NCERT Solutions Class 6th Maths Chapter -14 Practical Geometry Exercise – 14.1

NCERT Solutions Class 6th Maths Chapter -14 Practical Geometry

TextbookNCERT
Class  6th
Subject Mathematics
Chapter 14th
Chapter NamePractical Geometry
CategoryClass 6th Mathematics 
Medium English
SourceLast Doubt

NCERT Solutions For Class 6 Maths Chapter 14 Exercise – 14.1 consists of exercise wise solved questions of Practical Geometry topic. These pdf can be downloaded easily and could be used for learning anywhere and at any time. To get an idea of the types of questions asked from Practical Geometry and the methods of solving these problems in the right way, students have to refer to NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry without fail.

NCERT Solutions Class 6th Maths Chapter -14 Practical Geometry

Chapter -14

Practical Geometry

Exercise – 14.1

Question 1. Draw a circle of radius 3.2 cm.

Solution:
Step I: Mark a point O as a centre.
Step II: Open the compass up to the given radius 3.2 cm.

Step III : Put the needle of the compass at the centre O.
Step IV : Holding the top of the compass take one full round with pencil. The figure thus obtained is the required circle of radius 3.2 cm.

Question 2. With the same centre O, Draw two circles of radius 4 cm and 2.5 cm.

Solution:
Step I : Take centre O and open the compass up to 4 cm.

Step II : Draw a circle keeping the needle fixed at O.
Step III : 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.

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?

Solution:

(i) Draw a circle with centre 0 with suitable radius.
(ii) AB and CD are any two diameters.
(iii) On joining the end points of the diameters, we get a quadrilateral ACBD.
(iv) We note that OA = OB = OC = OD [Same radius]
and AC = DB, AD = BC
∠A = ∠C = ∠B = ∠D = 90°
Thus ACBD is a rectangle.
Again if the diameters are perpendicular to each other then on measuring, we get
AC = DB = AD = BC
Thus, ACBD 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.

Solution:
Draw a circle with centre 0 and a suitable radius.
Here
(a) A is on the circle.
(b) B is in the interior of the circle.
(c) C is in the exterior of the circle.

Question 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 AB¯ and CD¯ are at right angles.

Solution:
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In the given figure two circles of equal radii intersect each other at C and D on measuring, we see that AB¯ and CD¯ intersect each other at right angles.

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