CBSE Class 6

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

These NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 Questions and Answers are prepared by our highly skilled subject experts.

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

Question 1.
Draw \(\overline{\mathrm{AB}}\) of length 7.3 cm and find its axis of symmetry.
Answer:
Axis of symmetry of line segment \(\overline{\mathrm{AB}}\) will be the perpendicular bisector of \(\overline{\mathrm{AB}}\). So, draw the perpendicular bisector of \(\overline{\mathrm{AB}}\).
Steps of construction:

  • Draw a line segment \(\overline{\mathrm{AB}}\) = 7.3 cm
  • Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C and D.
  • Join CD. Then CD is the axis of symmetry of the line segment AB.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 1

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

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

  • Draw a line segment \(\overline{\mathrm{AB}}\) = 9.5 cm
  • Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C and D.
  • Join CD. Then CD is the perpendicular bisector of \(\overline{\mathrm{AB}}\) .

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 2

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

  • Draw a line segment \(\overline{\mathrm{XY}}\) = 10.3 cm
  • Taking X and Y as centres and radius more than half of AB, draw two arcs which intersect each other at C and D.
  • Join CD. Then CD is the required perpendicular bisector of \(\overline{\mathrm{XY}}\) .

Now:
(a) Take any point P on the bisector drawn. With the help of divider we can check that \(\overline{\mathrm{PX}}\) = \(\overline{\mathrm{PY}}\)
(b) If M is the mid-point of \(\overline{\mathrm{PX}}\) and \(\overline{\mathrm{MX}}\) – 1/2\(\overline{\mathrm{XY}}\)
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 3

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

Question 4.
Draw a line segment of length 12.8 cm. Using compasses, divide it into four equal parts. Verify by actual measurement.
Answer:
Steps of construction:
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 4

  • Draw a line segment AB = 12.8 cm
  • Draw the perpendicular bisector of \(\overline{\mathrm{AB}}\) which cuts it at C. Thus, C is the midpoint of \(\overline{\mathrm{AB}}\).
  • Draw the perpendicular bisector of \(\overline{\mathrm{AC}}\) which cuts it at D. Thus D is the midpoint of.
  • Again, draw the perpendicular bisector of \(\overline{\mathrm{CB}}\) which cuts it at E. Thus, E is the mid-point of \(\overline{\mathrm{CB}}\).
  • Now, point C, D and E divide the line segment \(\overline{\mathrm{AB}}\) in the four equal parts.
  • By actual measurement, we find that \(\overline{\mathrm{AD}}=\overline{\mathrm{DC}}=\overline{\mathrm{CE}}=\overline{\mathrm{EB}}\) = 3.2 cm

Question 5.
With \(\overline{\mathrm{PQ}}\) of length 6.1 cm as diameter,
draw a circle.
Answer:
Steps of construction:

  • Draw a line segment \(\overline{\mathrm{PQ}}\) = 6.1 cm.
  • Draw the perpendicular bisector of PQ which cuts, it at O. Thus O is the mid-point of \(\overline{\mathrm{PQ}}\).
  • Taking O as centre and OP or OQ as radius draw a circle where diameter is the line segment \(\overline{\mathrm{PQ}}\).

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 5

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

Question 6.
Draw a circle with centre C and radius 3.4 cm. Draw any chord \(\overline{\mathrm{AB}}\). Construct the perpendicular bisector \(\overline{\mathrm{AB}}\) and examine if it passes through C.
Answer:
Steps of construction:

  • Draw a circle with centre C and radius 3.4 cm.
  • Draw any chord \(\overline{\mathrm{AB}}\).
  • Taking A and B as centres and radius more than half of \(\overline{\mathrm{AB}}\), draw two arcs which cut each other at P and Q.
  • Join PQ. Then PQ is the perpendicular bisector of \(\overline{\mathrm{AB}}\).
  • This perpendicular bisector of \(\overline{\mathrm{AB}}\) passes through the centre C of the circle.

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

Question 7.
Repeat Q6, if \(\overline{\mathrm{AB}}\) happens to be a diameter.
Answer:
Steps of construction:

  • Draw a circle with centre C and radius 3.4 cm.
  • Draw its diameter \(\overline{\mathrm{AB}}\).
  • Taking A and B as centres and radius more than half of it, draw two arcs which intersect each other at P and Q.
  • Join PQ. Then PQ is the perpendicular bisector of \(\overline{\mathrm{AB}}\) .
  • We observe that this perpendicular bisector of \(\overline{\mathrm{AB}}\) passes through the centre C of the circle.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 7

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

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?
Answer:
Steps of construction:

  • Draw the circle with O and radius 4 cm.
  • Draw any two chords \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) in this circle.
  • Taking A and B as centres and radius more than half AB, draw two arcs which intersect each other at E and F.
  • Join EF. Thus EF is the perpendicular bisector of chord \(\overline{\mathrm{CD}}\).
  • Similarly draw GH the perpendicular bisector of chord \(\overline{\mathrm{CD}}\).
  • These two perpendicular bisectors meet at O, the centre of the circle.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 8

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 \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\). Let them meet at P. Is PA = PB?
Answer:
Steps of construction:

  • Draw any angle with vertex O.
  • Take a point A on one of its arms and B on another such that OA = OB.
  • Draw perpendicular bisector of \(\overline{\mathrm{OA}}\) and OB.
  • Let them meet at P. Join PA and PB.
  • With the help of divider, we check that \(\overline{\mathrm{PA}}=\mathrm{PB}\)

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5 9

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

These NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.4 Questions and Answers are prepared by our highly skilled subject experts.

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

Question 1.
Draw any line segment \(\overline{\mathrm{AB}}\). Mark any point M on it. Through M, draw a perpendicular to \(\overline{\mathrm{AB}}\). (Use ruler and compasses)
Answer:
Steps of construction:

  • With M as centre and a convenient radius, draw an arc intersecting the line AB at two points C and B.
  • With C and D as centres and a radius greater than MC, draw two arcs, which cut each other at P.
  • Join PM. Then PM is perpendicular to AB through the point M.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.4 1

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

Question 2.
Draw any line segment \(\overline{\mathrm{PQ}}\). Take any point R not on it. Through R, draw a perpendicular to \(\overline{\mathrm{PQ}}\). (Use ruler and set- square)
Answer:
Steps of construction:

  • Place a set-square on \(\overline{\mathrm{PQ}}\) such that one arm of its right angle aligns along \(\overline{\mathrm{PQ}}\).
  • Place a ruler along the edge opposite to the right angle of the set-square.
  • Hold the ruler fixed. Slide the set square along the ruler till the point R touches the other arm of the set square.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.4 2

  • Join RM along the edge through R meeting \(\overline{\mathrm{PQ}}\) at M. Then RM ⊥ PQ.

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

Question 3.
Draw a line 1 and a point X on it. Through X,draw a line segment \(\overline{\mathrm{XY}}\) perpendicular to 1. Now draw a perpendicular to \(\overline{\mathrm{XY}}\) to Y. (use ruler and compasses)
Answer:
Steps of construction:

  • Draw a line ’l’ and take point X on it.
  • With X as centre and a convenient radius, draw an arc intersecting the line ’l’ at two points A and B.
  • With A and B as centres and a radius greater than XA, draw two arcs, which cut each other at C.
  • Join AC and produce it to Y. Then XY is perpendicular to ’l’
  • With D as centre and a convenient radius, draw an arc intersecting XY at two points C and D.
  • With C and D as centres and radius greater than YD, draw two arcs which cut each other at F.
  • Join YF, then YF is perpendicular to XY at Y.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.4 3

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

These NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.3 Questions and Answers are prepared by our highly skilled subject experts.

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

Question 1.
Draw any line segment \(\overline{\mathrm{PQ}} .\) Without measuring \(\overline{\mathrm{PQ}} \), construct a copy of \(\overline{\mathrm{PQ}} .\) Answer:
Steps of construction:
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.3 1

  • Given \(\overline{\mathrm{PQ}}\) whose length is not known.
  • Fix the compasses pointer on P and the pencil end on Q. The opening of the instrument now gives the length of \(\overline{\mathrm{PQ}}\).
  • Draw any line ‘l’. Choose a point A on ‘l’’. Without changing the compasses setting, place the pointer on A.
  • Draw an arc that cuts ‘l’ at a point, say B.
    Hence, \(\overline{\mathrm{AB}}\) is the copy of \(\overline{\mathrm{PQ}}\) .

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

Question 2.
Given some line segment \(\overline{\mathrm{AB}}\) , whose length you do not know, construct \(\overline{\mathrm{PQ}}\) such that the length of \(\overline{\mathrm{PQ}}\) is twice that of \(\overline{\mathrm{AB}}\).
Answer:
Steps of construction:
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.3 2

  • Given \(\overline{\mathrm{AB}}\) whose length is not known.
  • Fix the compasses pointer on A and the pencil end on B. The opening of the instrument now gives the length of \(\overline{\mathrm{AB}}\).
  • Draw any line ‘l’. Choose a point P on ‘l’. Without changing the compasses setting, place the pointer on Q.
  • Draw an arc that cuts ‘l’ at a point R.
  • Now place the pointer on R and without changing the compasses setting, draw another arc that cuts ‘l’ at a point Q.
    Hence, \(\overline{\mathrm{PQ}}\) is the required line segment whose length is twice that of AB.

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

These NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.2 Questions and Answers are prepared by our highly skilled subject experts.

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

Question 1.
Draw a line segment of length 7.3 cm, using a ruler.
Answer:
Steps of construction:
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.2 1

  • Place the zero mark of the ruler at a point A.
  • Mark a point B at a distance of 7.3 cm from A.
  • Join AB.
    Hence, \(\overline{\mathrm{AB}}\) is the required line segment of length 7.3 cm.

Question 2.
Construct a line segment of length 5.6 cm using ruler and compasses.
Answer:
Steps of construction:
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.2 2

  • Draw a line ’l’. 1 Mark a point A on this line.
  • Place the compasses pointer on zero mark of the ruler. Open it to place the pencil point up to 5.6 cm mark.
  • Without changing the opening of the compasses. Place the pointer on A and cut an arc ’l’ at B.
    \(\overline{\mathrm{AB}}\) is the required line segment of length 5.6 cm.

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

Question 3.
Construct \(\overline{\mathrm{AB}}\) of length 7.8 cm. From this, cut off \(\overline{\mathrm{AC}}\) of length 4.7 cm. Measure \(\overline{\mathrm{BC}}\).
Answer:
Steps of construction:
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.2 3

  • Place the zero mark of the ruler at A.
  • Mark a point B at a distance 7.8 cm from A.
  • Again, mark a point C at a distance 4.7 from A.
    Hence, by measuring \(\overline{\mathrm{BC}}\), we find that BC = 3.1 cm.

Question 4.
Given \(\overline{\mathrm{AB}}\) of length 3.9 cm, construct \(\overline{\mathrm{PQ}}\) such that the length of \(\overline{\mathrm{PQ}}\) is twice that of \(\overline{\mathrm{AB}}\). Verify by measurement.
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.2 4
(Hint: construct \(\overline{\mathrm{PX}}\) such that length of \(\overline{\mathrm{PX}}\) = length of \(\overline{\mathrm{AB}}\); then cut off \(\overline{\mathrm{XQ}}\) such that \(\overline{\mathrm{XQ}}\) also has the length of \(\overline{\mathrm{AB}}\).)
Answer:
Steps of construction:

  • Draw a line ’l’.
  • Construct \(\overline{\mathrm{PX}}\) such that length of \(\overline{\mathrm{PX}}\) = length of \(\overline{\mathrm{AB}}\)
  • Then cut of \(\overline{\mathrm{XQ}}\) such that \(\overline{\mathrm{XQ}}\) also has the length of \(\overline{\mathrm{AB}}\).
    NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.2 5
  • Thus the length of \(\overline{\mathrm{PX}}\) and the length of \(\overline{\mathrm{XQ}}\) added together make twice the length of \(\overline{\mathrm{AB}}\)

Verification:
Hence, by measurement we find that PQ = 7.8 cm = 3.9 cm + 3.9 cm =
\(\overline{\mathrm{AB}}+\overline{\mathrm{AB}}\) 2 × \(\overline{\mathrm{AB}}\)

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

Question 5.
Given AB of length 7.3 cm and \(\overline{\mathrm{CD}}\) of length 3.4 cm, construct a line segment \(\overline{\mathrm{XY}}\) such that the length of \(\overline{\mathrm{XY}}\) is equal to the difference between the lengths of \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) Verify by measurement.
Answer:
Steps of construction:

  • Draw a line ‘l’ and take a point X on it.
  • Construct \(\overline{\mathrm{XZ}}\) such that length \(\overline{\mathrm{XZ}}\) = length of \(\overline{\mathrm{AB}}\) = 7.3 cm
  • Then cut off \(\overline{\mathrm{ZY}}\) = length of \(\overline{\mathrm{CD}}\) = 3.4 cm
  • Thus the length of \(\overline{\mathrm{XY}}\) = length of \(\overline{\mathrm{AB}}\) – length of \(\overline{\mathrm{CD}}\)
    NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.2 6
    Verification:
    Hence, by measurement we find that length of \(\overline{\mathrm{XY}}\) = 3.9 cm = 73. cm -3.4 cm = \(\overline{\mathrm{AB}}-\overline{\mathrm{CD}}\)

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NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1

These NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Exercise 14.1

Question 1.
Draw a circle of radius 3.2 cm.
Answer:
NCERT Solutions for Class 6 Maths Chapter 1 Number Systems Ex 14.1 1
Steps of construction:
(a) Open the compass for the required radius of 3.2 cm.
(b) Make a point with a sharp pencil where we want the centre of circle to be.
(c) Name it O.
(d) Place the pointer of compasses on O.
(e) Turn the compasses slowly to draw the circle.
Hence, it is the required circle.

Question 2.
With the same centre O, draw two circles of radii 4 cm and 2.5 cm.
Answer:
NCERT Solutions for Class 6 Maths Chapter 1 Number Systems Ex 14.1 2
Steps of construction:
(a) Marks a point ‘O’ with a sharp pencil where we want the centre of the circle.
(b) Open the compasses 4 cm.
(c) Place the pointer of the compasses on O.
(d) Turn the compasses slowly to draw the circle.
(e) Again open the compasses 2.5 cm and place the pointer of the compasses on O.
(f) Turn the compasses slowly to draw the second circle.
Hence, it is the required figure.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1

Question 3.
Draw a circle and any two of its diameters. If you join the ends of these diameters, what is the figure obtained? What figure is obtained if the diameters are perpendicular to each other? How do you check your answer?
NCERT Solutions for Class 6 Maths Chapter 1 Number Systems Ex 14.1 3
Answer:
(i) By joining the ends of two diameters, we get a rectangle. By measuring, we find AB = CD = 3 cm, BC = AD = 2 cm, i.e., pairs of opposite sides are equal and also ∠A = ∠B = ∠C = ∠D = 90°,
i.e. each angle is of 90°.
Hence, it is a rectangle.
NCERT Solutions for Class 6 Maths Chapter 1 Number Systems Ex 14.1 4

(ii) If the diameters are perpendicular to each other, then by joining the ends of two diameters, we get a square. By measuring, we find that AB = BC = CD = DA = 2.5 cm, i.e., all four sides are equal. Also ∠A = ∠B = ∠C = ∠D = 90°, i.e. each angle is of 90°.
Hence, it is a square.

Question 4.
Draw any circle and mark points A, B and C such that:
(a) A is on the circle.
(b) B is in the interior of the circle.
(c) C is in the exterior of the circle.
NCERT Solutions for Class 6 Maths Chapter 1 Number Systems Ex 14.1 5
Answer:
(i) Mark a point ‘O’ with sharp pencil where we want centre of the circle,
(ii) Place the pointer of the compasses at ‘O’. Then move the compasses slowly to draw a circle.
(a) Point A is on the circle.
(b) Point B is in interior of the circle.
(c) Point C is in the exterior of the circle.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1

Question 5.
Let A, B be the centres of two circles of equal radii; draw them so that each one of them passes through the centre of the other. Let them intersect at C and D. Examine whether \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) are at right angles.
Answer:
Draw two circles of equal radii taking A and B as their centre such that one of them passes through the centre of the other. They intersect at C and D. Join AB and CD.
Yes, AB and CD intersect at right angle as ∠COB is 90°.
NCERT Solutions for Class 6 Maths Chapter 1 Number Systems Ex 14.1 6

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NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3

These NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Exercise 1.3

Question 1.
Estimate each of the following using general rule:
(a) 730 + 998
(b) 796-314
(c) 12,904 + 2,888
(d) 28,292-21,496
Make ten more such examples of addition, subtraction and estimation of their outcome.
Answer:
(a) 730 round off to 700
998 round off to 1000
Estimated sum = 1700

(b) 796 round off to 800
314 round off to 300
Estimated sum = 500

(c) 12904 round off to 13000
2888 round off to 3000
Estimated sum = 16000

(d) 28292 round off to 28000
21496 round off to 21000
Estimated difference = 7000

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3

Question 2.
Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens):
(a) 439 + 334 + 4317
(b) 1,08,737-47,599
(c) 8325 – 491
(d) 4,89,348-48,365
Make four more such examples
Answer:
(a) Rounding off to nearest hundreds
439 round off to 400
334 round off to 300
4,317 round off to 4,300
Estimated sum = 5,000
Again, Rounding off to nearest tens
439 → 440
334 → 330
4,317 → 4,320
∴ Closer estimate = 440 + 330 + 4,320 = 5,090

(b) Rounding off to Areas tens
= 440 + 330 + 4320 = 5090
108737 round off to 108700
47599 round off to 47600
Estimated difference = 61100
Again, Rounding off to nearest tens
1,08,734 → 1,08,730
47,599 → 47,600
∴ Closer estimate=1,08,730 – 47,600 = 61,130

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3

(c) Rounding off to nearest tens
= 108730-47600 = 61130
8325 round off to 8300
491 round off to 500
Estimated difference = 7800
Again, Rounding off to nearest tens
8,325 → 8,330
491 → 490
∴ Closer estimate = 8,330 – 490 = 7,800

(d) Rounding off to nearest tens
= 8330 – 490 = 7840
489348 round off to 489300
48365 round off to 48400
Estimated difference = 440900
Again, Rounding off to nearest tens
4,89,348→ 4,89,300
48,365 → 48,370
Closer estimate=4,89,350-48,370=4,40,980

Question 3.
Estimate the following products using general rule:
(a) 578 x 161
(b) 5281 x 3491
(c) 1291 x 592
(d) 9250 x 29
Make four more such examples
Answer:
(a) 578 x 161
578 round off to 600
161 round off to 200
The estimated product
= 600 x 200 = 1,20,000

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3

(b) 5281 x 3491
5281 round of to 5,000
3491 round off to 3,500
The estimated product
= 5,000 x 3,500 = 1,75,00,000

(c) 1291 x 592
1291 round off to 1300
592 round off to 600
The estimated product
= 1300 x 600 = 7,80,000

(d) 9250 x 29
9250 round off to 10,000
29 round off to 30
The estimated product
= 10,000 x 30 = 3,00,000

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