NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev

Mathematics (Maths) Class 6

Created by: Praveen Kumar

Class 6 : NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev

The document NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev is a part of the Class 6 Course Mathematics (Maths) Class 6.
All you need of Class 6 at this link: Class 6

Exercise 14.1
Question 1: Draw a circle of radius 3.2 cm.
Answer 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.

NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev

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 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 D.
(f) Turn the compasses slowly to draw the second circle.

NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev

Hence, it is the required figure.


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?
Answer 3:
(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 = 90o  i.e., each angle is of 90o.
Hence, it is a rectangle.

NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev

(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 = 90o, i.e., each angle is of 90o.
Hence, it is a square.

NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev


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.
Answer 4:
(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(Part - 1) - Practical Geometry Class 6 Notes | EduRev


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 NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRevare at right angles.
Answer 5: 
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 90o .

NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev


Exercise 14.2
Question 1: Draw a line segment of length 7.3 cm, using a ruler.
Answer 1:
Steps of construction:

NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev

(i) Place the zero mark of the ruler at a point A.
(ii) Mark a point B at a distance of 7.3 cm from A.
(iii) Join AB.
Hence, NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev 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 2: Steps of construction:

NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev

(i) Draw a line 'l'. l Mark a point A on this line.
(ii) Place the compasses pointer on zero mark of the ruler. Open it to place the pencil point up to 5.6 cm mark.
(iii) Without changing the opening of the compasses. Place the pointer on A and cut an arc 'l'  at B.NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev is the required line segment of length 5.6 cm.


Question 3:
Construct NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev of length 7.8 cm. From this, cut off  NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRevof length 4.7 cm. Measure NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev
Answer:
Steps of construction:

NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev(i) Place the zero mark of the ruler at A.
(ii) Mark a point B at a distance 7.8 cm from A.
(iii) Again, mark a point C at a distance 4.7 from A.
Hence, by measuring  NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev, we find that BC = 3.1 cm.

Question 4: Given NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev of length 3.9 cm, construct NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev such that the length of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev is twice that of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev. Verify by measurement.

NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev(Hint: construct NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev such that length of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev = length of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev ; then cut off NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev such that NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev also has the length of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev.)
Answer 4:
Steps of construction:
NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev

(i) Draw a line 'l'.
(ii) Construct NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev such that length of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev = length of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev
(iii) Then cut of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev such that NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev also has the length of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev.
(iv) Thus the length of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev and the length of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev added together make twice the length of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev
Verification: Hence, by measurement we find that PQ = 7.8 cm
= 3.9 cm + 3.9 cm
NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev

Question 5: Given NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev of length 7.3 cm and NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev of length 3.4 cm, construct a line segment NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev such that the length of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev is equal to the difference between the lengths of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev Verify by measurement.
Answer 5:
Steps of construction:
(i) Draw a line 'l' and take a point X on it.
(ii) Construct NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev such that length NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev = length of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev = 7.3 cm
(iii) Then cut off NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev = length of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev = 3.4 cm
(iv) Thus the length of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev = length of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev – length of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev

NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev

Verification:
Hence, by measurement we find that length of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev
= 3.9 cm
= 73. cm – 3.4 cm
= NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev - NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev

Exercise 14.3
Question 1:
Draw any line segment NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev. Without measuring NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev, construct a copy of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev.
Answer 1:
Steps of construction:

NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev

(i) Given NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev  whose length is not known.
(ii) Fix the compasses pointer on P and the pencil end on Q. The opening of the instrument now gives the length of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev .
(iii) Draw any line 'l'. Choose a point A on 'l'. Without changing the compasses setting, place the pointer on A.
(iv) Draw an arc that cuts 'l' at a point, say B.
Hence, NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev  is the copy of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev .

Question 2:
Given some line segment NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev , whose length you do not know, construct NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev such that the length of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev is twice that of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev.
Answer 2:
Steps of construction:

NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev

(i) Given NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev whose length is not known.
(ii) Fix the compasses pointer on A and the pencil end on B. The opening of the instrument now gives the length of NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev.
(iii) Draw any line 'l'. Choose a point P on 'l'. Without changing the compasses setting, place the pointer on Q.
(iv) Draw an arc that cuts 'l' at a point R.
(v) Now place the pointer on R and without changing the compasses setting, draw another arc that cuts 'l' at a point Q.
Hence, NCERT Solutions(Part - 1) - Practical Geometry Class 6 Notes | EduRev is the required line segment whose length is twice that of AB.

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