NCERT Solutions for Class 6 Maths

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.3.

• Practical Geometry Class 6 Ex 14.1
• Practical Geometry Class 6 Ex 14.2
• Practical Geometry Class 6 Ex 14.4
• Practical Geometry Class 6 Ex 14.5
• Practical Geometry Class 6 Ex 14.6

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.3

Question 1.
Draw any line segment \(\overline { PQ }\). Without measuring \(\overline { PQ }\) . construct a copy of \(\overline { PQ }\).
Solution :
Step 1. Given \(\overline { PQ }\) whose length is not known.
Step 2. Fix the compasses pointer on P and the pencil end on Q. The opening of the instrument now
gives the length of \(\overline { PQ }\).
Step 3. Draw any line l. Choose a point A on /. Without changing the compasses setting, place the pointer on A.
Step 4. Swing an arc that cuts l at a point, say, B. Now \(\overline { AB }\) is a copy of \(\overline { PQ }\).

Question 2.
Given some line segment \(\overline { AB }\), whose length you do not know, construct \(\overline { PQ }\) such that the length of \(\overline { PQ }\) is twice that of \(\overline { AB }\).
Solution :
Step 1. Given \(\overline { AB }\) whose length is not known.
Step 2. Fix the compasses pointer on A and the pencil end on B. The opening of the instrument now gives the length of \(\overline { AB }\).
Step 3. Draw any line l. Choose a point P on l. Without changing the compasses setting, place the pointer on P.
Step 4. Strike an arc that cuts l at a point, say, X.
Step 5. Now fix the compasses pointer on X. Strike an arc away from P that cuts l at a point, say, Q. Now, the length of \(\overline { PQ }\) is twice that of \(\overline { AB }\).

We hope the NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.3 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.3, 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 14.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.2.

• Practical Geometry Class 6 Ex 14.1
• Practical Geometry Class 6 Ex 14.3
• Practical Geometry Class 6 Ex 14.4
• Practical Geometry Class 6 Ex 14.5
• Practical Geometry Class 6 Ex 14.6

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.2

Question 1.
Draw a line segment of length 7.3 cm. using a ruler.
Solution :
Using ruler, we mark two points A and B which are 7.3 cm apart. Join A and B and get AB. \(\overline { AB }\) is a line segment of length 7.3 cm.

Question 2.
Construct a line segment of length 5.6 cm using ruler and compasses.

Solution :
Step 1. Draw a line l. Mak a point A on line l.
Step 2. Place the compasses pointer on the ∠ero mark on the ruler. Open it to place the pencil point upto the 5.6 cm mark.
Step 3. Without changing the opening of the compasses, place the pointer on A and swing an arc to cut / at B.
Step 4. AB is a line segment of required length.

Question 3.
Construct \(\overline { AB }\) of length 7.8 cm. From this cut off \(\overline { AC }\) of length 4.7 cm. Measure \(\overline { BC }\).
Solution :
Steps of Construction
Step 1. Draw a line l. Mark a point A on line l.
Step 2. Place the compasses pointer on the ∠ero mark on the ruler. Open it to place the pencil point upto the 7.8 cm mark.

Step 3. Without changing the opening of the compasses, place the pointer on A and swing an arc to cut / at B.
Step 4. \(\overline { AB }\) is a line segment of length 7.8 cm.
Step 5. Place the compasses pointer on the ∠ero mark on the ruler. Open it to place the pencil point upto 4.7 cm mark.
Step 6. Without changing the opening of the compasses, place the pointer on A and swing an arc to cut / at C.
Step 7. \(\overline { AC }\) is a line segment of length 4.7 cm. On measurement. \(\overline { BC }\) =3.1 cm.

Question 4.
Given \(\overline { AB }\) of length 3.9 cm, construct \(\overline { PQ }\) such that the length of \(\overline { PQ }\) is twice that of AB Verify by measurement.

(Hint : Construct \(\overline { PX }\) such that length of PX = length of \(\overline { AB }\); then cut off \(\overline { XQ }\) such that \(\overline { XQ }\) also has the length of \(\overline { AB }\)).
Solution :
Steps of Construction
Step 1. Draw a line I. Mark a point P on line l.
Step 2. Place the compasses pointer on the A mark of the given line segment \(\overline { AB }\) . Open it to place the pencil point upto B mark of the given line segment \(\overline { AB }\).

Step 3. Without changing the opening of the compasses, place the pointer on P and swing an arc to cut l at X.
Step 4. Again without changing the opening of the compasses place the compasses pointer on the X mark of line / and swing an arc to cut / at Q.
Step 5. \(\overline { PQ }\) is a line segment of length twice that of \(\overline { AB }\). Please verify yourself by measurement.

Question 5.
Given \(\overline { AB }\) of length 7.3 cm and \(\overline { CD }\) of length 3.4 cm, construct a line segment \(\overline { XY }\) such that the length of \(\overline { XY }\) is equal to the difference between the lengths of \(\overline { AB }\) and \(\overline { CD }\) . Verify by measurement.
Solution :
Steps of Construction
Step 1. Draw a line l. Mark a point X on line l.
Step 2. Place the compasses pointer on the A mark of the given line segment \(\overline { AB }\). Open it to place the pencil point upto B mark of the given line segment \(\overline { AB }\).
Step 3. Without changing the opening of the compasses, place the pointer of compasses on X and swing an arc to cut / at ∠.
Step 4. Place the compasses pointer on the C mark of the given line segment \(\overline { CD }\). Open it to place the pencil point upto D mark of the given line segment \(\overline { CD }\).
Step 5. Without changing the opening of the compasses, place the pointer of compasses on ∠ and swing an arc towards X to cut l at Y.
Step 6. \(\overline { XY }\) is a required line segment of length = the difference between the lengths of \(\overline { AB }\) and \(\overline { CD }\).

Please verify yourself by measurement.

We hope the NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.2 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.3.

NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.3

Question 1.
Find the number of lines of symmetry in each of the following shapes. How will you check your answers?


Solution :
(a)

Number of lines of symmetry = 4
(b)

Number of lines of symmetry = 1
(c)

Number of lines of symmetry = 2
(d)

Number of lines of symmetry = 2
(e)

Number of lines of symmetry = 1
(f)

Number of lines of symmetry = 2

Question 2.
Copy the following drawing on squared paper. Complete each one of them such that the resulting figure has the two dotted lines as two lines of symmetry :

How did you go about completing the picture?
Solution :




Using the given lines of symmetry, we go about completing the picture.

Question 3.
In each figure alongside, a letter of the alphabet is shown along with a vertical line. Take the mirror image of the letter in the given line. Find which letters look the same after reflection (i.e., which letters look the same in the image) and which do not. Can you guess why?

Solution :
The letter A looks the same after reflection but not the letter B. The reason is that in reflection, the sense of direction changes. In the given letters, the letters O, M, N, H, T, V, and X look the same after reflections because these letters have a vertical line of symmetry.

We hope the NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.3 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.2.

NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.2

Question 1.
Find the number of lines of symmetry for each of the following shapes :

Solution :

Question 2.
Copy the triangle in each of the following figures, on squared paper. In each case, draw the line(s) of symmetry if any and identify the type of triangle. (Some of you may like to trace the figures and try paper-folding first!)

Solution :

Question 3.
Complete the following table :

Solution :

Question 4.
Can you draw a triangle which has
(a) exactly one line of symmetry?
(b) exactly two lines of symmetry?
(c) exactly three lines of symmetry?
(d) no lines of symmetry?
Sketch a rough figure in each case.
Solution :
(a) Yes; an isosceles triangle

(b) No
(c) Yes; an equilateral triangle

(d) Yes, an equilateral triangle

Question 5.
On a squared paper, sketch the following:
(a) A triangle with a horizontal line of symmetry’ but no vertical line of symmetry.
(b) A quadrilateral with both horizontal and vertical lines of symmetry.
(c) A quadrilateral with a horizontal line of symmetry but no vertical line of symmetry.
(d) A hexagon with exactly two lines of symmetry.
(e) A hexagon with six lines of symmetry.
(Hint: It will be helpful if you first draw the lines of symmetry and then complete the figures.)
Solution :


Question 6.
Trace each figure and draw the lines of symmetry, if any :



Solution :

no line symmetry


Question 7.
Consider the letters of English alphabets A to Z. List among them the letters which have
(a) vertical lines of symmetry (like A)
(b) horizontal lines of symmetry (like B)
(c) no lines of symmetry (like Q)

Solution :
(a) A, H, I, M, O, T, U, V, W, X, Y
(b) B, C, D, E, H, I, K, O, X
(c) F, G, J, L, N, P, Q, R, S, Z

Question 8.
Given here are figures of a few folded sheets and designs drawn about the fold. In each case, draw a rough diagram of the complete figure that would be seen when the design is cut off.

Solution :

We hope the NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.2 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 13 Symmetry Ex 13.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.2.

• Ratio and Proportion Class 6 Ex 12.1
• Ratio and Proportion Class 6 Ex 12.3

NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.2

Question 1.
Determine if the following are in proportion:
(a) 15, 45, 40, 120
(b) 33, 121, 9, 96
(c) 24, 28, 36, 48
(d) 32, 48, 70, 210
(e) 4, 6, 8, 12
(f) 33, 44, 75, 100
Solution :





Question 2.
Write True (T) or False (F) against each of the following statements :
(a) 16: 24: : 20: 30
(b) 21: 6: 35: 10
(c) 12: 18: 28: 12
(d) 8: 9: : 24: 27
(e) 5.2: 3.9: : 3: 4
(f) 0.9: 0.36: : 10: 4
Solution :




Question 3.
Are the following statements true₹ I
(a) 40 Persons : 200 Persons =₹ 15: ₹75 ;
(b) 7.5 litre : 15 litre = 5 kg : 10 kg 1
(c) 99 kg : 45 kg = f 44 : ₹20
(d) 32 m : 64 m = 6 sec: 12 sec
(e) 45 km: 60 km = 12 hours : 15 hours.
Solution :


Since the two ratios are equal, therefore, the given statement is true.

Since the two ratios are equal, therefore, the given statement is true.

Since, the two ratios are equal, therefore, the given statement is true.


Since the two ratios are equal, therefore the given statement is true.

Since the two ratios are not equal, therefore the given statement is false.

Question 4.
Determine if the following ratios form a proportion. Also, write the middle terms and extreme terms where the ratios form a proportion.
(a) 25 cm: 1 m and 7 40 : 7160
(b) 39 litre : 65 litre and 6 bottle : 10 bottle
(c) 2 kg : 80 kg and 25 g : 625 g
(d) 200 ml: 2.5 litre and ₹4 ; ₹ 50.
Solution :
(a)
∴ 1 m= 100 cm

Since the two ratios are equal, therefore, the given ratios are in proportion. Middle terms are 1 m and ₹ 40. Extreme terms are 25 cm and ₹ 160.

Since the two ratios are equal, therefore, the given ratios are in proportion. Middle terms are 65 liters and 6 bottles. Extreme terms are 39 liters and 10 bottles.


Since the two ratios are not equal, therefore, the given ratios are not in proportion.

Since the two ratios are equal, therefore, the given ratios are in proportion. Middle terms are 2.5 litres and ₹ 4. Extreme terms are 200 ml and ₹ 50.

We hope the NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.2 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 12 Ratio and Proportion Ex 12.2, drop a comment below and we will get back to you at the earliest.