RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A

RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 7 Decimals Ex 7A.

Other Exercises

Question 1.
Solution:
(i)Fifty eight point six three = 58.63
(ii)One hundred twenty four point four two five = 124.425
(iii)Seven point seven six = 7.76
(iv)Nineteen point eight = 19.8
(v)Four hundred four point zero four four = 404.044
(vi)Point one seven three = 173
(v)Point zero one five = .015 Ans.

Question 2.
Solution:
(i) 14.83
Place value of 1 = 10,
Place value of 4 = 4,
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A Q2.1
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A Q2.2
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A Q2.3

Question 3.
Solution:
(i) 67.83 = (6 x 10) + (7 x 1)
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A Q3.1

Question 4.
Solution:
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A Q4.1

Question 5.
Solution:
(i) 7.5, 64.23, 0.074 = 7.500, 64.230, 0.074
(Here, at the most 0.074 has 3 places)
(ii) 0.6, 5.937, 2.36, 4.2 = 0.600, 5.937, 2.360, 4.200
(Here, 5.937 has at most 3 places)
(iii) 1.6, 0.07, 3.58, 2.9 = 1.60, 0.07, 3.58, 2. 90
(Here, at the most are two places)
(iv) 2.5. 0.63, 14.08, 1.637 = 2.500, 0.630. 14.080, 1.637 Ans.
(Here, at the most are three places)

Question 6.
Solution:
Making like decimals where ever it is necessary,
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A Q6.1

Question 7.
Solution:
First of all making them in like decimals,
(i) 5.8, 7.2, 5.69, 7.14, 5.06
=> 5.80, 7.20, 5.69, 7.14, 5.06
Arranging in ascending order,
5:06 <5.69 <5.80 <7.14 <7.20
=> 5.06 < 5.69 < 5.8 < 7.14 < 7.2 Ans.
(ii) 0.6, 6.6, 6.06, 66.6, 0.06
=>0.60, 6.60, 6.06, 66.60, 0.06
Arranging in ascending order,
0.06 < 0.60 < 6.06 < 6.60 < 66.60
=> 0.06 < 0.6 < 6.06 < 6.6 < 66.6 Ans.
(iii) 6.54, 6.45, 6.4, 6.5, 6.05
=> 6.54, 6.45, 6.4, 6.5, 6.05
Arranging in ascending order,
6. 05 < 6.40 < 6.45 < 6.50 < 6.54
=> 6.05 < 6.4 < 6.45 < 6.5 < 6.54 Ans.
(iv) 3.3,3.303, 3.033, 0.33, 3.003
=> 3.300, 3.303, 3.033, 0.330, 3.003
Arranging in descending order,
0.330 < 3.003 < 3.033 < 3.300 < 3.303
=> 0.33 < 3.003 < 3.033 < 3.3 < 3.303 Ans.

Question 8.
Solution:
Making them in like decimals and them comparing
(i) 7.3, 8.73, 73.03, 7.33, 8.073
=> 7.300, 8.730, 73.030, 7.330, 8.073
Arranging in descending order
73.030 > 8.730 > 8.073 > 7.330 > 7.300
=> 73.03 > 8.73 > 8.073 > 7.33 > 7.3 Ans.
(ii) 3.3, 3.03, 30.3, 30.03, 3.003
=> 3.300, 3.030, 30.300, 30.030, 3.003
Arranging in descending order
30.300> 30.030 >3.300 >3.030 > 3.003
=> 30.3 > 30.03 > 3.3 > 3.03 > 3.003 Ans.
(iii) 2.7, 7.2, 2.27, 2.72, 2.02, 2.007
=> 2.700, 7.200, 2.270, 2.720, 2.020, 2.007
Arranging in descending order
7. 200 > 2.720 > 2.700 > 2.270 > 2.020 > 2.007
=> 7.2 > 2.72 > 2.7 > 2.27 > 2.02 > 2.007 Ans.
(iv) 8.88, 8.088, 88.8, 88.08, ,8.008
=> 8.880, 8.088, 88.800, 88.080, 8.008
Arranging in descending order,
88.800 > 88.080 > 8.880 > 8.088 > 8.008
=> 88.8 > 88.08 > 8.88 > 8.088 > 8.008

 

Hope given RS Aggarwal Solutions Class 6 Chapter 7 Decimals Ex 7A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6B

RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6B

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 6 Simplification Ex 6B.

Other Exercises

OBJECTIVE QUESTIONS
Mark against the correct answer in each of the following :

Question 1.
Solution:
(c) 8 + 4 ÷ 2 x 5
= 8 + 4 x \(\\ \frac { 1 }{ 2 } \) x 5
= 8 + 10
= 18.

Question 2.
Solution:
(b) 54 ÷ 3 of 6 + 9 = 54 ÷ 18 + 9
= 54 x \(\\ \frac { 1 }{ 18 } \) + 9
= 3 + 9
= 12.

Question 3.
Solution:
(b) 13 – (12 – 6 ÷ 3)
= 13 – (12 – 2)
= 13 – 10
= 3

Question 4.
Solution:
(a) 1001 ÷ 11 of 13
= 1001 ÷ 143
= 1001 x \(\\ \frac { 1 }{ 143 } \)
= 7.

Question 5.
Solution:
(b) 133 + 28 ÷ 7 – 8 x 2
= 133 + 4 – 16
= 137 – 16
= 121.

Question 6.
Solution:
(a) 3640 – 14 ÷ 7 x 2
= 3640 – 2 x 2
= 3640 – 4
= 3636.

Question 7.
Solution:
(b) 100 x 10 – 100 + 2000 ÷ 100
= 1000 – 100 + 20
= 920.

Question 8.
Solution:
(b) \(27-\left[ 18-\left\{ 16-\left( 5-\overline { 4-1 } \right) \right\} \right] \)
= 27 – [18 – {16 – (5 – 4 + 1)}]
= 27 – [18 – {16 – 5 + 4 – 1}]
= 27 – [18 – 16 + 5 – 4 + 1]
= 27 – 18 + 16 – 5 + 4 – 1
= 23

Question 9.
Solution:
\(32-\left[ 48\div \left\{ 36-\left( 27-\overline { 16-9 } \right) \right\} \right] \)
= 32 – [48 ÷ {36 – (27 – 16 + 9)}]
= 32 – [48 ÷ {36 – 27 + 16 – 9}]
= 32 – [48 ÷ 16]
= 32 – 3
= 29

Question 10.
Solution:
(a) 8 – [28 ÷ {34 – (36 – 18 ÷ 9 x 8)}]
\(8-\left[ 28\div \left\{ 34-\left( 36-18\times \frac { 1 }{ 9 } \times 8 \right) \right\} \right] \)
= 8 – [28 ÷ {34 – (36 – 16)}]
= 8 – [28 ÷ {34 – 20}]
= 8 – {28 ÷ 14}
= 8 – 2
= 6.

Hope given RS Aggarwal Solutions Class 6 Chapter 6 Simplification Ex 6B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A

RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 6 Simplification Ex 6A.

Other Exercises

Simplify

Question 1.
Solution:
21 – 12 ÷ 3 x 2
= 21 – 4 x 2
= 21 – 8
= 13. Ans

Question 2.
Solution:
16 + 8 ÷ 4 – 2 x 3
= 16 + 2 – 2 x 3
= 16 + 2 – 6
= 18 – 6
= 12. Ans.

Question 3.
Solution:
13 – (12 – 6 ÷ 3)
= 13 – (12 – 2)
= 13 – (10)
= 13 – 10
= 3 Ans.

Question 4.
Solution:
19 – [4 + {16 – (12 – 2)}]
= 19 – [4 + {16 – 10}]
= 19 – [4 + 6]
= 19 – 10
= 9. Ans

Question 5.
Solution:
36 – [18 – {14 – (15 – 4 ÷ 2 x 2)}]
= 36 – [18 – {14 – (15 – 2 x 2)}]
= 36 – [18 – {14 – (15 – 4)}]
= 36 – [18 – {14 – 11}]
= 36 – [18 – 3]
= 36 – 15
= 21. Ans.

Question 6.
Solution:
\(27-[18-\{ 16-(5-\overline { 4-1 } )\} ] \)
= 27 – [18 – {16 – (5 – 3)}]
= 27 – [18 – {16 – 2}]
= 27 – [18 – 14]
= 27 – 4
= 23. Ans.

Question 7.
Solution:
\(4\frac { 4 }{ 3 } \div \frac { 3 }{ 5 } of5+\frac { 4 }{ 5 } \times \frac { 3 }{ 10 } -\frac { 1 }{ 5 } \)
RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A 7.1

Question 8.
Solution:
\(\left( \frac { 2 }{ 3 } +\frac { 4 }{ 9 } \right) of\frac { 3 }{ 5 } \div 1\frac { 2 }{ 3 } \times 1\frac { 1 }{ 4 } -\frac { 1 }{ 3 } \)
RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A 8.1

Question 9.
Solution:
\(7\frac { 1 }{ 3 } \div \frac { 2 }{ 3 } of2\frac { 1 }{ 5 } +1\frac { 3 }{ 8 } \div 2\frac { 3 }{ 4 } -1\frac { 1 }{ 2 } \)
RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A 9.1

Question 10.
Solution:
\(5\frac { 1 }{ 7 } -\left\{ 3\frac { 3 }{ 10 } \div \left( 2\frac { 4 }{ 5 } -\frac { 7 }{ 10 } \right) \right\} \)
RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A 10.1

Question 11.
Solution:
\(9\frac { 3 }{ 4 } \div \left[ 2\frac { 1 }{ 6 } +\left\{ 4\frac { 1 }{ 3 } -\left( 1\frac { 1 }{ 2 } +1\frac { 3 }{ 4 } \right) \right\} \right] \)
RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A 11.1
RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A 11.2

Question 12.
Solution:
\(4\frac { 1 }{ 10 } -\left[ 2\frac { 1 }{ 2 } -\left\{ \frac { 5 }{ 6 } -\left( \frac { 2 }{ 5 } +\frac { 3 }{ 10 } -\frac { 4 }{ 15 } \right) \right\} \right] \)
RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A 12.1

Question 13.
Solution:
\(1\frac { 5 }{ 6 } +\left[ 2\frac { 2 }{ 3 } -\left\{ 3\frac { 3 }{ 4 } \left( 3\frac { 4 }{ 5 } \div 9\frac { 1 }{ 2 } \right) \right\} \right] \)
RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A 13.1

Question 14.
Solution:
\(4\frac { 4 }{ 5 } \div \left\{ 2\frac { 1 }{ 5 } -\frac { 1 }{ 2 } \left( 1\frac { 1 }{ 4 } -\overline { \frac { 1 }{ 4 } -\frac { 1 }{ 5 } } \right) \right\} \)
RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A 14.1

Question 15.
Solution:
\(7\frac { 1 }{ 2 } -\left[ 2\frac { 1 }{ 4 } \div \left\{ 1\frac { 1 }{ 4 } -\frac { 1 }{ 2 } \left( \frac { 3 }{ 2 } -\overline { \frac { 1 }{ 3 } -\frac { 1 }{ 6 } } \right) \right\} \right] \)
RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6A 15.1

Hope given RS Aggarwal Solutions Class 6 Chapter 6 Simplification Ex 6A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G

RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 5 Fractions Ex 5G.

Other Exercises

Objective Questions :
Tick the correct answer in each of the following :

Question 1.
Solution:
(c) ∴ canceling the common factor 2, we get \(\\ \frac { 3 }{ 5 } \)

Question 2.
Solution:
(c) ∴ multiplying numerator and denominator by 4, we get \(\\ \frac { 8 }{ 12 } \)

Question 3.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 3.1

Question 4.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 4.1

Question 5.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 5.1

Question 6.
Solution:
(c) each of the fractions has the same denominator.

Question 7.
Solution:
(d) none of these has greater denominator than its numerator.

Question 8.
Solution:
(a) its denominator is greater than its numerator.

Question 9.
Solution:
(b) their numerators are same and 4 < 5 , \(\frac { 3 }{ 4 } >\frac { 3 }{ 5 } \)

Question 10.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 10.1

Question 11.
Solution:
(b) In \(\frac { 4 }{ 5 } ,\frac { 2 }{ 7 } ,\frac { 4 }{ 9 } ,\frac { 4 }{ 11 } \) numerator is same then the smallest denominator’s fraction is greater.

Question 12.
Solution:
(a) Denominators are same, then fraction of smallest numerator will be smallest.

Question 13.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 13.1

Question 14.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 14.1

Question 15.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 15.1

Question 16.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 16.1

Question 17.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 17.1

Question 18.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 18.1

Question 19.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 19.1

Question 20.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5G 20.1

Hope given RS Aggarwal Solutions Class 6 Chapter 5 Fractions Ex 5G are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

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.