NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4.

BoardCBSE
TextbookNCERT
ClassClass 7
SubjectMaths
ChapterChapter 1
Chapter NameIntegers
ExerciseEx 1.4
Number of Questions Solved7
CategoryNCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4

Question 1.
Evaluate each of the following:

(a) (-30)+ 10
(b) 50 + (-5)
(c) (-36) +(-9)
(d) (-49) + (49)
(e) 13 + [(- 2) + 1]
(f) 0 + (-12)
(g) (-31) + [(-30) + (-1)]
(h) [(-36)+ 12]+3
(i) [(- 6) + 5] + [(- 2) + 1].

Solution:

(a) (- 30) + 10 = – 3
(b) 50 +(-5) = – 10
(c) (-36) +(-9) = 4
(d) (- 49) + (49) = – 1
(e) 13 + [(- 2) + 1] = 13 + (- 1) = – 13
(f) 0 + (- 12) = 0
(g) (- 31) + [(- 30) + (- 1)] = (- 31) + (- 31) = 1
(h) [(- 36) + 12] + 3 = (- 3) + 3 = – 1
(i) [(- 6) + 5] + [(- 2) + 1] = (- 1) + (- 1) = 1.

Question 2.
Verify that
a + (b + c) ≠ (a + b) + (a ÷ c)
for each of the following values of a, b and c.
(a) a = 12, b = – 4, c = 2
(b) a = (- 10), b = 1, c = l.
Solution:
(a) a + (b + c) = 12 ÷ [(- 4) + 2] = 12 + (- 2) = – 6
(a ÷ b) + (a ÷ c) = 12 ÷ (- 4) + 12 ÷ 2 = -3 + 6 = 3
So, a + (b + c) ≠ (a + b) + (a + c)

(b) a ÷ (b + c) = (- 10) + (1 + 1) = (- 10) + 2 = – 5
a ÷ b + a ÷ c = (- 10) ÷ 1 + (- 10) ÷ 1 = (- 10) + (- 10) = – 20
So, a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c).

Question 3.
Fill in the blanks:

(a) 369 ÷ …….. = 369
(b) -75 ÷ …….. = – 1
(c) (- 206) ÷ ……. = 1
(d) -87 ÷ …….. = 87
(e) ……. ÷ 1 = -87
(f) ……. ÷ 48 = -1
(g) 20 ÷ …… = -2
(h) …… ÷ (4) = – 3.

Solution:

(a) 369 ÷ 1 = 369
(b) – 75 ÷ 75 = -1
(c) (- 206) ÷ (- 206) = 1
(d) – 87 ÷ – 1 = 87
(e) – 87 ÷ 1 = – 87
(f) – 48 ÷ 48 = – 1
(g) 20 ÷ (-10) = – 2
(h) – 12 ÷ (4) = – 3.

Question 4.
Write five pairs of integers (a, b) such that a + b = -3. One such pair is (6, -2) because 6 +(-2) = (-3).
Solution:
Five pairs of integers (a, b) such that a + b = -3 are (- 6, 2), (-9, 3), (12,- 4), (21, -7), (-24, 8)
Note: We may write many such pairs of integers.

Question 5.
The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour until mid-night, at what time would the temperature be 8°C degrees below zero? What would be the temperature at mid night?
Solution:
Difference in temperatures +10 °C and -8
= [10 – (- 8)] °C = (10 + 8)° C = 18 °C
Decrease in temperature in one hour = 2°C
Number of hours taken to have temperature 8 °C below zero \(=\frac { Total\quad decrease }{ Decrease\quad in\quad one\quad hour } \)
\(=\frac { 18 }{ 2 }\)
So, at 9 P.M., the temperature will be 8 °C below zero
Temperature at mid-night = 10 °C – (2 x 12) °C
= 10°C – 24 °C = -14 °C

Question 6.
In a class test (+3) marks are given for every correct answer and (- 2) marks are given for every incorrect answer and no marks for not attempting any question.
(i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly?
(ii) Mohini scores – 5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?
Solution:
(i) Let ‘x’ be the number of incorrect questions attempted by Radhika.
According to the question, we get
(+ 3) × 12 + x × (-2) = 20
⇒ 36 – 2x = 20
⇒ 2x = 36 – 20
⇒ x = \(\frac { 16 }{ 2 } \) = 8
Therefore, Radhika attempted 8 incorrect questions.

(ii) Let ‘x’ be the number of incorrect question attempted by Mohini.
According to the question, we get
(+ 3) × 7 + x × (- 2) = – 5
⇒ 21 – 2x = -5
⇒ 2x = 21 + 5
⇒ x = \(\frac { 26 }{ 2 } \) = 13
Therefore, Mohini attempted 13 incorrect questions.

Question 7.
An elevator descends into a mine shaft at the rate of 6m/min. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 m.
Solution:
Difference in heights at two positions = 10 m – (-350 m) = 360 m
Rate of descent = 6 m/minute
∴ Time taken \(=\left( 360 \right) \div \left( 6 \right)\) minutes = 60 minutes = 1 hour
Hence, the elevator will take 1 hour to reach = 350 m.

We hope the NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4 helps you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4, drop a comment below and we will get back to you at the earliest.

 

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3.

BoardCBSE
TextbookNCERT
ClassClass 7
SubjectMaths
ChapterChapter 1
Chapter NameIntegers
ExerciseEx 1.3
Number of Questions Solved9
CategoryNCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3

Question 1.
Find each of the following products:
(a) 3 × (- 1)
(b) (- 1) × 225
(c) (-21) × (- 30)
(d) (- 316) × (- 1)
(e) (- 15) × 0 × (- 18)
(f) (- 12) × (- 11) × (10)
(g) 9 × (-3) × (-6)
(h) (- 18) ×(-5)× (- 4)
(i) (- 1) × (-2) × (-3) × 4
(j) (- 3) × (- 6) × (-2) × (- 1).
Solution:
(a) 3 x (- 1) = – (3 x 1) = – 3
(b) (- 1) x 225 = – (1 x 225) = – 225
(c) (- 21) x (- 30) = 21 x 30 = 630
(d) (- 316) x (- 1) = 316 x 1 = 316
(e) (- 15) x 0 x (- 18) = [(- 15) x 0]  x  (- 18) = 0 x (- 18) = 0
(f) (- 12) x (- 11) x (10) = [(- 12) x (- 11)] x (10) = (132) x (10) = 1320
(g) 9 x (- 3) x (- 6) = [9 x (- 3)] x (- 6) = (- 27) x (- 6) = 162
(h) (- 18) x (- 5) x (- 4) = [(- 18) x (- 5)] x (- 4) = 90 x (- 4) = – 360
(i) (- 1) x (- 2) x (- 3) x 4 = [(- 1) x (- 2)] x [(- 3) x 4] = (2)x (- 12) = -24
(j) (- 3) x (- 6) x (- 2) x (- 1) = [(- 3) x (- 6)] x [(- 2) x (- 1)] = (18) x (2) = 36

Question 2.
Verify the following:
(a) 18 × [7 + (- 3)] = [18 × 7] + [18 × (- 3)]
(b) (-21)×[(-4) + (-6)] = [(-21) × (-4)] + [(-21) × (-6)
Solution:
(a) 18 × [7 + (- 3)] = [18 × 7] + [18 × (- 3)]
L.H.S. = 18 × [7 + (- 3)]
= 18 × L(7 – 3)] = 18 × (4) = 18 × 4 = 72
R.H.S. = [18 × 7] + [18 × (- 3)]
= 126 + [- (18 × 3)] = 126 + (- 54) = 126 – 54 = 72
So, 18 × [7 + (- 3)]
= [18 × 7] + [18 × (- 3)]

(b) (- 21) × [(- 4) + (- 6)] = [(- 21) × (- 4)] + [(- 21) × (- 6)]
L.H.S. = (- 21) × [(- 4) + (- 6)]
= (- 21) × (- 10)
= 21 × 10 = 210
R. H.S. = [(- 21) × (- 4)] + [(- 21) × (- 6)]
= (21 × 4) + (21 × 6)
= 84 + 126 = 210
So, (- 21) × [(- 4) + (- 6)]
= [(- 21) × (- 4)] + [(- 21) × (- 6)].

Question 3.
(i) For any integer a, what is (-1)×a equal to?
(ii) Determine the integer whose product with (- 1) is
(a) – 22
(b) 37
(c) 0.
Solution:
(i) For any integer a, (-1) x a = -a.
(ii) We know that the product of any integer and (-1) is the additive inverse of an integer.
The integer whose product with (-1) is
(a) additive inverse of -22, t. e., 22.
(b) additive inverse of 37, i.e., -37.
(c) additive inverse of 0, i.e., 0.

Question 4.
Starting from (- 1) × 5, write various products showing some pattern to show (- 1) × (-1) – 1.
Solution:
(- 1) × 5 = – 5
(- 1) × 4 = – 4 [= (- 5) + 1]
(- 1) × 3 = – 3 [= (- 4) + 1]
(- 1) × 2 = – 2 [= (- 3) + 1]
(- 1) × 1 = – 1 [= (- 2) + 1]
(- 1) × 0 = 0 [= (- 1) + 1]
(- 1) × (- 1) = 1 [= 0 + 1]

Question 5.
Find the product, using suitable properties:
(a) 26 × (- 48) + (- 48) × (- 36)
(b) 8 × 53 × (- 125)
(c) 15×(-25)×(-4)×(- 10)
(d) (-41) × 102
(e) 625 × (-35) + (- 625) × 65
(f) 7 × (50 -2)
(g) (-17) × (-29)
(h) (- 57) ×(-19)+ 57.
Solution:
(a) We have, 26 x (-48) + (- 48) x (- 36)
= (- 48) x 26 + (- 48) x (- 36)
= (- 48) x [26 + (- 36)]
= (- 48) x (26 – 36)
=(- 48) x (- 10)= 480
(b) We have,
8 x 53 x (- 125) = [8 x (- 125)] x 53
= (- 1000) x 53 = – 53000
(c) We have,
15 x (- 25) x (- 4) x (- 10)
=15 x [(- 25) x (-4)] x (- 10)
= 15 x (100) x (- 10)
= (15 x 100) x (- 10)
= 1500 x (- 10) = – 15000
(d) We have,
(- 41) x 102 = (- 41) x (100 + 2)
= (- 41) x 100 + (- 41) x 2 = -4100 – 82 = – 4182
(e) We have, 625 x (- 35) + (- 625) x 65
= 625 x (- 35) + (625) x (- 65)
= 625 x [(- 35)+ (- 65)]
= 625 x (- 100) = – 62500
(f) 7 x (50 – 2) = 7 x 50 – 7 x 2
= 350 -14 =336
(g) (-17) x (- 29) = (-17) x [(- 30) + 1]
= (- 17) x (- 30) + (- 17) x 1 = 510 – 17 = 493
(h) (- 57) x (-19)+ 57 =57 x 19 + 57 x 1
= 57 x (19 +1)
= 57 x 20 = 1140

Question 6.
A certain freezing process requires that room temperature be lowered from 40°C at the rate of 5° C every hour. What will be the room temperature 10 hours after the process begins?
Solution:
Room temperature 10 hours after the process begins
= 40°C – 10 × 5°C
= 40°C – 50°C
= – (50 – 40)°C = – 10°C

Question 7.
In a class test containing 10 questions, 5 marks are awarded for every correct answer and (-2) marks are awarded for every incorrect answer and 0 for questions not attempted.
(i) Mohan gets four correct and si× incorrect answers. What is his score?
(ii) Reshma gets five correct answers and five incorrect answers, what is her score?
(iii) Heena gets two correct and five incorrect answers out of seven questions she attempts. What is her score?
Solution:
(i) Mohan gets for four correct answers 4 × 5 = 20 marks
He also gets for si× incorrect answers. 6 × (- 2) = – 12 marks.
Therefore, Mohan’s score = 20 + (- 12) = 20-12 = 8 marks.

(ii) Reshma gets for five correct answers 5 × 5 = 25 marks
She also gets for five incorrect answers 5 × (- 2) = – 10 marks Therefore, Reshma’s score = 25 + (- 10) = 25-10 = 15 marks.

(iii) Heena gets for two correct answers
2 × 5 = 10 marks.
She also gets for five incorrect answers 5 × (- 2) = – 10 marks
She didn’t attempt three questions. For these, she gets 3×0 = 0 marks
Therefore, Heena’s score = 10 + (- 10) + 0 = 10 – 10 + 0 = 0 marks.

Question 8.
A cement company earns a profit of ₹ 8 per bag of white cement sold and a loss of ₹ 5 per bag of grey cement sold.
(a) The company sells 3,000 bags of white cement and 5,000 bags of grey cement in a month. What is its profit or loss?
(b) What is the number of white cement bags it must sell to have neither profit nor loss if the number of grey bags sold is 6,400 bags.
Solution:
(a) The company sells 3,000 bags of white cement. So her profit = 3,000 × 8 = ₹ 24,000
Also, the company sells 5,000 bags of grey cement. So her loss = 5,000 × 5 = ₹ 25,000
Since 25,000 > 24,000
Therefore, the company is at a loss and the loss is = 25000 – 24000 = ₹ 1000

(b) Let ‘×’ be the number of white cement bags sold.
According to the question, we get
x × 8 = 6400 × 5
⇒ x = \(\frac { 6400\times 5 }{ 8 }\) = 800 × 5 = 4,000 bags.
Therefore, 4,000 bags of white cement must be sold to have neither profit nor loss.

Question 9.
Replace the blank with an integer to make it a true statement.

  1. (a) (- 3) × …….. = 27
  2. (b) 5 × …….. = -35
  3. (c) …….. × (- 8) = – 56
  4. (d) …….. × (- 12) = 132.

Solution:

  1. (a) (-3) x (- 9) = 27
  2. (b) 5 x (-7) = (-35)
  3. (c) 7 x (-8) = (-56)
  4. (d) (-11) x (-12) = 132

We hope the NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.2

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.2.

BoardCBSE
TextbookNCERT
ClassClass 7
SubjectMaths
ChapterChapter 1
Chapter NameIntegers
ExerciseEx 1.2
Number of Questions Solved4
CategoryNCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.2

Question 1.
Write down a pair of integers whose:

(a) the sum is -7
(b) the difference is -10
(c) the sum is 0.

Solution:

(a) (-15) and 8
(b) 15 and 25.
(c) (-49) and 49

Question 2.

(a) Write a pair of negative integers whose difference gives 8.
(b) Write a negative integer and a positive integer whose sum is -5.
(c) Write a negative integer and a positive integer whose difference is -3.

Solution:

(a) (-10) and (-18)
(b) (-10) and 5
(c) (-1) and 2

Question 3.
In a quiz, team A scored – 40, 10,0 and team B scored 10, 0, – 40 in three successive rounds. Which team scored more? Can we say that we can add integers in any order?
Solution:
Total scores of team A = (- 40) + 10 + 0 = – 40 + 10 + 0 = – 30
and, total scores of team B = 10 + 0 + (- 40) = 10 + 0 – 40 = – 30
Since the total scores of each team are equal.
∴ No team scored more than the other but each has an equal score.
Yes, integers can be added in any order and the result remains unaltered.
For example, 10 + 0 + (-40) = -30 = -40 + 0 + 10

Question 4.
Fill in the blanks to make the following statements true:

  1. (-5) + (-8) = (+8) + (……)
  2. -53 + …… = -53.
  3. 17 + …… = 0
  4. [13 + (-12)] + (…… ) = 13 + [(-12) + (- 7)]
  5. (- 4) + [15 + (- 3)] = [(-4) + 15] + …….

Solution:

  1. (-5) + (-8) = (-8) + (- 5)
  2. -53 + 0 = -53
  3. 17 + (- 17) = 0
  4. [13 + (- 12)] + (- 7) = 13 + [(- 12) + (-7)]
  5. (- 4) + [15 + (- 3)] = [(- 4) + 15] + (- 3).

 

We hope the NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.2, drop a comment below and we will get back to you at the earliest.

 

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6.

BoardCBSE
TextbookNCERT
ClassClass 6
SubjectMaths
ChapterChapter 14
Chapter NamePractical Geometry
Exercise Ex 14.6
Number of Questions Solved9
CategoryNCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6

Question 1.
Draw ∠POQ of measure 75° and find its line of symmetry.
Solution :
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 25
Step 1. Draw a ray \(\overline { OQ }\).
Step 2. Place the center of the protractor at O and the ∠ero edge along \(\overline { OQ }\).
Step 3. Start with 0 near Q. Mark point P at 75°
Step 4. Join \(\overline { OP }\). Then, ∠POQ = 75°.
Step 5. With O as center and using compasses, draw an arc that cuts both rays of ∠POQ. Label the points of intersection as F and Q’.
Step 6. With Q’ as center, draw (in the interior of ∠POQ) an arc whose radius is more than half the length Q’F.
Step 7. With the same radius and with F as center, draw another arc in the interior of ∠POQ. Let
the two arcs intersect at R. Then, \(\overline { OR }\) is the bisector of ∠POQ which is also the line of symmetry of ∠POQ as ∠POR = ∠ROQ.

Question 2.
Draw an angle of measure 147° and construct its bisector.
Solution :
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 26
Step 1. Draw \(\overline { OQ }\) of any length.
Step 2. Place the center of the protractor at O and the ∠ero edge along \(\overline { OQ }\).
Step 3. Start with 0 near Q. Mark a point P at 147°.
Step 4. Join OP. Then, ∠POQ = 147°.
Step 5. With O as center and using compasses, draw an arc that cuts both rays of ∠POQ. Label the points of intersection as F and Q’.
Step 6. With Q’ as center, draw (in the interior of ∠POQ) an arc whose radius is more than half the length Q’F.
Step 7. With the same radius and with F as center, draw another arc in the interior of ∠POQ. Let the two arcs intersect at R. Then, \(\overline { OR }\) is the bisector of ∠POQ.

Question 3.
Draw a right angle and construct its bisector.
Solution :
Step 1. Draw a ray OQ.
Step 2. Place the center of the protractor at O and the ∠ero edge along \(\overline { OQ }\).
Step 3. Start with 0 near Q. Mark point P at 90°.
Step 4. Join \(\overline { OP }\). Then, ∠POQ = 90°.
Step 5. With O as center and using compasses, draw an arc that cuts both rays of ∠POQ. Label the points of intersection as F and Q’.
Step 6. With Q’ as center, draw (in the interior of ∠POQ) an arc whose radius is more than half th length Q’F.
Step 7. With the same radius and with P center, draw another arc in the interior of ∠POQ the two arcs intersect at R. Then, \(\overline { OR }\) is the bisector of ∠POQ.
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 27
Question 4.
Draw an angle of measure 153° and divide it into four equal parts.
Solution :
Step 1. Draw a ray \(\overline { OQ }\).
Step 2. Place the center of the protractor at O and the ∠ero edge along \(\overline { OQ }\).
Step 3. Start with 0 near Q. Mark a point P at 153°.
Step 4. Join OP. Then, ∠POQ = 153°.
Step 5. With O as the center and using compasses, draw an arc that cuts both rays of ∠POQ. Label the points of intersection as F and Q.
Step 6. With Q’ as the center, draw (in the interior of ∠POQ) an arc whose radius is more than half the length Q’F.
Step 7. With the same radius and with F as a center, draw another arc in the interior of ∠POQ. Let the two arcs intersect at R. Then, \(\overline { OR }\) is the bisector of ∠POQ.
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 28
Step 8. With O as a center and using compasses, draw an arc that cuts both rays of ∠ROQ. Label the points of intersection as B and A.
Step 9. With A as a center, draw (in the interior of ∠ROQ) an arc whose radius is more than half the length AB.
Step 10. With the same radius and with B as a center, draw another arc in the interior of ∠ROQ. Let the two arcs intersect at S. Then, \(\overline { OS }\) is the bisector of ∠ROQ.
Step 11. With O as a center and using compasses, draw an arc that cuts both rays of ∠POR. Label the points of intersection as D and C.
Step 12. With C as a center, draw (in the interior of ∠POR) an arc whose radius is more than half the length CD.
Step 13. With the same radius and with D as centre, draw another arc in the interior of ∠POR. Let the two arcs intersect at T. Then, \(\overline { OT }\) is the bisector of ∠POR. Thus, \(\overline { OS }\), \(\overline { OR }\) and \(\overline { OT }\) divide ∠POQ = 153° into four equal parts.

Question 5.
Construct with ruler and compasses, angles of following measures:
(a) 60°
(b) 30°
(c) 90°
(d) 120°
(e) 45°
(f) 135°.
Solution :
(a) Construction of an angle of measure 60°
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 29
Step 1. Draw a line PQ and mark a point O on it.
Step 2. Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line PQ at a point say A.
Step 3. With the pointer at A (as center), now draw an arc that passes through O.
Step 4. Let the two arcs intersect at B. Join OB. We get ∠BOA whose measure is 60°.

(b) Construction of an angle of measure 30°
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 30
Step 1. Draw a line PQ and mark a point O on it.
Step 2. Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line PQ at a point say A.
Step 3. With the pointer at A (as center), now draw an arc that passes through O.
Step 4. Let the two arcs intersect at B. Join OB. We get ∠BOA whose measure is 60°.
Step 5. With O as a center and using compasses, draw an arc that cuts both rays of ∠BOA. Label the points of intersection as D and C.
Step 6. With C as a center, draw (in the interior of ∠BOA) an arc whose radius is more than half the length CD.
Step 7. With the same radius and with D as a center, draw another arc in the interior of ∠BOA. Let the two arcs intersect at E. Then, \(\overline { OE }\) is the bisector of ∠BOA, i.e., ∠BOE = ∠EOA = 30°.

(c) Construction of an angle of measure 90°
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 31
Step 1. Draw any line PQ and take a point O on it.
Step 2. Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line at A.
Step 3. Without disturbing the radius on the compasses, draw an arc with A as a center which cuts the first arc at B.
Step 4. Again without disturbing the radius on the compasses and with B as a center, draw an arc which cuts the first arc at C.
Step 5. Join OB and OC.
Step 6. With O as a center and using compasses, draw an arc that cuts both rays of ∠COB. Label the points of intersection as D and E.
Step 7. With E as a center, draw (in the interior of ∠COB) an arc where the radius is more than half the length ED.
Step 8. With the same radius and with D as a center, draw another arc in the interior of ∠COL. Let
the two arcs intersect at F. Join \(\overline { OF }\). Then, \(\overline { OF }\) is the bisector of ∠COB, i.e., ∠COF = ∠FOB. Now, ∠ FOQ = 90°.

(d) Construction of an angle of measure 120°
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 32
Step 1. Draw any line PQ and take a point O on it.
Step 2. Place the pointer of the compass es at O and draw an arc of convenient radius which puts the line at A. • ‘
Step 3. Without disturbing the radius on the compasses, draw an arc with A as a center which cuts the first arc at B.
Step 4. Again without disturbing the i alius > on the compasses and with B as a center, draw an arc which cuts the first arc at C. .
Step 5. Join OC. Then, ∠COA is the required angle whose measure is 120°.

(e) Construction of an angle of measure 45°
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 33
Step 1. Draw any line PQ and take a point O on it.
Step 2. Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line at A.
Step 3. Without disturbing the radius on the compasses, draw an arc with A as a center which cuts the first arc at B.
Step 4. Again without disturbing the radius on the compasses and with B as a center, draw an arc which cuts the first arc at C.
Step 5. Join OB and OC.
Step 6. With O as a center and using compasses, draw an arc that cuts both rays of ∠COB. Label the points of intersection as D and E.
Step 7. With E as a center, draw (in the interior of ∠COB) an arc whose radius is more than half the length ED.
Step 8. With the same radius and with D as a center, draw another arc in the interior of ∠COB. Let the two arcs intersect at F. Join OF. Then, ∠FOQ = 90°.
Step 9. With O as a center and using compasses, draw an arc that cuts both rays to ∠FOQ. Label the points of the intersection as G and H.
Step 10. With H as a center, draw (in the interior of. ∠FOQ) an arc whose radius is more than half the length HG.
Step 11. With the same radius and with G as a center, draw another arc in the interior of ∠FOQ. Let the two arcs intersect at I. Join OI. Then, OI is the bisector of ∠FOH, i.e., ∠FOI = ∠IOH. Now,
∠FOI = ∠IOH = 45°.

(f) Construction of an angle of measure 135° it.
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 34
Step 1. Draw any line PQ and take a point O on it.
Step 2.
Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line at A.
Step 3. Without disturbing the radius on the compasses, draw an arc with A as a center which cuts the first arc at B.
Step 4. Again without disturbing the radius on the compasses and with B as a center, draw an arc which cuts the first arc at C.
Step 5. Join OB and OC.
Step 6. With O as a center and using compasses, draw an arc that cuts both rays of ∠COB. Label the points of intersection as D and E.
Step 7. With E as the center, draw (in the interior of ∠COB) an arc whose radius is more than half the length ED.
Step 8. With the same radius and with D as the center, draw another arc in the interior of ∠COB. Let the two arcs intersect at F. Join \(\overline { OF }\). Then, \(\overline { OF }\) is the bisector of ∠COB, i.e., ∠COF = ∠FOB. Now, ∠FOQ = 90°.
Step 9. With O as a center and using compasses, draw an arc that cuts both rays of ∠POF. Label the points of intersection as G and H.
Step 10. With H as center draw (in the interior of ∠POF) an arc whose radius is more than half the length HG.
Step 11. With the same radius and with G as a center, draw another arc in the interior of ∠POF. Let
the two arcs intersect at I. Join 01. Then, \(\overline { OI }\) is the bisector of ∠POF, i.e., ∠GOI = ∠IOF. Now, ∠IOQ = 135°.

Question 6.
Draw an angle of measure 45° and bisect it.
Solution :
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 35
Step 1. Draw any line PQ and take a point O on it.
Step 2. Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line at A. ’
Step 3. Without disturbing the radius on the compasses, draw an arc with A as a center which cuts the first arc at B.
Step 4. Again without disturbing the radius on the compasses and with B as a center, draw an arc which cuts the first arc at C.
Step 5. Join OB and OC.
Step 6. With O as a center and using compasses, draw an arc that cuts both rays of ∠COB. Label the points of intersection as D and E.
Step 7. With E as a center, draw (in the interior of ∠COB) an arc whose radius is more than half the length ED.
Step 8. With the same radius and with D as a center, draw another arc in the interior of ∠COB. Let
the two arcs intersect at F. Join OF. Then ∠FOQ = 90°.
Step 9. With O as a center and using compasses, draw an arc that cuts both rays of ∠FOQ. Label the points of intersection on G and H.
Step 10. With G as a center, draw in the interior of ∠FOQ) an arc whose radius is more than half the length HG.
Step 11. With the same radius and with G as a center, draw another arc in the interior of ∠FOQ. Let the two arcs intersect at I. Join OI. Then OI is the bisector of ∠FOQ, i.e., ∠FOI = ∠IOH. Now, ∠FOI = ∠IOH = 45°.
Step 12. With O as a center and using compasses, draw an arc that cuts both rays of ∠IOH. Label the points of intersection as J and K.
Step 13. With K as a center, draw (in the interior of ∠IOH) an arc whose radius is more than half the length KJ.
Step 14. With the same radius and with J as a center, draw another arc in the interior of ∠IOH. Let the two arcs intersect at L. Join OL. Then OL is the bisector of ∠IOH, i.e., ∠IOL = ∠LOK \(22\frac { 1^{ \circ } }{ 2 }\) .

Question 7.
Draw an angle of measure 135° and bisect it.
Solution :
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 36
Step 1. Draw any line PQ and take a point O on.
Step 2. Place the pointer of the compasses at’ and draw an arc of convenient radius which cul line at A.
Step 3. Without disturbing the radius on the compasses, draw an arc with A as a center which cuts the first arc at B.
Step 4. Again without disturbing the radius on the compasses and with B as a center, draw an arc which cuts the first arc at C.
Step 5. Join OB and OC.
Step 6. With O as a center and using compasses, draw an arc that cuts both rays of ∠COB. Label the points of intersection as D and E.
Step 7. With E as a center, draw (in the interior of ∠COB) an arc whose radius is more than half the length ED.
Step 8. With the same radius and with D as a center, draw another arc in the interior of ∠COB. Let
the two arcs intersect at F. Join \(\overline { OF }\). Then \(\overline { OF }\) is the bisector of ∠COB, i.e., ∠COF = ∠FOB. Now, ∠FOQ = 90°.
Step 9. With O as a center and using compasses, draw an arc that cuts both rays of ∠POF. Label the points of intersection as G and H.
Step 10. With G as a center, draw in the interior of ∠POF an arc whose radius is more than half the length HG.
Step 11. With the same radius and with H as a centre, draw another arc in the interior of ∠POF. Let
the two arcs intersect at I. Join \(\overline { OI }\). Then \(\overline { OI }\) is the bisector of ∠POF, i.e., ∠POI = ∠IOF. Now, ∠IOQ = 135°. .
Step 12. With O as centre and using compasses, draw an arc that cuts both rays of ∠IOQ. Label the points of intersection as J and K.
Step 13. With K as centre, draw (in the interior of ∠IOQ) an arc whose radius is more than half the length KJ.
Step 14. With the same radius and with J as centre, draw another arc in the interior of ∠IOQ. Let the two arcs intersect at L. Join \(\overline { OL }\). Then \(\overline { OL }\) is the bisector of ∠IOQ, i.e., ∠IOL = ∠LOQ.

Question 8.
Draw an angle of 70°. Make a copy of it using only a straight edge and compasses.
Solution :
Steps of construction
1. Construct an angle ABC = 70°.
2. Take a line z and mark a point D on it.
3. Fix the compasses pointer on B and draw an arc which cuts the sides of ∠ABC at D and E.
4. Without changing the compasses setting, place the pointer on P and draw an arc which cuts ∠ at Q.
5. Open the compasses equal to length DE.
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 37
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 38
6. Without disturbing the radius on compasses, place its pointer at Q and draw an arc which cuts the previous arc at R.
7. Join PR and draw ray PR. It gives ∠RPQ which is the required angle whose measure is equal to the measure of ∠ABC.

Question 9.
Draw an angle of 40°. Copy its supplementary angle.
Solution :
Steps of construction
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 39
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 40
1. Draw ∠CAB = 40°.
2. Draw a line I and mark a point P on it.
3. Place the pointer of the compasses on A and draw an arc which cuts extended BA at E and AC at F.
4. Without changing the radius on compasses, place its pointer at P and draw an arc which cuts l at Q.
5. Open the length of compasses equal to EF.
6. Without disturbing the radius on compasses, place its pointer at Q and draw an arc which cuts the previous arc at R.
7. Join QR and draw ray QR. It gives ∠RQS which is the required angle whose measure

We hope the NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5.

BoardCBSE
TextbookNCERT
ClassClass 6
SubjectMaths
ChapterChapter 14
Chapter NamePractical Geometry
Exercise Ex 14.5
Number of Questions Solved9
CategoryNCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5

Question 1.
Draw \(\overline { AB }\) of length 7.3 cm and find its axis of symmetry.
Solution :
Step 1. Draw a line segment \(\overline { AB }\) of length 7.3 cm.
Step 2. With A ascentere, using compasses, drawthe circle. The radius of this circle should be more than half of the length of \(\overline { AB }\).
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 16
Step 3. With the same radius and with B as a centre, draw another circle using compasses. Let it cut the previous circle at C and D.
Step 4. Join CD. Then, \(\overline { CD }\) is the axis of symmetry of \(\overline { AB }\).

Question 2.
Draw a line segment of length 9.5 cm and construct its perpendicular bisector.
Solution :
Step 1. Draw a line segment \(\overline { AB }\) of length 9.5 cm.
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 17
Step 2. With A as a centre, using compasses, draw a circle. The radius of this circle should be more than half of the length of \(\overline { AB }\).
Step 3. With the same radius and with B as a centre, draw another circle using compasses. Let it cut the previous circle at C and D.
Step 4. Join CD. Then \(\overline { CD }\) is the perpendicular bisector of the line segment \(\overline { AB }\).

Question 3.
Draw the perpendicular bisector of \(\overline { XY }\) whose length is 10.3 cm.
(a) Take any point P on the bisector drawn. Examine whether PX = PY.
(b) If M is the midpoint of \(\overline { XY }\), what can you say about the lengths MX and XY ?
Solution :
Step 1. Draw a line segment \(\overline { XY }\) of length 10.3 cm.
Step 2. With X as a centre, using compasses, draw a circle. The radius of this circle should be more than half of the length of \(\overline { XY }\).
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 18
Step 3. With the same radius and with Y as a centre, draw another circle using compasses. Let it cuts the previous circle at A and B.
Step 4. Join AB. Then \(\overline { AB }\) is the perpendicular bisector of the line segment \(\overline { XY }\).
(a) On examination, we find that PX = PY.
(b) We can say that the lengths of MX is half of the length of XY.

Question 4.
Draw a line segment of length 12.8 cm. Using compasses, divide it into four equal parts. Verify by actual measurement.
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 19
Solution :
Step 1. Draw a line segment \(\overline { AB }\) of length 12.8 cm.
Step 2. With A as centre, using compasses, draw two arcs on either side of AB. The radius of this arc should be more than fialf of the length of \(\overline { AB }\).
Step 3. With the same radius and with B as centre, draw another arcs using compasses. Let it cut the previous arcs at C and D.
Step 4. Join \(\overline { CD }\). It cuts \(\overline { AB }\) at E. Then \(\overline { CD }\) is the perpendicular bisector of the line segment \(\overline { AB }\).
Step 5. With A as centre, using compasses, draw a circle. The radius of this circle should be more than half of the length of AE.
Step 6. With the same radius and with E as centre, draw another circle using compasses; Let it cut the previous circle at F and G.
Step 7. Join \(\overline { FG }\). It cuts \(\overline { AE }\) at H. Then \(\overline { FG }\) is the perpendicular bisector of the line segment \(\overline { AE }\).
Step 8. With E as centre, using compasses, draw a circle. The radius of this circle should be more than half of the length of EB.
Step 9. With the same radius and with B as centre, draw another circle using compasses. Let it cut the previous circle at I and J.
Step 10. Join \(\overline { IJ }\). It cuts \(\overline { EB }\) at K. Then \(\overline { IJ }\) is the perpendicular bisector of the line segment \(\overline { EB }\). Now, the points H, E and K divide AB into four equal parts, i.e., \(\overline { AH }\) = \(\overline { HE }\) = \(\overline { EK }\) = \(\overline { KB }\) By measurement, \(\overline { AH }\) = \(\overline { HE }\) = \(\overline { EK }\) = \(\overline { KB }\) = 3.2 cm.

Question 5.
With \(\overline { PQ }\) of length 6.1 cm as diameter draw a circle.
Solution :
Step 1. Draw a line segment \(\overline { PQ }\) of length 6.1 cm.
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 20
Step 2. With P as centre, using compasses, draw a circle. The radius of this circle should be more than half of the length of \(\overline { PQ }\).
Step 3. With the same radius and with Q as centre, draw another circle using compasses. Let it cut the previous circle at A and B.
Step 4. Join \(\overline { AB }\). It cuts \(\overline { PQ }\) at C. Then \(\overline { AB }\) is
the perpendicular bisector of the line segment PQ .
Step 5. Place the pointer of the compasses at C and open the pencil upto P.
Step 6. Turn the compasses slowly to draw the circle.

Question 6.
Draw a circle with centre C and radius, 3.4 cm. Draw any chord \(\overline { AB }\). Construct the perpendicular bisector of \(\overline { AB }\) and examine if it passes through C.
Solution :
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 21
Step 1. Draw a point with a sharp pencil and mark it as C.
Step 2. Open the compasses for the required radius 3.4 cm, by putting the pointer on 0 and opening the pencil upto 3.4 cm.
Step 3. Place the pointer of the compasses at C. Step 4. Turn the compasses slowly to draw the
circle.
Step 5. Draw any chord \(\overline { AB }\) of this circle.
Step 6. With A as centre, using compasses, draw a circle. The radius of this circle should be more than half of the length of \(\overline { AB }\).
Step 7. With the same radius and with B as centre, draw another circle using compasses. Let it cut the previous circle at D and E.
Step 8. Join \(\overline { DE }\) . Then \(\overline { DE }\) is the perpendicular bisector of the line segment \(\overline { AB }\). On examination, we find that it passes through C.

Question 7.
Repeat Question 6, if \(\overline { AB }\) happens to be a diameter.
Solution :
Step 1. Draw a point with a sharp pencil and mark it as C.
Step 2. Open the compasses for the required radius 3.4 cm, by putting the pdinter of compasses on 0 of the scale and opening the pencil upto 3.4 cm.
Step 3. Place the pointer of the compasses at C.
Step 4. Turn the compasses slowly to draw the circle.
Step 5. Draw any diameter \(\overline { AB }\).
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 22
Step 6. With A as centre, using compasses, draw arcs on either side. The radius of this arc should be more than half of the length of \(\overline { AB }\).
Step 7. With the same radius and with B as centre, draw another arcs using compasses. Let it cut the previous arcs at D and E.
Step 8. Join \(\overline { DE }\) . Then \(\overline { DE }\) is the perpendicular bisector of the line segment \(\overline { AB }\). On examination, we find that it passes through C.

Question 8.
Draw a circle of radius 4 cm. Draw any two of its chords. Construct the perpendicular bisectors of these chords. Where do they meet?
Solution :
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 23
Step 1. Draw a point with a sharp pencil and mark it as O.
Step 2. Open the compasses for the required radius of 4 cm. by putting the pointer on 0 and opening the pencil upto 4 cm.
Step 3. Place the pointer of the compasses at O.
Step 4. Turn the compasses slowly to draw the circle.
Step 5. Draw any two chords \(\overline { AB }\) and \(\overline { CD }\) of this circle.
Step 6. With A as centre, using compasses, draw two arcs on either side of AB. The radius of this arc should be more than half of the length of \(\overline { AB }\).
Step 7. With the same radius and with B as centre, draw another two arcs using compasses. Let it cut the previous circle at E and F.
Step 8. Join \(\overline { EF }\). Then \(\overline { EF }\) is the perpendicular bisector of the chord \(\overline { AB }\).
Step 9. With C a< centre, using compasses, draw two arcs on either side of CD. The radius of this arc should be more than half of the length of \(\overline { CD }\).
Step 10. With the same radius and with D as centre, draw another two arcs using compasses. Let it cut the previous circle at G and H.
Step 11. Join \(\overline { GH }\). Then \(\overline { GH }\) is the perpendi¬cular bisector of the chord \(\overline { CD }\). We find that the perpendicular bisectors \(\overline { EF }\) and \(\overline { GH }\) meet at O, the centre of the circle.

Question 9.
Draw any angle with vertex O. Take a point A on one of its arms and B on another such that OA- OB. Draw the perpendicular bisectors of \(\overline { OA }\) and \(\overline { OB }\). Let them meet at P. Is PA = PB?
Solution :
Step 1. Draw any angle POQ with vertex O.
Step 2. Take a point A on the arm OQ and another point B on the arm OP such that \(\overline { OA }\) = \(\overline { OB }\).
Step 3. With O as centre, using compasses, draw a circle. The radius of this circle should be more than half of the length of \(\overline { OA }\).
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry 24
Step 4. With the same radius and with A as centre, draw another circle using compasses. Let it cut the previous circle at C and D.
Step 5. Join \(\overline { CD }\). Then \(\overline { CD }\) is the perpendicular bisector of the line segment \(\overline { OA }\).
Step 6. With O as centre, using compasses, draw a circle. The radius of this circle should be more than half of the length of \(\overline { OB }\).
Step 7. With the same radius and with B as centre, draw another circle using compasses. Let it cut the previous circle at E and F.
Step 8. Join \(\overline { EF }\). Then \(\overline { EF }\) is the perpendicular bisector of the line segment OB. The two perpendicu¬lar bisectors meet at P.
Step 9. Join \(\overline { PA }\) and \(\overline { PB }\). We find that \(\overline { PA }\) = \(\overline { PB }\).

 

We hope the NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5, drop a comment below and we will get back to you at the earliest.