NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry

Answer :

(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 for Class 6 Maths Chapter 14 Practical Geometry

Answer :

Take centre O and open the compass up to 4 cm.

Draw a circle keeping the needle fixed at O.

Take the same centre O and open the compass up to 2.5 cm, and draw another circle.

The figure shows the required two circles with the same centre.

Q.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 14 Practical Geometry

Answer :

(i) By joining the ends of two diameters, we get a rectangle. By measuring, we find AB = CD and BC = AD i.e., pairs

of opposite sides are equal and also \(\angle \) A = \(\angle \) B = \(\angle \) C = \(\angle \) D = 90 ,

, i.e. each angle is of 90 .

Henc,e it is a rectangle.

(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 = 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.

Q.5 Let A, B be the centres of the 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\(\overline{AB} \) and \(\overline{CD} \) are at right angles.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry

Answer :

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry

Answer :

(i) Draw a line '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 pointup 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. AB is the required line segment of length 5.6 cm.

Q.3 Construct \(\overline{AB} \) of length 7.8 cm. From this, cut off \(\overline{AC} \) of length 4.7 cm. Measure \(\overline{BC} .\)

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry

Answer :

(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.

(iv) By measuring BC , we find that BC = 3.1 cm

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

(Hint : Construct \(\overline{PX} \) such that the length of \(\overline{PX} \) = length of \(\overline{AB} \) then cut off \(\overline{XQ} \) such that \(\overline{XQ} \) also has the length of \(\overline{AB} \) .

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry

Answer :

Q.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 XY is equal to the difference between the length of \(\overline{AB} \) and \(\overline{CD} \) . Verify the measurement.

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry

Answer :

(i) Construct \(\overline{AB} \) = 7.3 cm and \(\overline{CD} \) = 3.4 cm.

(ii) Take a point P on the given line l.

(iii) Construct \(\overline{PR} \) such that \(\overline{PR} \) = \(\overline{AB} \) = 7.3 cm.

(IV) Construct \(\overline{RQ} \) = \(\overline{CD} \) = 3.4 cm such that PQ = \(\overline{AB} \) – \(\overline{CD} \) .

Verification : On measuring, we observe that \(\overline{PQ} \) = 3.9 cm = 7.3 cm - 3.4 cm.
= \(\overline{AB} \) – \(\overline{CD} \)

Thus, \(\overline{PQ} \) = \(\overline{AB} \) – \(\overline{CD} \).

Q.1 Draw any line segment \(\overline{PQ} \) . Without measuring \(\overline{PQ} \) , construct a copy of \(\overline{PQ} \) .

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry

Answer :

I : Draw \(\overline{PQ} \) of unknown length.

II : Draw a line l and mark a point A on it.

III: Open the compass equal to PQ.

IV : Place the needle of the compass at A and mark a point B on l.

Thus, \(\overline{AB} \) is a copy of \(\overline{PQ} \).

Q.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 \(\overline{AB} \) .

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry

Answer :

I : Draw \(\overline{AB} \)of any suitable length.

II : Place the needle of the compass at A and the other pencil end at B.

III : Draw a line l and take a point P on it.

IV: With the same opening of the compass, place the needle at P and mark another point Q on l.

Thus \(\overline{PQ} \) is the required line segment

whose length is twice the length of \(\overline{AB} \) i.e \(\overline{PQ} \) = 2\(\overline{AB} \) .

Q.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).

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry

Answer :

(i) Place a set-square on PQ such that one arm of
its right angle aligns along PQ

(ii) Place a ruler along the edge opposite to the right
angle of the set-square.

(iii) Hold the ruler fixed. Slide the set square along
the ruler till the point R touches the other arm of
the set square.

(iv) Join RM along the edge through R meeting PQ
at M. Then RM perpendicular PQ.

Q.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 length of MX and MY?

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry

Answer :

On measuring, \(\overline{MX} \)= \(\overline{MY} \) = 12 XY = 5.15 cm.

There are total 33 questions present in ncert solutions for class 6 maths chapter 14 practical geometry

There are total 0 long question/answers in ncert solutions for class 6 maths chapter 14 practical geometry

There are total 6 exercise present in ncert solutions for class 6 maths chapter 14 practical geometry