CBSE Class 6

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.

• Practical Geometry Class 6 Ex 14.1
• Practical Geometry Class 6 Ex 14.2
• Practical Geometry Class 6 Ex 14.3
• Practical Geometry Class 6 Ex 14.4
• Practical Geometry Class 6 Ex 14.6

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 \(\overline { AB }\).

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.

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 \(\overline { XY }\).

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.

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.

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 :

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 \(\overline { AB }\).

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 :

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 \(\overline { OA }\).

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.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.4 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.4.

• Practical Geometry Class 6 Ex 14.1
• Practical Geometry Class 6 Ex 14.2
• Practical Geometry Class 6 Ex 14.3
• 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 14.4

Question 1.
Draw any line segment \(\overline { AB }\) . Mark any point M on it. Through M draw a perpendicular to \(\overline { AB }\). (use ruler and compasses).
Solution :
Step 1. Given a point M on any line \(\overline { AB }\).

Step 2. With M as centre and a convenient radius, construct a part circle (arc) intersecting the line segment \(\overline { AB }\) at two points C and D.
Step 3. With C and D as centres and a radius greater than CM, construct two arcs which cut each other at N.
Step 4. Join \(\overline { MN }\). Then \(\overline { MN }\) is perpendicular to \(\overline { AB }\) at M, i.e., \(\overline { MN }\) \(\overline { AB }\).

Question 2.
Draw any’line segment \(\overline { PQ }\). Take any point R not on it. Through R draw a perpendicular to \(\overline { PQ }\). (use ruler and set-square).
Solution :
Step 1. Let \(\overline { PQ }\) be the given line segment and R be a point not on it.

Step 2. Place a set-square on \(\overline { PQ }\) such that one arm of the right angle aligns along \(\overline { PQ }\).
Step 3. Place a ruler along the edge opposite of the right angle.
Step 4. Hold the ruler fixed. Slide the set-square along the ruler all the point R touches the arm of the set-square.
Step 5. Join RS along the edge through R, meeting \(\overline { PQ }\) at S. Now \(\overline { RS }\) \(\overline { PQ }\).

Question 3.
Draw a line l and a point X on it. Through X, draw a line segment \(\overline { XY }\) perpendicular to l. Now draw a perpendicular to XY at Y. (use ruler and compasses)
Solution :
Step 1. Given a point X on a line l.
Step 2. With X as centre and a convenient radius, construct a part circle (arc) intersecting the line l at two points A and B.
Step 3. With A and B as centres and a radius greater than AX, construct two arcs which cut each other at Y.

Step 4. Join \(\overline { XY }\). Then \(\overline { XY }\) is perpendicular to l atX, i.e., \(\overline { XY }\) l.

We hope the NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.4 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.4, 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 14.3 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.