NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7.

• Playing with Numbers Class 6 Ex 3.1
• Playing with Numbers Class 6 Ex 3.2
• Playing with Numbers Class 6 Ex 3.3
• Playing with Numbers Class 6 Ex 3.4
• Playing with Numbers Class 6 Ex 3.5
• Playing with Numbers Class 6 Ex 3.6

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 3
Chapter Name	Playing with Numbers
Exercise	Ex 3.7
Number of Questions Solved	1
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7

Question 1.
Renu purchases two bags of fertilizer of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertilizer e×act number of times.
Solution :
Factors of 75 are 1, 3, 5, 15, 25 and 75.
Factors of 69 are 1, 3, 23 and 69.
∴ Common factors of 75 and 69 are 1 and 3.
Highest of these common factors is 3.
∴ H.C.F. of 75 and 69 is 3.
Hence, the maximum value of weight which can measure the weight of the fertilizer e×act number of times is 3 kg.

Question 2.
Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm, and 77 cm, respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?
Solution :

∴ L.C.M. of 63, 70 and 77
= 2 × 3 × 3 × 5 × 7 × 11 = 6930.
Hence, the minimum distance each should cover so that all can cover the distance in complete steps is 6930 cm.

Question 3.
The length, breadth, and height of a room are 825 cm, 675 cm, and 450 cm, respectively. Find the longest tape which can measure the three dimensions of the room exactly.
Solution :
Factors of 825 are 1, 3, 5, 11, 15, 25, 33, 55,75,165, 275 and 825.
Factors of 675 are 1, 3, 5,9,15, 25, 27,45,75, 135, 225 and 675.
Factors of 450 are 1,2,3,5,6,9,10,15,18,25, 30,45, 50, 75, 90, 150, 225 and 450.
∴ Common factors of 825, 675 and 450 are 1,3,5,15, 25 and 75.
Highest of these common factors is 75.
Hence, the length of the longest tape which can measure the three dimensions of the room exactly is 75 cm.

Question 4.
Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.
Solution :

∴ L.C.M. of 6, 8 and 12 = 2 × 2 × 2 × 3 = 24. Multiples of 24 are 24,48,72,96,120,144,
Hence, the smallest 3-digit number which is exactly divisible by 6, 8 and 12 is 120.

Question 5.
Determine the largest 3-digit number exactly divisible by 8, 10 and 12.
Solution :

∴ L.C.M. of 8,10 and 12 = 2×2×2×3×5
= 120.
Multiples of 120 are :
120 × 1 = 120,120 × 2 = 240,120 × 3 = 360,120 × 4 = 480,120 × 5 = 600,120 × 6 = 720,120 × 7 = 840,
120 × 8 = 960,120 × 9 = 1080,
Hence, the largest 3-digit number exactly divisible by 8, 10 and 12 is 960.

Question 6.
The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 am at what time will they change simultaneously again?
Solution :

∴ L.C.M. of 48,72 and 108 = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432.
432 seconds = 7 min 12 seconds.
Hence, they will change simultaneously again after 7 min 12 seconds from 7 a.m.

Question 7.
Three tankers contain 403 liters, 434 liters and 465 liters of diesel respectively. Find the ma×imum capacity of a container that can measure the diesel of the three containers e×act a number of times.
Solution :
Factors of 403 are 1, 13, 31 and 403. Factors of 434 are 1, 2, 7, 14, 31, 62, 217 and 434.
Factors of 465 are 1, 3, 5, 15, 31, 93, 155 and 465.
Common factors of 403,434 and 465 are 1 and 31.
Highest of these common factors is 31.
∴ H.C.F. of 403. 434 and 465 is 31.
Hence, the maximum capacity of the container that can measure the diesel of the three containers an e×act number of times is 31 litres.

Question 8.
Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.
Solution :

∴ L.C.M. of 6, 15 and 18 = 2 × 3 × 3 × 5 = 90.
Hence, the required number is 90 + 5 i.e., 95.

Question 9.
Find the smallest four digit number which is divisible by 18, 24 and 32.
Solution :

∴ L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288.
Multiples of 288 are :
288 × 1 = 288, 288 × 2 = 576, 288 × 3 = 864, 288×4= 1152,
Hence, the smallest four digit number which is divisible by 18, 24 and 32 is 1152.

Question 10.
Find the L.C.M. of the following numbers:
(a) 9 and 4
(b) 12 and 5
(c) 6 and 5
(d) 15 and 4.
Observe a common property in the obtained L.C.M.s. Is L.C.M. the product of two numbers in each case?
Solution :
(a) 9 and 4

∴ L.C.M. of 9 and 4 = 2 × 2 × 3 × 3 = 36 (= 9 × 4).

(b) 12 and 5

∴ L.C.M. of 12 and 5 = 2×2×3×5 = 60 (= 12 × 5).

(c) 6 and 5

∴ L.C.M. of 6 and 5 = 2×3×5 = 30 (= 6 × 5).

(d) 15 and 4

∴ L.C.M. of 15 and 4 = 2 × 2 × 3 × 5 = 60 (=15×4).
We observe a common property in the obtained L.C.M.’s that L.C.M. is the product of two numbers in each case.

Question 11.
Find the L.C.M. of the following numbers in which one number is the factor of the other.
(a) 5, 20
(b) 6, 18
(c) 12, 48
(d) 9, 45.
What do you observe in the results obtained?
Solution :
(a) 5, 20
Prime factorisations of 5 and 20 are as follows: 5 = 5
20 = 2 × 2 × 5 ∴ L.C.M. of 5 and 20
=2×2×5 = 20.

(b) 6, 18
Prime factorisations of 6 and 18 are as follows:
6 = 2×3 18 = 2×3×3
∴ L.C.M. of 6 and 18 = 2×3×3 = 18.

(c) 12, 48
Prime factorisations of 12 and 48 are as follows:
12 = 2 × 2 × 3
48 = 2×2×2×2×3
∴ L.C.M. of 12 and 48 = 2 × 2 × 2 × 2 × 3 = 48.

(d) 9, 45
Prime factorisations of 9 and 45 are as follows:
9 = 3×3
45 = 3 × 3 × 5
∴ L.C.M. of 9 and 45 = 3 × 3 × 5 = 45.
In the results obtained, we observe that L.C.M. of the two numbers in which one number is the factor of the other is the greater number.

We hope the NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Exz 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.

• 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 14.5

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 14
Chapter Name	Practical Geometry
Exercise	Ex 14.6
Number of Questions Solved	9
Category	NCERT Solutions

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

Question 1.
Draw ZPOQ of measure 75° and find its line of symmetry.
Solution:
Steps of Construction:

1. Draw a ray OP.
2. Draw ∠POR = 60° and ∠POS = 90°.
3. Draw OQ, the bisector of ∠ROS.
   Then,
   

Question 2.
Draw an angle of measure 147° and construct its bisector.
Solution:
Steps of Construction:

1. Draw a ray OA.
2. Place the protractor on OA such that its centre falls on the initial point O and 0 -180 line lies along OA.
   
3. Mark a point B on the paper against the mark of 147° on the protractor.
4. Remove the protractor and draw OB. Then, the ∠AOB so obtained is the required angle such that ZAOB = 147°.

To construct its bisector:

1. With centre O and a convenient radius draw an arc cutting sides OA and OB at P and Q respectively.
2. With centre P and radius > PQ, draw an arc.
3. With centre Q and the same radius, as in the previous step, draw another arc intersecting the arc drawn in the previous step at R.
4. Join OR and produce it to form ray OC.
   Then, the ∠AOC so obtained is the bisector of ∠AOB.

Question 3.
Draw a right angle and construct its bisector.
Solution:
Steps of Construction:
To draw an angle of 90°:

1. Draw a ray OA.
2. With O as centre and any convenient radius, draw an arc cutting at P.
3. With P as centre and same radius, draw an arc cutting the arc of step 2 at Q.
4. With Q as centre and the same radius as in steps 2 and 3, draw an arc cutting the arc drawn in step 3 at R.
5. With Q as centre and the same radius, draw an arc.
6. With R as centre and the same radius, draw an arc cutting the arc drawn in step 5 at B.

To draw its bisector:

1. With P as centre and radius > PT, draw an arc in the interior of ∠AOC.
2. With T as centre and the same radius, an in step 1, draw another arc intersecting the arc in step 1 at D.
3. Join OD and produce it to any point E.
   Then, ∠AOE so obtained is the bisector of ∠AOC.

Question 4.
Draw an angle of measure 153° and divide it into four equal parts.
Solution:
Steps of Construction:

1. With the help of the protractor, draw ∠AOB = 153°.
2. With centre O and any convenient radius, draw an arc cutting OA and OB at P and Q respectively.
3. With centre P and radius > PQ, draw an arc in the interior of ∠AOB.
4. With centre Q and the same radius, as in step 3, draw another arc intersecting the arc in step 3 at B₁.
5. Join OB₁ and produce it to any point C.
   Then, ∠AOC = x ∠AOB i.e., bisector of ∠AOB.
6. Draw OD, the bisector of ∠AOC. Then ∠AOD = ∠DOC.
7. Draw OE, the bisector of ∠COB. Then, ∠COE = ∠COB.
   Combining these results, we have
   ∠AOD = ∠DOC = ∠COE = ∠EOB.
   Thus, ∠AOB is divided into four equal parts by the rays OD,OC and OE.

Question 5.
Construct with ruler and compasses, angles of following measures:
(a) 60°
(b) 30°
(c) 90°
(d) 45°
(f) 135°
Solution:
(a) Steps of Construction:

1. Draw a ray OA.
2. With centre O and any radius, draw an arc PQ with the help of compasses, cutting the ray at P.
3. With centre P and the same radius draw an arc cutting the arc PQ at R.
4. Join OR and produce it to obtain ray OB.
   Then, ∠AOB so obtained is of 60°.

(b) Steps of Construction:

1. Draw a ray OA.
2. With centre O any radius draw an arc PT with the help of compasses, cutting ray OA at P.
3. With centre P and the same radius draw an arc cutting the arc PT at Q.
4. Join OQ and produce it to obtain ray OB.
   Then, ∠AOB = 60°.
5. With centre P and radius > PQ, draw an arc in the interior of ∠AOB.
6. With centre Q and the same radius, as in step 5, draw another arc intersecting the arc in step 5 at R.
7. Join OR and produce it on any point C.
   Then, ∠AOC = 30°

(c) Step of Construction:

1. Draw a ray OA.
2. With O as centre and any convenient radius, draw an arc cutting at P.
3. With P as centre and same radius, draw an arc cutting the arc of step 2 at Q.
4. With Q as centre and the same radius as in steps 2 and 3, draw an arc cutting the arc drawn in step 3 at R.
5. With Q as centre and the same radius, draw an arc.
6. With R as centre and the same radius, draw an arc cutting the arc drawn in step 5 at B.
7. Join O to B and produce it to any point C.
   Then, ∠AOC =90°

(d) Steps of Construction:

1. Draw a ray OA.
2. With O as centre and any convenient radius, draw an arc cutting OA at P.
3. With P as centre and the same radius draw an arc cutting the first arc at Q.
4. With Q as centre and the same radius, draw an cutting the arc drawn in step 2 at R.
5. Join AR and produce it to any point C.
   Then, ∠AOC so obtained is of 120°.

(e) Steps of Construction:

1. Draw ∠AOC = 90° by following the steps given in part (iii) above,
2. Draw OE, the bisector of ∠AOC.
   
   Then, ∠AOD so obtained is the required angle of 45°.

(f) Steps of Construction:

1. Draw ∠AOC = 120°
   ∠AOB =150°.
2. Draw OD, bisector of ∠COB.
   Then,
   

Or

Steps of Construction:

 

1. Draw a line AB and mark a point O on it.
2. With centre O and any convenient radius draw a semi-circle cutting OA and OB at P and Q respectively.
3. With Q as centre and same radius, draw an arc cutting the semi-circle as R.
4. With R as centre and same radius, draw an arc cutting the semi-circle of step 2 at S.
5. With R as centre and same radius draw an arc.
6. With S as centre and same radius draw an arc cutting the arc drawn in step 5 at T. Join OT and produce it to D such that ∠BOD – ∠AOD = 90°.
7. Draw OE, the bisector of ∠AOD.
   

Question 6.
Draw an angle of measure 45° and bisect it.
Solution:
Steps of Construction:

1. Draw ∠AOB = 90° by the steps given in question 5 (c).
   
2. Draw OC, the bisector of ∠AOB. Then, ∠AOC = 45°.
3. Draw OD, the bisector of ∠AOC. Then, ∠AOD = ∠DOC.

Question 7.
Draw an angle of measure 135° and bisect it.
Solution:
Steps of Construction:

1. Draw ∠EOB = 135° by the steps given in question 5 (f).
   
2.  Draw OF, the bisector of ∠EOB.
   Then, ∠BOF = ∠FOE.

Question 8.
Draw an angle of 70°. Make a copy of it using only a straight edge and compasses.
Solution:
Steps of Construction:

1. Draw an angle 70° with protractor, i. e. ∠POQ = 70°
2. Draw a ray
3. Place the compasses at O and draw an arc to cut the ray of ∠POQ at I and M.
4. Use the same compasses, setting to draw an arc with A as centre, cutting AB at X.
5. Set your compasses setting to the length LM with the same radius.
   
6.  Place the compasses pointer at X and draw the arc to cut the arc drawn earlier at Y.
7. Join A7.
   Thus, ∠YAX =70°

Question 9.
Draw an angle of 40° copy its supplementary angle.
Solution:
Steps of construction:

1. Draw an angle of 40° with the help of protractor, naming ∠ AOB.
2. Draw a line PQ.
   
3. Take any point M on PQ.
4. Place the compasses at O and draw an arc to cut the rays of ∠ AOB at L and N.
5. Use the same compasses setting to draw an arc O as centre, cutting MQ at X.
6.  Place the compasses at X and draw the arc to cut the arc drawn earlier Y.
7. Join MY.
   Thus, ∠QMY = 40° and ∠PMY is supplementary of it.

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 Exz 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

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 14
Chapter Name	Practical Geometry
Exercise	Ex 14.5
Number of Questions Solved	9
Category	NCERT Solutions

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

Question 1.
Draw of length 7.3 cm and find its axis of symmetry.
Solution:
Steps of Construction:

1. Draw a line segment AB = 7.3 cm.
2. With centre A and radius > AB, draw arcs one on each side of AB.
3. With centre B and the same radius as before, draw arcs cutting the previously drawn arcs at C and D respectively.
4. Join CD intersecting AB at M. Then M bisects the line segment AB as shown.
   
   The line segment so obtained is the required axis of symmetry.

Question 2.
Draw a line segment of length 9.5 cm and construct its perpendicular bisector.
Solution:
Steps of Construction:

1. Draw a line segment AB = 9.5 cm.
2. With centre A and radius > AB, draw arcs one on each side of AB. . A
3. With centre B and the same radius as before, draw arcs cutting the previously drawn arcs at C and D respectively.
4. Join CD. Then the line segment CD is the required perpendicular bisector of AB.

Question 3.
Draw the perpendicular bisector of whose length is 10.3 cm.
(a) Take any point P on the bisector drawn. Examine whether PX = PY.
(b) If M is the mid-point of , what can you say about the lengths MX and XY?
Solution:
Steps of Construction:

1. Draw a line segment XY =10.3 cm.
   
2. With centre X and radius > XY, draw arcs one on each side of XY.
3. With centre Y and the same radius as before, draw arcs cutting the previously drawn arcs at A and B respectively.
4. oin .AB intersecting XY at M. Then, AB is the perpendicular bisector of XY.
   (a) Mark any point P on AB, the perpendicular bisector. On measuring, we find that PX = PY.
   (b) Since M is the mid-point of the segment XY. Therefore,
   MX = XY = x 10.3 cm
   = 5.15 cm

Question 4.
Draw a line segment of length 12.8 cm. Using compasses, divide it into four equal parts. Verify by actual measurement.
Solution:
Steps of Construction:

1. Draw a line segment AB = 12.8 cm.
2. With centre A and radius > AB, draw arcs one on each side of AB.
3. With centre B and the same radius as before, draw arcs cutting the previously drawn arcs at C and D respectively.
4. Join CD intersecting AB at M.
5. Further find the mid-points M₁ and M₂ of AM and MB respectively proceeding in the same way qs-before.
   ∴ AM₁ = M₁M = MM₂ = M₂B, On measuring, we find that each part = 3.2 cm.

Question 5.
With of length 6.1 cm as diameter draw a circle.
Solution:
Steps of Construction:

1. Draw a line segment PQ = 6.1 cm.
2. Bisect the segment PQ by drawing its perpendicular bisector. Let M be its mid-point.
3. M as centre and radius = MP draw a circle.
   The circle so obtained is the required circle.

Question 6.
Draw a circle with centre C and radius 3.4 cm. Draw any chord . Construct the perpendicular bisector of and examine if it passes through C.
Solution:
Steps of Construction:

1. Mark a point C on the plane of a paper.
2.  With C as centre and radius 3.4 cm, draw a circle.
3. Let AB be any chord to this circle.
4. Draw PQ, the perpendicular bisector of chord AB.
   Clearly, this perpendicular bisector passes through C, the centre of the circle.

Question 7.
Repeat Question 6, if AB happens to he a diameter.
Solution:
If AB happens to be a diameter then C will be the mid-point of the diameter AB.

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:
Steps of Construction:

1. Mark a point O on the plane of paper.
2. With O as centre, draw a circle of radius 4 cm.
3. Let AB and CD be any two chords of this circle.
4. Draw PQ and RS the perpendicular bisectors of chords AB and CD respectively.
   Clearly, these perpendicular bisectors pass through 0, 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 and . Let them meet at P. Is PA = PB?
Solution:
Steps of Construction:

1. Draw any angle XOY.
2. Take a point A on OX and a point B on OY such that OA = OB.
3. Draw CD and EF, the perpendicular bisectors of OA and OB respectively. Let them meet at P.
   On measuring, we find that PA = 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 Exz 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

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 14
Chapter Name	Practical Geometry
Exercise	Ex 14.4
Number of Questions Solved	3
Category	NCERT Solutions

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.4

Question 1.
Draw any line segment ,. Mark any point M on it. Through M, draw a perpendicular to ,. (use ruler and compasses)
Solution:
Steps of Construction:

1. Draw a line segment AB, and mark any point M on it.
2. With centre M and any radius, cut off MX and MY of equal lengths on both sides of M.
3. With centre X and any radius > MX, draw an arc.
4. With centre Y and the same radius draw another arc, cutting the previously drawn arc at P
5. Join MP
   Then, the segment PM so obtained is the required perpendicular.

Question 2.
Draw any line segment ,. Take any point R not on it. Through R, draw a perpendicular to ,. (use ruler and set-square)
Solution:
Steps of Construction:

1. Let PQ be the line and R is any point not lying on PQ.
2. Place the set-square so that the base AB of the set-square lies exactly on the line PQ.
3. Hold the set-square fixed and place a ruler so that its edge position lies along the side AC of the set-square.
4. Holding the ruler fixed, slide the set-square along the ruler till the point R coincides with the point B of the set-square.
5. Keeping the set-square fixed in this position, draw a line RT along the edge BC of the set-square through R.
   
   Thus, RT is the required perpendicular line to the line PQ passing through R.

Question 3.
Draw a line l and a point X on it. Through X, draw a line segment perpendicular to l. Now draw a perpendicular to XY at Y. (use ruler and compasses)
Solution:
Steps of Construction:

1. Draw a line l and mark any point X on it. ,
2. With centre X and any radius, cut off XA = XB on both sides of X.
3. With centre A and any radius > XA, draw an arc.
4. With centre B and the same radius draw another arc, cutting the previously drawn arc at Y.
5. Join XY. Then XY is perpendicular to line 1.
6. By proceeding as above draw a perpendicular YZ to XY.

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 Exz 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

Board	CBSE
Textbook	NCERT
Class	Class 6
Subject	Maths
Chapter	Chapter 14
Chapter Name	Practical Geometry
Exercise	Ex 14.3
Number of Questions Solved	2
Category	NCERT Solutions

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

Question 1.
Draw any line segment . Without measuring , construct a copy of .
Solution:
Steps of Construction:
Let PQ be the given line segment and l be a given line.

1. Mark a point 0 on the line l.
2. Open out the compasses and adjust so that the steel end is on A and the pencil end is A on B.
3. Transfer the compasses to the line l without disturbing their opening so that the steel end is on 0.
4. With the pencil end make a small stroke on the line l to cut it at the point P.
   Then, the fine segment OP so obtained = the given line segment AB.

Question 2.
Given some line segment , whose length you do not know, construct such that the length of is twice that of .
Solution:
Steps of Construction:
Let AB be the given line segment.

1. Draw any line l and make a point 0 on it.
2. Open out the compasses in such way that the steel end is on A and the pencil end is on B.
3. Transfer the compasses without disturbing their opening to the line l so that the steel end is on O.
4. With the pencil end make a small stroke on -the line l to cut it at the point P.
5. Repeat the steps 3 and 4 with same opening having P as the initial point and Q as the terminal point.
   Then, the segment OQ = OP + PQ = AB + AB= 2 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.