) = 2 cm Extend \(\overline{\mathbf{C B}}\) up to D such that \(\overline{\mathbf{C D}}\) = 5 cm (\(\overline{\mathbf{C B}}\) = 2 cm, \(\overline{\mathbf{B D}}\) = 3 cm)
∴ \(\overline{\mathbf{A C}}\) = 3cm, \(\overline{\mathbf{C B}}\) = 2cm, \(\overline{\mathbf{B D}}\) = 3cm
∴ \(\overline{\mathrm{AC}}=\overline{\mathrm{BD}}\) = 3cm.

Question 3.
Draw an angle ∠AOB as m∠AOB = 40°. Draw an angle ∠BOC such that ∠AOC = 90°
Check whether m∠AOB + m∠BOC = m∠AOC.
Sol. Draw an angle ∠AOB = 40° and ∠AOC = 90° on the same ray \(\overrightarrow{\mathrm{OA}}\)
Now, measure ∠AOB = 40° and ∠BOC = 50°
∴ m∠AOB + m∠BOC = 40° + 50 = 90°
m∠AOB + m∠BOC = 90° = m∠AOC
∴ m∠AOB + m∠BOC = m∠AOC.

Question 4.
Draw an angle ∠XYZ as m∠XYZ = 62°. Measure the exterior angle ∠XYZ.
Solution:

Draw angle ∠XYZ = 62° (Interior)
Now, measure the exterior angle ∠XYZ.
∴ Exterior angle of ∠XYZ = 298°

Question 5.
Match the following.
1. Set square — A) to measure angles
2. Protractor — B) to measure the lengths of line segments
3. Divider — C) to draw parallel lines
Sol. 1) Set square — C) to draw parallel lines
2) Protractor — A) to measure angles
3) Divider — B) to measure the lengths of line segments

Question 6.
List out the letters of English alphabet (capital letters) which consist of right angles.
Solution:
The letters which consists of right angles in English alphabet are

Question 7.
Measure the angles ∠AQP, ∠CPR, ∠BRQ.

Find m∠AQP, m∠CPR, m∠BRQ.
Solution:
In the given figure, measure the angles ∠AQR, ∠CPR, ∠BRQ
∠AQP = 120°
∠CPR = 120°
and ∠BRQ = 120°
∴ m∠AQP = 120°; m∠CPR = 120°; m∠BRQ = 120°

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.4 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 8th Lesson Basic Geometric Concepts Ex 8.4

Question 1.

Measure all the angles in the above figures.

∠1 = 70°
∠3 = 70°
∠5 = 70°
∠7 = 70°

∠2 = 110°
∠4 = 110°
∠6 = 110°
∠8= 110°

∠a = 60°
∠c = 60°
∠e = 50°
∠g = 50°

∠b = 120°
∠d = 120°
∠f = 130°
∠h = 130°

Question 2.
Sum of which two angles is 180° in each figure ?
Solution:
From above question,
i) ∠1 + ∠2 = 180°

i) ∠1 + ∠2 = 180°
∠2 + ∠3 = 180°
∠1 + ∠4 = 180°
∠3 + ∠4 = 180°

∠5 + ∠6 = 180°
∠6 + ∠7 = 180°
∠7 + ∠8 = 180°
∠5 + ∠8 = 180°

∠1 + ∠8 = 180°
∠4 + ∠5 = 180°
∠2 + ∠7 = 180°
∠3 + ∠6 = 180°

ii) ∠a + ∠b = 180°
∠b + ∠c = 180°
∠c + ∠d = 180°
∠a + ∠d = 180°

∠e + ∠f = 180°
∠f + ∠g = 180°
∠g + ∠h = 180°
∠e + ∠h = 180°

Question 3.
In the given figure measure ∠FOG and draw the same in your note book.

Solution:

Question 4.
In the given figure measure the angles ∠AOB, ∠BOC.

Solution:

∠AOB = 110°
∠BOC = 60°
∠AOC = 50°

Question 5.
Write some acute, obtuse and reflexive angles atleast 2 for each.
Solution:
Acute angles : 10°, 30°, 45°, 60°, 89° (< 90°)
Obtuse angles : 110°, 150°, 160°, 172°, 178° (90° < obtuse < 180°) Reflex angles : 210°, 270°, 300°, 345°, 359° (reflex > 180°)

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 8th Lesson Basic Geometric Concepts Ex 8.3

Question 1.
Given ” \(\overline{\mathrm{AB}} / / \overline{\mathrm{CD}}\), l ⊥ m”. Which are perpendicular? Which are parallel?
Solution:
are parallel lines (// is the symbol for parallel),
l, m are perpendicular lines (⊥ is the symbol for perpendicular).

Question 2.
Write the set of parallels and perpendiculars in the given by using symbols.

Solution:
a) \(\overline{\mathrm{AB}}, \overline{\mathrm{DC}}\) are parallel lines, we denote this by writing \(\overline{\mathrm{AB}} / / \overline{\mathrm{DC}}\) and can be read as \(\overline{\mathrm{AB}}\) is parallel to \(\overline{\mathrm{DC}}\)
b) \(\overline{\mathrm{AD}}, \overline{\mathrm{BC}}\) are parallel lines. We denote this by writing \(\overline{\mathrm{AD}} / / \overline{\mathrm{BC}}\) and can be read as \(\overline{\mathrm{AD}}\) is parallel to \(\overline{\mathrm{BC}}\)
c) \(\overline{\mathrm{AQ}}, \overline{\mathrm{PC}}\) are parallel lines. We denote this by writing \(\overline{\mathrm{AQ}} / / \overline{\mathrm{PC}}\) and can be read as \(\overline{\mathrm{AQ}}\) is parallel to \(\overline{\mathrm{PC}}\).
d) \(\overline{\mathrm{AB}}, \overline{\mathrm{AD}}\) are perpendicular lines. We denote this by writing \(\overline{\mathrm{AB}} \perp \overline{\mathrm{AD}}\) and can be read as \(\overline{\mathrm{AB}}\) is perpendicular to \(\overline{\mathrm{AD}}\).
e) \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}\) are perpendicular lines. We denote this by writing \(\overline{\mathrm{AB}} \perp \overline{\mathrm{BC}}\) and can be read as \(\overline{\mathrm{AB}}[latex] is perpendicular to [latex]\overline{\mathrm{BC}}\) .
f) \(\overline{\mathrm{BC}}, \overline{\mathrm{CD}}\) are perpendicular lines, we denote this by writing \(\overline{\mathrm{BC}} \perp \overline{\mathrm{CD}}\) and can be read as \(\overline{\mathrm{BC}}\) is perpendicular to \(\overline{\mathrm{CD}}\).
g) \(\overline{\mathrm{CD}}, \overline{\mathrm{AD}}\) are perpendicular lines, we denote this by writing \(\overline{\mathrm{CD}} \perp \overline{\mathrm{AD}}\) and can be read as \(\overline{\mathrm{CD}}\) is perpendicular to \(\overline{\mathrm{AD}}\).

ii) \(\overline{\mathrm{PX}}, \overline{\mathrm{QR}}\) are perpendicular lines, we denote this by writing \(\overline{\mathrm{PX}} \perp \overline{\mathrm{QR}}\) and can be read as \(\overline{\mathrm{PX}}\) is perpendicular to \(\overline{\mathrm{QR}}\).

iii) a) \(\overline{\mathrm{LM}}, \overline{\mathrm{KN}}\) are parallel lines. We denote this by writing \(\overline{\mathrm{LM}} / / \overline{\mathrm{KN}}\) and can be read as \(\overline{\mathrm{LM}}\) is parallel to \(\overline{\mathrm{KN}}\).
b) \(\overline{\mathrm{MN}}, \overline{\mathrm{LK}}\) are parallel lines. We denote this by writing \(\overline{\mathrm{MN}} / / \overline{\mathrm{LK}}\) and can be read as \(\overline{\mathrm{MN}}\) is parallel to \(\overline{\mathrm{LK}}\).
c) \(\overline{\mathrm{ON}}, \overline{\mathrm{LP}}\) are parallel lines. We denote this by writing \(\overline{\mathrm{ON}} / / \overline{\mathrm{LP}}\) and can be read
as \(\overline{\mathrm{ON}}\) is parallel to \(\overline{\mathrm{LP}}\)
d) \(\overline{\mathrm{LM}}, \overline{\mathrm{ON}}\) are perpendicular lines, we denote this by writing \(\overline{\mathrm{ON}} \perp \overline{\mathrm{LM}}\) and can be read as \(\overline{\mathrm{ON}}\) is perpendicular to \(\overline{\mathrm{LM}}\).
e) \(\overline{\mathrm{LP}}, \overline{\mathrm{KN}}\) are perpendicular lines, we denote this by writing \(\overline{\mathrm{LP}} \perp \overline{\mathrm{KN}}\) and can be read as \(\overline{\mathrm{LP}}\) is perpendicular to \(\overline{\mathrm{KN}}\).

Question 3.
From the given figure find out the Intersecting lines and concurrent lines.

Solution:
i) Intersecting lines : (l, m); (l, n); (n, o); (m, o); (l, o); (m, n)
ii) Intersecting lines: (p, q); (p, r); (p, s); (q, r); (q, s)
Concurrent lines : (p, q, s)

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 6 Basic Arithmetic InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 6th Lesson Basic Arithmetic InText Questions

Check Your Progress (Page No. 84)

Question 1.
Express the terms 45 and 70 by using ratio symbol.
Solution:
Given terms are 45 and 70
Ratio = 45 : 70
It can be read as 45 is to 70.

Question 2.
Write antecedent in the ratio 7:15.
Solution:
Given ratio 7 : 15
In the ratio first term is called antecedent.
In 7 : 15 antecedent is 7.

Question 3.
Write the consequent in the ratio 8 : 13.
Solution:
Given ratio 8 : 13
In the ratio second term is called consequent.
In 8 : 13 consequent is 13.

Question 4.
Express the ratio 35 : 55 in the simplest form.
Solution:
Given ratio 35 : 55 (or)
To write the ratio in the simplest form we have to divide by the common factor of two terms 35 and 55.
Common factor is 5.
Now divide by 5,
\(\frac{35}{55}=\frac{35 \div 5}{55 \div 5}=\frac{7}{11}\)
Simplest form of \(\) is \(\frac{7}{11}\)

Question 5.
In the given figure, find the ratio of
i) Shaded part to unshaded parts.
ii) Shaded part to total parts,
iii) Unshaded parts to total parts.

Solution:
i) In the given figur.e,
Number of shaded parts = 1
Number of unshaded parts = 3
Ratio = shaded parts : unshaded parts = 1:3

ii) Number of shaded parts = 1
Total parts = 4
Ratio = shaded parts : total parts = 1:4

iii) Number of unshaded parts = 3
Total parts = 4
Ratio = unshaded parts : total parts = 3:4

Question 6.
Express the following in the form of ratio.
a) The length of a rectangle is triple its breadth. ‘
b) In a school, the workload of teaching 19 sections has been assigned to 38 teachers.
Solution:
a) Let breadth of rectangle = x or one part = 1 part
length of rectangle = triple the breadth
= 3 x x = 3x = 3 parts
Ratio = l : b = x : 3x =\(\frac{1 x}{3 x}=\frac{1}{3}\) = 1:3

b) Given number of sections = 19
Number of teachers = 38
Ratio = 19 : 38 = \(\frac{19}{38}=\frac{1}{2}\) = 1 : 2

(Page No. 88)

Question 1.
Which ratio is larger in the following pairs ?
(a) 5 : 4 or 9 : 8
(b) 12 : 14 or 16 : 18
(c) 8: 20 or 12: 15
(d)4:7 or 7:11
Solution:
a) 5 : 4 or 9 : 8
Write the given ratios as fractions, we have 5 : 4 = \(\frac{5}{4}\) and 9 : 8 = \(\frac{9}{8}\)
Now find the LCM of the denominators of 4 and 8 is 8.
Make the denominator of the each fraction equal to 8.
We have \(\frac{5}{4} \times \frac{2}{2}=\frac{10}{8}\) and \(\frac{9}{8} \times \frac{1}{1}=\frac{9}{8}\)
Clearly we know that 10 > 9
∴ \(\frac{10}{8}>\frac{9}{8}\) (or) 10 : 8 > 9 : 8
10 : 8 is equal to 5 : 4
Therefore the larger ratio is 5 : 4.

b) 12 : 14 or 16:18
12 : 14 = \(\frac{12}{14}=\frac{6}{7}\) and 16 : 18 = \(\frac{16}{18}=\frac{8}{9}\)
Now find the LCM of the denominators of 7 and 9 is 63.
Make the denominator of the each fraction equal to 63.
we have \(\frac{6}{7} \times \frac{9}{9}=\frac{54}{63}\) and \(\frac{8}{9} \times \frac{7}{7}=\frac{56}{63}\)
Clearly, we know that 54 < 56
∴ \(\frac{54}{63}<\frac{56}{63}\) (or) 54:63 < 56:63
56 : 63 is equal to 16 : 18 (or) 8 : 9
∴ The larger ratio is 16 : 18.

c) 8 : 20 or 12 : 15
Write the given ratios as fractions we have
8:20 = \(\frac{8}{20}=\frac{2}{5}\) and 12:15 = \(\frac{12}{15}=\frac{4}{5}\)
\(\frac{2}{5}\) and \(\frac{4}{5}\)

Clearly \(\frac{2}{5}\) < \(\frac{4}{5}\)
i.e., 2:5 < 4 : 5 (or) 8: 20 < 12: 15
Therefore the larger ratio is 12 : 15.

d) 4: 7 or 7: 11
Write the given ratios as fractions, we have 4 7
4 : 7 = \(\frac{4}{7}\) and 7:11 = \(\frac{7}{11}\) .
Now find the LCM of the denominators of 7 and 11 is 77.
Make the denominators of the each fraction equal to 77.
We have \(\frac{4}{7} \times \frac{11}{11}=\frac{44}{77}\) and \(\frac{7}{11} \times \frac{7}{7}=\frac{49}{77}\)
\(\frac{44}{7}\) and \(\frac{49}{77}\)
Clearly we know that 44 < 49
∴ \(\frac{44}{77}<\frac{49}{77}\) (or) 44 : 77 < 49 : 77
i.e.,4: 7 < 7 : 11
Therefore the larger ratio is 7 : 11

Question 2.
Find three equivalent ratios of 12 : 16.
Solution:
Given ratio is 12 : 16
Write the given ratio as fraction we have 12:16= \(\frac{12}{16}=\frac{3}{4}\)
Now, write equivalent fractions of \(\frac{3}{4}\)

i. e., 6 : 8 = 9 : 12 = 12 : 16 = 15 : 20 = 18 : 24
∴ Equivalent ratios of 12 : 16 are 6 : 8, 9 : 12, 12 : 16, 15 : 20 and 18 : 24.

(Page No. 90)

Question 1.
Check whether the following terms are in proportion ?
1) 5,6,7,8
2) 3,5,6,10
3) 4,8,7,14
4) 2,12,3,18
Solution:
1) Given, 5, 6, 7, 8
If a, b, c, d are in proportion i.e., a : b :: c : d
If 5, 6, 7, 8 are in proportion i.e., 5 : 6 : : 7 : 8
We know that, product of extremes = Product of means [a x d : b x c]
5 x 8 = 6 x 7
40 ≠ 42
So, 5, 6, 7, 8 are not in proportion.

2) Given, 3, 5, 6, 10
If a, b, c, d are in proportion i.e., a : b :: c : d
If 3, 5, 6, 10 are in proportion i.e., 3 : 5 :: 6 : 10
We know that, product of extremes = Product of means a x d = b x c
3 x 10 = 5 x 6
30 = 30
So, 3, 5, 6, 10 are in proportion.

3) Given, 4, 8, 7, 14.
If a, b, c, d are in proportion i.e., a : b : : c : d
If 4, 8, 7, 14 are in proportion i.e., 4 : 8 : : 7 : 14
We know that, product of extremes = Product of means a x d = b x c
4 x 14 = 8 x 7
56 = 56
So, 4, 8, 7, 14 are in proportion.

4) Given, 2, 12, 3, 18
If a, b, c, d are in proportion i.e., a : b :: c : d
If 2, 12, 3, 18 are in proportion i.e., 2 : 12 : : 3 : 18
We know that, product of extremes = Product of means [ a x d = b x c ]
2 x 18 = 12 x 3
36 = 36
So, 2, 12, 3, 18 are in proportion.

Let’s Explore (Page No. 92)

Question 1.
Read the table and fill in the boxes.

Prepare two similar problems and ask your friend to solve them
Solution:

(Page. No. 94)

Question 1.
Represent the following in other forms.

Solution:

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 8th Lesson Basic Geometric Concepts Ex 8.2

Question 1.
Measure the lengths of the given line segments.

Solution:
i) \(\overline{\mathrm{AB}}\) = 2.4 cm
ii) \(\overline{\mathrm{PQ}}\) = 1.5 cm
iii) \(\overline{\mathrm{KL}}\) = 1cm, \(\overline{\mathrm{LM}}\) = 1 cm
\(\overline{\mathrm{KM}}=\overline{\mathrm{KL}}+\overline{\mathrm{LM}}\) =1 + 1 = 2 cm

Question 2.
Draw the following line segments.
i) AB = 6.3 centimeters ii) MN = 3.6 centimeters

Question 3.
Draw PQ = 4.6 cm and extend upto R such that PR = 6 cm.
Solution:

\(\overline{\mathrm{PR}}=\overline{\mathrm{PQ}}+\overline{\mathrm{QR}}\) = 4.6 + 1.4 = 6 cm

Question 4.
Draw a line segment \(\overline{\mathrm{OP}}\) with certain length and mark a point Q on it.
Check whether \(\overline{\mathrm{OP}}-\overline{\mathrm{PQ}}=\overline{\mathrm{OQ}}\)
Solution:

Given, \(\overline{\mathrm{OP}}\) = 8 cm; \(\overline{\mathrm{PQ}}\) = 3 cm
\(\overline{\mathrm{OP}}-\overline{\mathrm{PQ}}\) = 8cm – 3cm = 5 cm = \(\overline{\mathrm{OQ}}\)
∴ \(\overline{\mathrm{OP}}-\overline{\mathrm{PQ}}=\overline{\mathrm{OQ}}\) = 5cm

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 5 Fractions and Decimals InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 5th Lesson Fractions and Decimals InText Questions

(Page No. 63)

Question 1.
Is it true to say that 3 × \(\frac{1}{5}=\frac{1}{5}\) x 3?
Solution:
3 × \(\frac{1}{5}=\frac{1}{5}\) × 3. Yes, it is true.
By using commutative property over multiplication a × b = b × a
3 × \(\frac{1}{5}=\frac{1}{5}\) × 3 = \(\frac{3}{5}\)

Check Your Progress (Page No. 63)

Find :
i) \(5 \times 3 \frac{2}{7}\)
ii) \(2 \frac{5}{9} \times 3\)
iii) \(2 \frac{4}{7} \times 3\)
iv) \(3 \times 1 \frac{3}{4}\)
Solution:

Let’s Explore (Page No. 64)

Question 1.
Observe the products of fractions.
Have you observed the products of any two fractions is always lesser or greater than each of its fraction, write conclusion.
\(\frac{1}{5} \times \frac{2}{3}=\frac{2}{15}\) (Product of two proper fractions)
Solution:
Product of any two proper fractions is always less than each of its fraction.
i. e., \(\frac{2}{15}<\frac{1}{5} \text { and } \frac{2}{15}<\frac{2}{3}\)

ii) \(\frac{3}{2} \times \frac{5}{4}=\frac{15}{8}\) (Product of two improper fractions) •
Solution:
The product of any two improper fractions is always greater than each of its fraction.
i.e, \(\frac{3}{2}<\frac{15}{8} \text { and } \frac{5}{4}<\frac{15}{8}\)

iii) \(\frac{2}{3} \times \frac{5}{3}=\frac{10}{9}\) (Product of proper and improper fractions)
Solution:
The product of a proper fraction and an improper fraction is always greater than its proper fraction and less than its improper fraction.
i.e., \(\frac{2}{3}<\frac{10}{9} \text { and } \frac{5}{3}>\frac{10}{9}\)

(Pg. No. 66)

Question 1.
i) 4 ÷ \(\frac{1}{8}\)
ii) 9 ÷ \(\frac{3}{4}\)
iii) 7 ÷ \(\frac{2}{3}\)
iv) 35 ÷ \(\frac{7}{3}\)
v) 4 ÷ \(\frac{15}{8}\)
Solution:

(Pg. No. 67)

Question 1.
Observation these products and fill in the blanks.

Solution:

Check Your Progress (Page No. 68)

Question 1.
Write the reciprocal of fractions in the given table.

Solution:

(Reciprocal of a fraction \(\frac{\mathrm{a}}{\mathrm{b}}\) is \(\frac{\mathrm{b}}{\mathrm{a}}\))

(Page No. 69)

Question 1.
Find
i) \(\frac{7}{9}\) ÷ 4
ii) \(\frac{3}{4}\) ÷ 9
iii) 4\(\frac{1}{2}\) ÷ 6
iv) \(\frac{1}{5}\) ÷ 3
Solution:

Check Your Progress (Page No. 73, 74 & 75)

Question 1.
Fill in the blanks.

Solution:

Question 2.
Write the place value of the circled digits.

Solution:

Question 3.
a) 700 + 40 + 2 + \(\frac{1}{10}+\frac{3}{100}+\frac{6}{1000}\)
Solution:
700 + 40 + 2 + 0.1 + 0.03 + 0.006 = 742.136

b) 9000 + 800 + 3 + 0.2 + 0.05 + 0.007
Solution:
9000 + 800 + 3 + 0.2 + 0.05 + 0.007 = 983.257

c) 6000 + 400 + 20 + 1 + \(\frac{2}{10}+\frac{5}{100}+\frac{9}{1000}\)
Solution:
6000 + 400 + 20 + 1 + 0.2 + 0.05 + 0.009 = 6421. 259

d) 400 + 5+ \(\frac{1}{10}+\frac{8}{100}\)
Solution:
400 + 5 + 0.1 +0.08 = 405.18

Question 4.
Expand the following into decimals and fractional forms,
a) 164.238
b) 968.054
Solution:
a) 164.238 = 100 + 60 + 4 + 0.2 + 0.03 + 0.008
= 100 + 60 + 4 + \(\frac{2}{10}+\frac{3}{100}+\frac{8}{1000}\)

b) 968.054
Solution:
= 900 + 60 + 8 + 0.0 + 0.05 + 0.004
= 900 + 60 + 8 + 0 + \(\frac{5}{100}+\frac{4}{1000}\)
= 900 + 60 + 8 + \(\frac{5}{100}+\frac{4}{1000}\)

Question 5.
Write fractions as decimals.
1. \(\frac{23}{10}\) = ………..
2. \(\frac{6}{100}\) = ………..
3. \(\frac{3}{8}\) = ………..
4. \(\frac{2}{25}\) = ………..
Solution:
1. \(\frac{23}{10}\) = 2.3
2. \(\frac{6}{100}\) = 0.06
3. \(\frac{3}{8}\) = 0.375
4. \(\frac{2}{25}\) = 0.08

Question 6.
Write decimals as fractions in simplest form.
1. 0.2 = ……………
2. 0.38 = ……………
3. 1.62 = ……………
4. 8.1 = ……………
Solution:
1. 0.2 = \(\frac{2}{10}\)
2. 0.38 = \(\frac{38}{100}\)
3. 1.62 = \(\frac{162}{100}\)
4. 8.1 = \(\frac{81}{10}\)

AP State Syllabus AP Board 6th Class Science Important Questions Chapter 8 How Fabrics are Made

AP State Syllabus 6th Class Science Important Questions 8th Lesson How Fabrics are Made

6th Class Science 8th Lesson How Fabrics are Made 2 Mark Important Questions and Answers

Question 1.
What are natural fibres? Give examples?
Answer:
The fibres which are deriving from plants and animals are called natural fibres.
Ex: cotton, wool, silk, jute, coir, etc.

Question 2.
What are Artificial fibres? Give examples?
Answer:
The fibres which are deriving from chemicals are artificial fibres or synthetic fibres. Ex : – Polyester, polythene, Nylon, Rayon, etc.

Question 3.
What is fibre?
Answer:
Small strand-like structures are called fibres.

Question 4.
What is weaving?
Answer:
Making fabric from yarn is called weaving.

Question 5.
How can you remove wrinkles from our dress?
Answer:
By ironing, we can remove wrinkles from our dress.

Question 6.
How are clothes useful to us?
Answer:
1) The clothes are used as a shield to protect ourselves from different weather conditions.
2) Along with protection, clothes can also be a symbol of beauty and status.

Question 7.
Which fabric is used in making banners and book bindings?
Answer:
Calico is a type of fabric used in making banners and book bindings.

Question 8.
Which fibre is known as golden fibres?
Answer:
Jute.

Question 9.
Which state is the highest produces of Jute in India?
Answer:
West Bengal.

Question 10.
Why should we use cloth bags instead of polythene bags?
Answer:
To protect our environment we should use cloth bags instead of polythene bags.

Question 11.
Which type of fabrics dries in a short time?
Answer:
Artificial fabrics dry in a short time.

Question 12.
Why seeds are removing from cotton balls?
Answer:
Cotton seeds are removing from cotton balls to make an even and uniform fabric.

Question 13.
Why we spin the fibres in order to make yarn?
Answer:
Fibres are very thin and weak we twist them together to make them strong thick and long.

Question 14.
Which material is used for making gunny bags and why?
Answer:
Jute fibres are used for making gunny bags. Because they are strong and can hold heavy loads.

Question 15.
Name the two types of looms?
Answer:
Handlooms and power loom.

Question 16.
Name the two simple devices used for spinning?
Answer:
Spindle (Takli) and Charaka.

Question 17.
Name the person who made the Charaka popular during the Independence movement?
Answer:
Mahatma Gandhi.

Question 18.
Where is the coconut industry well developed in India?
Answer:
The coconut industry is mainly well developed in the states of Kerala* Tamilnadu, Karnataka, Andhra Pradesh and Odisha.

Question 19.
Which area is famous for Kalamkari Textiles?
Answer:
Machilipatnam and Pedana are famous for Kalamkari textiles.

Question 20.
Which city is famous for the carpet industry?
Answer:
Machilipatnam.

6th Class Science 8th Lesson How Fabrics are Made 4 Mark Important Questions and Answers

Question 1.
While purchasing your dress what doubts would you want to clarify from the shopkeepers?
Answer:

• What type of washing does it need?
• Will it absorb the sweat of the body?
• Does the cloth provide free airflow to the body?
• What is the durability of cloth?

Question 2.
How can different fabrics be used?
Answer:

• Our purpose and the priority of fabric together determine which type of fabric to be used.
• Coarse fabrics can be used for mopping and making gunny bags but not for making clothes.
• Some fabrics have to be considered for choosing curtain fabrics.
• The calico fabric used to make banners and bookbinding.

Question 3.
Where is the handloom industry well developed in our state?
Answer:

• The handloom industry is very well developed in our state.
• Places like Venkatagiri, Narayanapeta, Dharmavaram, Mangalagiri and Kothakota are famous for handloom industries.
• Pedana and Machilipatnam are famous for kalamkari.
• Machilipatnam is also famous for the carpet industry.

Question 4.
Where is child labour working? Why are they forced into labour? How can you eradicate child labour?
Answer:

• In agricultural works, where cotton is widely grown, their child labourers are working.
• To pick up maturing cotton bolls from cotton plants, children work in field as child labourers.
• Some parents to get additional income through their children are putting them as child labourers.
• Some organizations are working against child labourers and sending them back to schools.

Question 5.
What are the uses of coir?
Answer:

• The coconut coir industry is one of the rural industry in India.
• The coir is still used for agricultural and domestic purposes, and controlling landslide or soil erosion.
• Brown coir is used in brushes, doormats, mattresses and for making sacks.

Question 6.
What are different clothes used in ancient times by human beings?
Answer:

• Human beings in ancient times use leaves and skins of animals as clothes.
• Clothes were also made from metal Warriors used to wear a metal jacket during wars.
• You can see clothes like these in historical museums are in television shows.

Question 7.
What are the uses of the gunny bags?
Answer:

• Paddy, Chilli and other commercial crops are packed in gunny bags.
• All bags of those types are made up of coarse Jute fabric.
• These bags are suitable for carrying heavy materials.

Question 8.
How do you prepare strong yarn from cotton boll? Describe the activity yam perform?
Answer:

• The yarn that we make from cotton wool is not strong enough to be used for weaving.
• To get strong yarn from fibre, a spindle (takli) an instrument for spinning has been used since olden days.
• Charaka is also used to make yarn.
• The process of making yarn from fibre is called spinning.
  

Question 9.
How yarn is used to make fabric?
Answer:

• The yarn that is prepared from the fibre is used to make fabric.
• Strands of yarn are arranged in vertical and horizontal rows in a Loom to weave fabric.
• The spinning of yarn on large scale is now done by using machines.
• Two sets of yarn arranged together to make the fabric is called weaving.
• Weaving is done on looms, the looms that work with the help of man are called Handlooms. Power looms are run by a machine.
  

Question 10.
How do you make cotton yarn?
Answer:

1. Cotton is usually picked by hands. Cotton wool is separated from seeds. This process is called ginning.
2. Cotton fibre is collected after removing the seeds from the cotton boll.
3. This cotton fibre is cleaned, washed and combed.
4. After combed its is spun to make cotton yarn.

Question 11.
How do you get jute yarn?
Answer:

1. Jute fibre is obtained from the stem of the jute plant.
2. The stem of the harvested plant is cut and immersed in water for some days.
3. When the stem is soaked in water it becomes rotten and easy to peel.
4. Then the fibres are separated from the stem to get jute yarn.

Question 12.
Where do we use jute yarn?
Answer:

1. Paddy, chilli and other commercial crops are packed in gunny bags.
2. All bags of these types are made up of coarse jute fabric.
3. These bags are suitable for carrying heavy material.
4. Jute fibre is obtained from stem of jute plant.

Question 13.
Where do we find the handloom industry in our state?
Answer:

1. The handloom industry is well developed in our state.
2. Places like Uppftda, Venkatagiri, Dharmavaram, Pondhuru, Chirala and Mangalagiri are famous for the handloom industry.
3. Kalamkari is a type of hand-printed cotton textile.
4. Machilipatnam, Pedana are famous for Kalamkari. Machilipatnam is also famous for its carpet industry.

Question 14.
Draw a flow chart to explain the process from fiber to fabric?
Answer:

Question 15.
Write about the coconut industry?
Answer:

1. The coconut coir industry is one of the rural industry in India.
2. It is located mainly in the states like Kerala, Tamilnadu, Karnataka, Andhra Pradesh and Odisha.
3. It provides a source of income to about 5 lakhs of artisans in rural areas.
4. Women constitute about 80% of the workforce in coir industry.

Question 16.
Draw a flow chart about coconut products?
Answer:

Question 17.
Write the uses of coir?
Answer:

1. Coir has come a long way from the ancient uses.
2. It is still used for agricultural and domestic purposes and to control landslide or soil erosion.
3. Coir is also used as a substrate to grow mushrooms.
4. Brown coir is used in brushes, doormats, mattresses and for making sacks.

Question 18.
What are the types of fibres?
Answer:

1. The fibres of some fabrics such as cotton, jute are obtained from plants. Silk and wool are obtained from animals. The fibres that are derived from plants and animals are natural fibres.
2. Nowadays, clothes are also made up of chemically developed yarn like polyester, terylene, nylon, acrylic etc. These are all called artificial fibres.

Question 19.
Why should we use cloth bags instead of plastic bags?
Answer:
We all use polythene bags for different purposes. Polythene is very difficult to decompose. To protect our environment. We should use cloth bags instead of polythene bags.

6th Class Science 8th Lesson How Fabrics are Made 8 Mark Important Questions and Answers

Question 1.
How do you make cotton yarn and fibres?
Answer:

• Collect some cotton balls from nearby houses or cotton fields.
• Remove the seeds from cotton balls and separate cotton.
• Take a small piece of cotton and observe it under a microscope.
• We will observe small tiny hairy structures called fibres.
• Removing of seeds from cotton is called ginning.
• These fibres are cleaned, washed and combed.
• These fibres are twisted together to make yarn.
• Now these yarns are dyed and coated with chemicals.
• Then they become strong enough to make fabrics.
  

Question 2.
How do you identify the fibres in fabric? Describe the procedure.
Answer:

• Take a fabric piece. With the help of a magnifying lense, observe how the fabric is
• Pull out threads one by one from the fabric.
  
• Take one thread, scratch its end and observe it through a magnifying lens.
• We can observe small tiny long structures called fibres.
• These fibres are twisted together to form yarn
• By using of these yarn on handlooms or powerlooms the workers are making the cotton fabrics.
  Fibre → Yarn → Fabric

Question 3.
What are the factors involved in the selection of a fabric?
Answer:

• The fabrics protect us from different weather conditions.
• Along with protection clothes are also be a symbol of beauty and status.
• Choice of fabric may vary from person to person.
• Some may like to wear clothes made up of light, thin, shiny fabrics.
• Another person may like to wear clothes that are brightly coloured and made of coarse fabrics.
• Fabrics for casual and formal wear may be different.
• Personal choice, the personality of the owner and the cost of the fabric are all important factors in the selection of perfect fabric.

Question 4.
How do you make a mat with coconut leaves?
Answer:

• Take coconut leaves or two colour paper strips.
• Cut and remove the middle vein of the leaf to get two halves.
• Now put these strips parallel to each other.
• Take one more strips and insert horizontally and alternately between the vertical strips.
• Finally, you will get a sheet-like structure. This is the way a mat is prepared.

Question 5.
What do you observe in gunny yarn? Compare Jute yarn with other yarns?
Answer:

• Collect gunny bags, pull out the threads from the bag and observe under magnifying lens.
• We will see strands of yarn.
• We can compare these fibres with cotton fibres.
• In the same way fibre is made from Red sorrel (Gongura) and bamboo.
• Hemp and flax also plant fibres that are used in making clothes but in smaller quantities as compared to cotton.
• Like cotton, jute yarn is also useful in making fabric.
• It is also called Golden fabric.
• Jute fabrics is not the same as cotton fabric. It is harder, stronger and rougher.

Question 6.
People dress up according to the season. The earth’s revolution is responsible for the changing seasons. Complete the following table.

Answer: