Class 6

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.

S.No.	Expression	Verbal Form
1.	y + 3	Three more than y
2.	2x – 1
3.	5z
4.	\(<table border=”2″>\)

Solution:

S.No.	Expression	Verbal Form
1.	y + 3	Three more than y
2.	2x – 1	One less than the double of x
3.	5z	5 times of z
4.	\(<table border=”2″>\)	Half of the m

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

Side	Area
1	1 × 1
2	2 × 2
3	3 × 3
4	4 × 4
………..	…………
s	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

p	\(\frac{\mathbf{p}}{3}\) = 4	Condition satisfied ? Yes/ No
3
6
9
12

Solution:

p	\(\frac{\mathbf{p}}{3}\) = 4	Condition satisfied ? Yes/ No
3	\(\frac{3}{3}\) ≠1 ≠ 4	No
6	\(\frac{6}{3}\) ≠2 ≠ 4	No
9	\(\frac{9}{3}\) ≠ 3 ≠ 4	No
12	\(\frac{12}{3}\) ≠ 4 ≠ 4	Yes

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:

Weight	Cost of Tomato	Cost of Potato
5 kg	₹ 75	₹ 60
1 kg	₹15	₹ 12
3 kg	₹ 45	₹ 36

(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.

Fraction	Decimal Number	” Read as
\(\frac{6}{10}\)	0.6	Zero point six
\(\frac{37}{100}\)	0.37	Zero point three seven
	0.721	Zero point seven two one
		Seventeen point two

Solution:

Fraction	Decimal Number	Read as
\(\frac{6}{10}\)	0.6	Zero point six
\(\frac{37}{100}\)	0.37	Zero point three seven
\(\frac{721}{1000}\)	0.721	Zero point seven two one
\(\frac{172}{10}\)	17.2	Seventeen point two

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.

S.No.	Season	Months	Climate	Clothes we wear
1.	Rainy
2.	Winter
3.	Summer

Answer:

S.No.	Season	Months	Climate	Clothes we wear
1.	Rainy	June – Sep	More humid, Rains	Polyester, PVC
2.	Winter	Oct – Jan	Very cool	Wool, Polyester, Nylon Etc
3.	Summer	Feb – May	Very hot	Cotton

 

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 4 Integers InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 4th Lesson Integers InText Questions

Check Your Progress?(Page No. 47)

Question 1.
Write any five positive integers.
Solution:
1, 2, 3, 4, 5, 6, 7,

Question 2.
Write any five negative integers.
Solution:
-1, -2, -3, -4, -5, -6,

Question 3.
Which number is neither positive nor negative?
Solution:
0 (zero)

Question 4.
Represent the following situations with integers,
(i) A gain of ₹ 500 ( )
(ii) Temperature is below 5°C ( )
Solution:
i) + 7 500
ii) – 5° C

Question 5.
Represent the following using either positive or negative numbers.
a) A bird is flying at a height of 25 meters above the sea level and a fish at a depth of 2 meters.
b) A helicopter is flying at a height of 60m above the sea level and a submarine is at 400m below sea level.
Solution:
a) Height of the flying bird 25 meters from th$ sea level = + 25 meters
Depth of the fish 2 meters from the sea level = – 2 meters
b) Height of the flying helicopter 60 meters from the sea level = + 60 m
Depth of the submarine 400 m from the sea level = – 400 m

Check Your Progress (Page No. 49)

Question 1.
Draw a vertical number line and represent -5,4,0,-6, 2 and 1 on it.
Solution:

Question 2.
Represent opposite integers of – 200 and + 400 on integer number line.
Solution:
Opposite integers means, additive inverse.
∴ Opposite integer (additive inverse) of – 200 is 200.
Opposite integer (additive inverse) of +400 is – 400.

Let’s Think (Page No. 50)

Question 1.
For any two integers, say 3 and 4, we know that 3 < 4.
Is it true to say -3 < -4? Give reason.
Solution:
On the number line, the value of a number increases as we move to right and decreases as we move to the left. As -3 lies right to -4 on the number line.
So, -3 < -4 is not true.

(Pg. No. 52)

Question 1.
What is additive inverse of 7 ?
Solution:
Additive inverse of 7 is -7.

Question 2.
What is additive inverse of -8 ?
Solution:
Additive inverse of -8 is 8.

Let’s Explore (Page.No. 52)

Question 1.
Find the value of the following using a number line.
i) (-3) + 5 ii) (-5) + 3
Make two questions on your own and solve them using the number line.
Solution:
i) (-3) + 5

On the number line, we first move 3 steps to the left of 0 to reach -3.
Then, we move 5 steps to the right of -3 and reach +2. So, (-3) + 5 = 2

ii) (-5) + 3

On the number line, we first move 5 steps to the left of 0 to reach -5. Then, we move 3 steps to the right of -5 and reach -2. So, (-5) + 3 = – 2

iii)(+6) + (-3)

On the number line, we first move 6 steps to the right of 0 to reach +6.
Then, we move 3 steps to the left of 6 and reach +3. So, (+6) + (-3) = 3

iv) (-4) + (-3)

On the number line, we first move 4 steps to the left of 0 to reach -4. Then, we move 3 steps to the left of -4 and reach -7.
So (-4) + (-3) = -7.

Question 2.
Find the solution of the following:
i) (+5) + (-5) (ii) (+6) + (-7) (iii) (-8) + (+2)
Ask your friend five such questions and solve them.
Solution:
i) (+5) + (-5) = 0

On the number line, we first move 5 steps to the right of 0 to reach +5.
Then, we move 5 steps to the left of +5 and reach 0.

ii) (+6) + (-7) = -1

On the number line, we first move 6 steps to the right of 0 to reach +6.
Then, we move 7 steps to the left of +6 and reach -1.

(iii) (-8) + (+2)

On the number line, we first move 8 steps to the left of 0 to reach -8. Then, we move 2 steps to the right of -8 and reach -6.

iv)(-4) + (+8) = +4

On the number line, we first move 4 steps to the left of 0 to reach -4. Then, we move 8 steps to the right of -4 and reach +4.

v) (+3) + (-4) = -1

On the number line, we first move 3 steps to the right of 0 to reach +3. Then, we move 4 steps to the left of +3 and reach -1.

vi) (+5) + (-6) = -1

On the number line, we first move 5 steps to the right of 0 to reach +5. Then, we move 6 steps to the left of +5 and reach -1.

vii) (+4) + (-4) = 0

On the number line, we first move 4 steps to the right of 0 to reach +4. Then, we move 4 steps to the left of +4 and reach 0.

viii) (-6) +(+4) =-2

On the number line, we first move 6 steps to the left of 0 to reach -6. Then, we move 4 steps to the right of -6 and reach -2.
So, (-6) + (+4) = -2

Let’s Explore (Page No. 55)

Question 1.
Take any two integers a and b. Check whether a+b is also an integer.
Case (i) : Consider two integers 3 and -2 (Positive and negative)
Sum = 3 + (-2) = +1 + 2 – 2 = +1 + 0 = +1 is also an integer.
Case (ii) : Consider two ihtegers 5 and 6 (Both are positive)
Sum = 5 +6 = + 11 is also an integer
Case (iii) : Consider two integers -4 and -6 (Both are negative)
Sum = -4 + (-6) = -4 -6 = -10 is also an integer
Case (iv) : Consider two integers -5 and 4 (Negative and positive)
Sum = -5 + 4 = -1 -4 + 4 = -1 + 0 = -1 is also an integer.
So, if a and b are integers, then their sum a + b is also an integer. Integers are closed under addition.

Question 2.
Check the following Properties on integers, a, b, c are any integers.
i) Closure Property under subtraction ‘
ii) Commutative Property under addition and subtraction (a + b = b + a ?, a – b = b – a?)
iii) Associative Property under addition and subtractioji.
(a + b) + c = a + (b + c) ? (a – b).- c = a – (b – c)?
Solution:
i) Closure Property under subtraction :
Case (i) : Consider two integers 4, -5 (positive and negative)
Then, difference a – b = 4 – (-5) = 4 + 5 = + 9is also an integer.
Case (ii) : Consider two integers 3, 8 (Both are positive)
Then, difference a – b = 3 – (+8) = 3-8
= +3 – 3 – 5 = 0 – 5 = -5is also an integer.
Case (iii) : Consider two integers -2, -6 (Both are negative) .
Then, difference a – b = -2 – (-6) = -2 + 6
= -2 + 2 + 4 = 0 + 4 = +4 is also an integer.

Case (iv) : Consider two integers -3, 2 (Negative and positive)
Then, difference a – b = -3 – (+2) = -3 -2 = -5 is also an integer.
So, if a and b are any two integers, then their difference a – b is also an integer. Integers are closed under subtraction.

ii) Commutative Property under addition and subtraction :
(a + b = b + a, a-b = b-a)
(A) Case (Q : Consider two integers-3 and 5 (Negative and positive)
Then, a + b = -3 + (+5)
= -3 + 5 = – 3 + 3 + 2 = 0 + 2 = + 2
b + a = +5 + (-3) = +2 + 3- 3 = +2 + 0 = + 2
∴ a + b = b + a

Case (ii) : Consider two integers +4 and +2 (Both are positive)
Then, a + b = +4 + (+2) = +4 + 2 = + 6 ‘b + a = +2 + (+4) = +2 + 4 = + 6
∴ a + b = b + a

Case (iii) Consider two integers -5 and -3 (Both are negative) Then, a + b = -5 + (-3) = -5 – 3 = -8 b + a = -3 + (-5) = – 3 – 5 = -8
∴ a + b = b + a

Case (iv) : Consider two integers +4 and -1 (Positive and negative)
Then, a + b = +4 + (*1) = +4 -1 = +3 + 1 -1 = +3 + 0 = +3 b + a = -1 + (+4) =-l + 4 = -l + l+ 3 = 0 + 3 = + 3
∴ a + b = b + a
So, integers are commutative under addition.

(B) Consider two integers -4 and +6 (Negative and positive)
Then, a – b = -4 – (+6) = -4 – 6 = -10 b-a = + 6-(-4) = +6 +4 = +10 -10*10
a – b ≠ b – a
So, integers are not commutative under subtraction.

iiO Associative property under addition and subtraction :
(a + b) + c = a + (b + c) ; (a – b) – c = a – (b – c)
(A) Case (i) : Consider any three integers 2, 4, -5
(a + b) + c = (2 + 4) + (-5) = 6 – 5 = +1 + 5- 5 = +1 + 0 = +1
a + (b + c) = 2 + (4 + (-5)) = 2 + (4 – 5) = 2 + (4 – 4 -1)
= 2 + (0-1) = + 2 – 1 = + 1 + 1 – 1 = +1 + 0 = +1
∴ (a + b) + c = a + (b + c)

Case (ii) : Consider any three integers +2, -5, +3
Then, (a + b) + c = [2 + (-5)] + 3 = [+2 -5] + 3 = +2 -2 -3 + 3 = 0 + 0 = 0
a + (b + c) = +2 + [( – 5)+3] = +2 + [-2 – 3 + 3] = +2 + (-2 + 0) = +2 -2 = 0
∴ (a + b) + c = a + (b + c)

Case (iii) : Consider any three integers 3, 4, 6
Then, (a + b) + c = [2 + (-5)] + 3 [+2-5] + 3
a + (b + c) = 3 + (4 + 6) = 3 + 10 = + 13
∴ (a + b) + c = a + (b + c)

Case (iv) : Consider any three integers -4, -2, +5
Then, (a + b) + c = [-4 + (-2)] + 5 = [-4 -2] + 5 = -6 + 5
= -1-5 + 5 = – 1 + 0 = – 1
a + (b + c) = -4 + [(-2) + 5] = -4 + [-2 + 5]
= _4 + [_2 + 2 +3] = -4 + 0 + 3 =-1-3+ 3
= -1 + 0 = -1
∴ (a + b) + c = a + (b + c)

Case (v) : Consider any three integers-3, 4, 1
Then, (a + b) + c = (-3 + 4) + 1 = (-3 + 3 + 1) + 1 = 0 + 1 + 1 = + 2
a + (b + c) = -3 + (4 + 1) = -3 + 5 = -3 + 3 + 2 = 0 + 2 = +2
∴ (a + b) + c = a + (b + c)

Case (vi) : Consider any three integers -2, 6, -7
Then, (a + b) + c = (-2 + 6) + (-7) = (-2 + 2 + 4) + (-7) = 0 + 4 – 7
= + 4 – 4 – 3 = 0 – 3 = -3
a + (b + c) = -2 + [6 + (-7)] = -2 + (6 – 7) = -2 + [+6 – 6 -1]
= -2 + (-1) = -2 -1 = -3
∴ (a + b) + c = a + (b + c)

Case (vii) : Consider any three integers +6, -3, -1
Then, (a + b) +c = [+6 + (-3)] + (-1) = (6 – 3) – 1 = (+3 +3 -3) -1
=+3-1 =+2 + 1 – 1 =+2 + 0 = + 2
a + (b + c) = +6 + [-3 + (-1)] = 6 + [-3 – 1] = 6 + (-4) .
= +2 + 4 – 4 = 2 + 0 = + 2
∴ (a + b) + c = a + (b + c)

Case(viii)
Consider any three integers -4, -1, -7
Then, (a + b) + c = [-4 + (-1)] + (-7) = (-4 -1) – 7 = -5 -7 = -12
a + (b + c) = -4 + [(-1) + (-7)] = -4 + [-1 -7] = -4 + (-8)
= -4 – 8 = -12
∴ (a + b) + c = a + (b + c)
From all the above cases we conclude that, integers are associative under addition.

(B) Consider any three integers +5, -4, 1
Then, (a – b) – c = (+5 – (-4)) – (+1) = (5 + 4) – 1 .
= + 9 – 1 = + 8 + 1 – 1 = + 8 + 0 = + 8
a – (b – c) = +5 – [- 4 – (+1)] = + 5 – [-4 – 1]
= + 5 – [ -5] =+ 5 + 5 = + 10
+ 8 ≠ +10
∴ (a – b) – c ≠ a – (b – c)
So, integers are not associative under subtraction.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 12 Data Handling InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 11th Lesson Data Handling InText Questions

Let’s Do (Page No. 159)

Question 1.
Take a die. Throw it and record the number. Repeat the activity 40 times and record the numbers. Represent the data in a frequency distribution table using tally marks.
Solution:

(Page No. 159)

Question 1.
In what way is the bar graph better than the pictograph ?
Solution:
Bar graphs are better indicative as they show exact numerical value. Also, to indicate negative values and positive values, it looks easier in a bar graph.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 3 HCF and LCM InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 3rd Lesson HCF and LCM InText Questions

(Page No. 35)

Question 1.
How does the Sieve of Eratosthenes work ?
Solution:
The following example illustrates how the Sieve of Eratosthenes, can be used to find all the prime numbers that are less than 100. .
Step 1: Write the numbers from 1. to 100 in ten rows as shown below.
Step 2: Cross out 1 as 1 is neither a prime nor a composite number.
Step 3: Circle 2 and cross out all the multiples of 2. (2, 4, 6, 8, 10, 12, ………… )
Step 4: Circle 3 and cross out all the multiples of 3. (3, 6, 9, 12, 15, 18,………….)
Step 5: Circle 5 and cross out all the multiples of 5. (5, 10. 15, 20, 25……………… )
Step 6: Circle 7 and cross out all multiples of 7. (7, 14, 21. 28, 35, ………………….. )

Circle all the numbers that are not crossed out and they are the required prime numbers less than 100.

First arrange the numbers ffom 1 to 100 in a table as shown above.
Enter 6 numbers in each row until the last number 100 is reached.
First we select a number and we strike off all the multiples of it.
Start with 2 which is greater than 1.
Round off number 2 and strike off entire column until the end.
Similarly strike off 4th column and 6th column as they are divisible by 2.
Now round off next number 3 and strike off entire column until end.
The number 4 is already gone.

Now round off next number 5 and strike off numbers in inclined fashion as shown in the figure (they are all divisible by 5). When striking off ends in some row, start again striking off with number in another end which is divisible by 5. New striking off line should be parallel to previous strike off line as. shown in the figure.
The number 6 is already gone.
Now round off number 7 and strike off numbers as we did in case of number 5.
8,9,10 are also gone. .
Stop at this point.
Count all remaining numbers. Answer will be 25.

Prime numbers :
There are 25 prime numbers less than 100.
These are:

What if we go above 100 ? Around 400 BC the Greek mathematician. Euclid, proved that there are infinitely many prime numbers.

Co-primes: Two numbers are said to be co-prime if they have no factors in common. Example: (2, 9), (25, 28)
Any two consecutive numbers always form a pair of co-prime numbers.

Example:
Co-prime numbers are also called relatively prime number to one another.
Example: 3, 5, 8, 47 are relatively prime to one another/co-prime to each other.

Twin primes: Two prime numbers are said to be twin primes, if they differ by 2.
Example: (3, 5), (5, 7), (11, 13), …etc.

Prime factorization: The process of expressing the given number as the product of prime numbers is called prime factorization.
Example: Prime factorization of 24 is
24 = 2 x 12
= 2 x 2 x 6
= 2 x 2 x 2 x 3, this way is unique.

Every number can be expressed as product of primes in a unique manner. We can factorize a given number in to product of primes in two methods. They are
a) Division method
b) Factor tree method

Common factors: The set of all factors which divides all the given numbers are called their common factors.
Example: Common factors to 24, 36 & 48 are 1, 2, 3, 4, 6 & 12
Factors of 24 = 1, 2, 3, 4, 6, 8, 12 & 24
Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18 & 36
Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24 & 48
Common factors to 24, 36 & 48 are 1, 2, 3, 4, 6 & 12
We can see that among their common factors 12 is the highest common factor. It is called H.C.F. of the given numbers. So H.C.F. of 24, 36 & 48 is 12.

H.C.F./G.C.D : The highest common factor or the greatest common divisor of given numbers is the greatest of their common factors.
H.C.F. of given two or more numbers can be found in two ways.
a) By prime factorization
b) By continued division
H.C.F. of any two consecutive numbers is always 1.
H.C.F. of relatively prime/co-prime numbers is always 1.
H.C.F. of any two consecutive even numbers is always 2.
H.C.F. of any two consecutive odd numbers is always 1.

Common multiples:
Multiples of 8: 8, 16, 24, 32, 40, 48,
Multiples of 12: 12, 24, 36, 48, … .
Multiples.common to 8 & 12: 24, 48; 72, 96, …
Least among the common multiple is 24. This is called L.C.M. of 8 & 12. The number of common multiples of given two or more numbers is infinite, as such greatest common multiple cannot be determined.

L.C.M.: The least common multiple of two or more numbers is the smallest natural number among their common multiples.
L.C.M. of given numbers can be found by the
a) Method of prime factorization.
b) Division method.
L.C.M. of any two consecutive numbers is always equal to their product.
L.C.M. of 8 &9 is 8 x 9 = 72
L.C.M. of co-prime numbers is always equal to their product.
L.C.M. of 8 & 15 is 8 x 15 = 120

Relation between the L.C.M. & H.C.F:
For a given two numbers Nj & N2 , the product of the numbers is equal to the product of their L.C.M.(L) & H.C.F.(H)
N₁ x N₂ = L x H

Check Your Progress (Page No. 29)

Question 1.
Are the numbers 900, 452, 9534, 788 divisible by 2? Why?
Solution:
Yes. Because these numbers have 0, 2, 4 and 8 in their ones place. The numbers having 0, 2, 4, 6 and 8 in their ones place are divisible by 2.

Question 2.
Are the numbers 953, 457, 781, 325, 269 divisible by 2? Why?
Solution:
No. Because, these numbers have 3, 7, 1, 5 and 9 in their ones place. The numbers having 0, 2,
4, 6 and 8 in their ones place are only divisible by 2. So, these are not divisible by 2.

Question 3.
Are the numbers 452, 673, 259, 356 divisible by 2? Verify.
Solution:
452 and 356 have 2 and 6 in their ones place respectively.
So, they are divisible by 2.
673 and 259 have 3 and 9 in their ones place respectively.
So, they are not divisible by 2.

Question 4.
Check whether the following numbers are divisible by 3 (using rule). Verify by actual division.
(i) 12345 (ii) 61392 (iii)8747
Solution:
i)12345
1 + 2 + 3 + 4 + 5=15 is a multiple of 3.
If the sum of the digits of a number is the multiple of 3, then the number is divisible by 3.
So, 12345 is divisible by 3.

ii) 61392
6 + 1 + 3 + 9 + 2 = 21is a multiple,of 3.
If the sum of the digits of a number is a multiple of 3, then the number is divisible
by 3.

So, 61392 is divisible by 3.

iii) 8747
8 + 7 + 4 + 7 = 26is not a multiple of 3.
If the sum of the digits of a number is a multiple of 3, then the numbei is divisible by 3. So, 8747 is not divisible by 3.

So, 8747 is not divisible by 3.

Let’s Explore (Page No. 29)

Question 1.
Is 8430 divisible by 6? Why?
Given number is 8430.
The given number has zero in the ones place.
So, 8430 is divisible by 2. –
And the unit sum is8 + 4 + 3 + 0 = 15 is a multiple of 3.
So, 8430 is divisible by 3.
If a number is divisible by both 2 and 3, then only it is divisible by 6.
8430 is divisible by both 2 and 3.
Therefore 8430 is divisible by 6.

Question 2.
Take any three 4 digit numbers and check whether they are divisible by 6.
Solution:
Consider: i) 5632, ii) 6855, iii) 9600 are three 4 digit numbers.
i) 5632 has 2 in its ones place. So, 5632 is divisible by 2.
The unit sum is
5 + 6 + 3 + 2 = 16 is not a multiple of 3. So, 5632 is not divisible by 3.
If a number is divisible by both 2 and 3, then only it is divisible by 6.
5632 is qnly divisible by 2, but not divisible by 3.
So, 5632 is not divisible by 6.

ii) 6855 has 5 in the ones place. So, 6855 is not divisible by 2.
6 + 8 + 5 + 5 = 24 is a multiple of 3.
So, 6855 is divisible by 3.
If a number is divisible by both 2 and 3, then only it is divisible by 6.
6855 is not divisible by 2, but it is divisible by 3.
So, 6855 is not divisible by 6.

iii) 9600 has ‘0’ in the ones place. So, 9600 is divisible by 2.
9 + 6 + 0 + 0 = 15 is a multiple of 3.
So, 9600 is divisible by 3. .
If a number is divisible by both 2 and 3, then only it is divisible by 6, 9600 is divisible by 1 both 2 and 3.
So, 9600 is divisible by 6.

Question 3.
Can you give an example of a number which is divisible by 6 but not by 2 and 3? Why?
Solution:
No. We can’t give any example, because if any number is divisible by both 2 and 3, then only it is divisible by 6. Otherwise it is not possible.

Check Your Progress (Page No. 30 & 31))

Question 1.
Test whether 6669 is divisible by 9. ,
Solution:
Given number is 6669.
Sum of the digits = 6 + 6 + 6 + 9 = 27 is divisible by 9
If the sum of the digits of a number is divisible by 9 then, it is divisible by 9. ” 27 is divisible by 9. So, 6669 is divisible by 9.

Question 2.
Without actual division, find whether 8989794 is divisible by 9.
Solution:
Given number is 8989794.
Sum of the digits =8+9+8+9+7+9+4=54 – 1
If the sum of the digits of a number is divisible by 9.
Then, it is divisible by 9.
54 is divisible by 9. So, 8989794 is divisible by 9.

Question 3.
Are the numbers 28570, 90875 divisible by 5? Verify by actual division also.
Solution:
a) Given number be 28570.
The numbers with zero or five at ones place are divisible by 5
28570 has zero in its ones place. So, 28570 is divisible by 5.
Actual division :

So, 28570 is completely divisible by 5.

b) Given number is 90875
In 90875, ones place digit is 5. So, 90875 is divisible by 5.
Actual division :

In the given number 90875 the digit in the units place is 5,
‘ So, it is divisible by 5.
So, 90875 is divisible by 5.

Question 4.
Check whether the number 598, 864, 4782 and 8976 are divisible by 4. Use divisibility rule and verify by actual division.
Solution:
a) Given number is 598.
The number formed by the digits in tens and ones places of 598 is 98.
If the number formed by last two digits (Ones and Tens) of the number is divisible
by 4, then the number is divisible by 4.
98 is not divisible by 4. So, 598 is not divisible by 4.

b) Given number is 864.
The number formed by tens and ones places of 864 is 64.
If the number formed by last two digits (ones and tens)
of the number is divisible by 4, then the number is divisible by 4.
64 is divisible by 4. So, 864 is divisible by 4.

c) Given number is 4782.
The number formed by ones and tens places of 4782 is 82.
If the number formed by last two digits (ones and tens) of the number is divisible by 4, then the number is divisible by 4. 82 is not divisible by 4.

d) Given number is 8976.
The number formed by the digits in tens and ones places of 8976 is 76.
If the number formed by last two digits
(Tens and ones) of the number is divisible by 4. Then the number is divisible by 4.
76 is divisible by 4. So, 8976 is divisible by 4.

Question 1.
Fill the blanks and complete the table. (Page No. 32)
Solution:

Lets Explore (Page No.33)

Question 1.
1221 is a polindrome number, which on reversing its digits gives the same number. Thus, every polindrome number with even number of digits is always divisible by 11. Write polindrome number of 6 – digits.
Solution:

There are some polindrome number of 6 – difits.

Check Your Progress (Page No.34)

Question 1.
Find the factors of 60.
Solution:
60 = 1 x 60
60 = 2 x 30
60 = 3 x 20
60 = 4 x 15
60 = 5 x 12
60 = 6 x 10
∴ The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.

Question 2.
Do all the factors of a given number divide the number exactly? Find the factors of 30 and verify by division.
30 = 1 x 30
30 = 3 x 10
30 = 2 x 15
30 = 5 x 6
The factors of 30 are 1, 2, 3, 5, 6,10, 15 and 30.

Yes, the factors of a given numbers are divide the number exactly.

Question 3.
3 is a factor of 15 and 24. Is 3 a factor of their difference also?
Solution:
Difference = 24 – 15 = 9 is the multiple of 3.
Yes, 3 is a factor of difference of 15 and 24.

Let’s Explore (Page No. 35)

Question 1.
What is the smallest prime number?
Solution:
2

Question 2.
What is the smallest composite number?
Solution:
4

Question 3.
What is the smallest odd prime number?
Solution:
3

Question 4.
What is the smallest odd composite number?
Solution:
9

Question 5.
Write 10 odd and 10 even composite numbers.
Solution:
Odd composite numbers are 9, 15, 21, 25, 27, 33, 35, 39, 45, 49.
Even composite numbers are 4, 6, 8, 10, 12, 14, 16, 18, 20, 22.
Except 2, every even number is a composite number.

Let’s Explore (Page No. 36)

Question 1.
Can you guess a prime number which when on reversing its digits, gives another prime number? (Hint a 2 digit prime number)
Solution:
13 and 31; 17 and 71, 37 and 73 79 and 97.

Question 2.
311 is a prime number. Can you find the other two prime numbers just by rearrang-ing the digits?
Solution:
113, 131

Check Your Progress (Page No. 36)

Question 1.
From the following numbers identify different pairs of co-primes. 2, 3,4, 5,6,7, 8, 9 and 10.
Solution:
The numbers which have only 1 as the common factor are called co-primes, (or) Numbers having no common factors, other than 1 are called co-primes.
2,3; 2, 5; 2, 7; 2, 9; 3,4; 3, 5; 3, 7; 3, 8; 3,10; 4,5; 4, 7; 4, 9; 5, 6; 5, 7; 5,8; 5, 9; 6, 7; 8, 9 and 9,10. These are the different pairs of co-primes with 2, 3, 4, 5, 6, 7, 8, 9 and 10.
1) Any two primes always forms a pair of co-primes.
2) Any two consecutive numbers always form a pair of co-primes.
3) Any two primes cilways form a pair of co-primes. .

Question 2.
Write the pairs of twin primes less than 50.
Solution:
Two prime numbers are said to be twin primes, if they differ each other by 2.
Twin primes less than 50 are (3, 5); (5, 7); (11,13); (17, 19); (29, 31) and (41, 43).

(Page No. 36)

Question 1.
Find the HCF of 12, 16 and 28
Solution:

Thus 12 = 2 x 2 x 3
16 = 2 x 2 x 2 x 2
28 = 2 x 2 x 7
The common factor of 12, 16 and 28 ¡s 2 x 2 = 4.
Hence, H.C.F of 12, 16 and 28 is 4.

Let’s Explore (Page No. 40)

What is the HCF of any two
i) Consecutive numbers ?
ii) Consecutive even numbers ?
iii) Consecutive odd numbers? What do you observe? Discuss with your Mends.
Solution:
Consider the two consecutive number are 5, 6 and 9, 10.

We observed that HCF of any two consecutive numbers is always 1.

ii) Consider two consecutive even numbers are 8, 10 and 20, 22.

We observed that HCF of any two consecutive even numbers is always 2.

iii)Consider the consecutive odd numbers are 7, 9 and 13, 15.

We observed that HCF of any two consecutive odd numbers is always 1.

(Page No. 42)

Question 1.
Find LCM of (i) 3, 4 (ii) 10, 11 (iii) 10, 30 (iv) 12, 24 (v) 3, 12 by prime factorization method.
Solution:
i) Given numbers are 3, 4
Factors of 3 = 1 x 3
Factors of 4 = 2 x 2
LCM of 3, 4 = 1 x 3 x 2 x 2 = 12

ii) Given numbers are 10, 11
Factors of 10 = 2 x 5
Factors of 11 = 1 x 11
LCM of 10, 11 = 2 x 5 x 11 = 110

iii) Given numbers are 10, 30
Factors of 10 = 2×5
Factors of 30 = 2x3x5
LCM of 10, 30 = 2 x 3 x 5 = 30

iv) Given numbers are 12, 24
Factors of 12 = 2x2x3 ,
Factors of 24 =. 2 x 2 x 2 x 3
LCM of 12, 24 = 2 x 2 x 2 x 3 = 24

v) Given numbers are 3,12 Factors of 3 = 1×3
Factors of 12 =2 x 2 x 3,
LCM of 3, 12 = 3 x 2 x 2 = 12

(Page No. 43)

Question 1.
What is the LCM and HCF of twin prime numbers ?
Solution:
LCM = Product of the taken twin primes and HCF = 1

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 10th Lesson Practical Geometry Ex 10.2

Question 1.
Draw a line segment PQ = 5.8 cm and construct its perpendicular bisector using ruler and compasses.
Solution:

Steps of construction :

1. Draw a line segment \(\overline{\mathrm{PQ}}\) = 5.8 cm.
2. Set the compasses as radius with more than half of the length of \(\overline{\mathrm{PQ}}\).
3. With P as centre, draw arcs below and above the line segment.
4. With the same radius and Q as centre draw two arcs above and below the line segment to cut the previous arcs. Name the intersecting points of arcs as X and Y.
5. Join the points X and Y. So, the line I is the perpendicular bisector of PQ.
   Hence l is required perpendicular bisector of PQ which meets at A.

Question 2.
Ravi made a line segment of length 8.6 cm. He constructed a bisector of AB on C. Find the length of AC and BC.

Solution:

As it is a bisector, it divides the line segment into two equal parts.
Each equal part is half of AB (8.6 cm) = \(\frac{\mathrm{AB}}{2}=\frac{8.6}{2}\) = 4.3 cm
∴ AC = BC = 4.3 cm

Question 3.
Using ruler and compasses, draw AB = 6.4 cm. Locate its mid point by geometric construction.
Solution:

1. Draw a line segment \(\overline{\mathrm{PQ}}\) = 6.4 cm
2. Set the compasses as radius with more than half of the length of \(\overline{\mathrm{PQ}}\).
3. With A as centre, draw arcs below and above the line segment.
4. With the same radius and B as centre draw two arcs above and below the line segment to cut the previous arcs. Name the intersecting points of arcs as X and Y.
5. Join the points X and Y. So, the line l is the perpendicular bisector of AB.
   Hence l is the required perpendicular bisector of AB which meets at M.