*According to the latest update on the CBSE Syllabus 2023-24, this chapter has been removed.
By solving the NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Exercise 14.1, students learn about various problems related to shapes and their construction, which are important from the exam point of view. Various instruments that are used in the construction of shapes and steps followed in the construction of a circle when its radius is known are the topics discussed in this exercise. For a better hold on these concepts, students can solve the exercise-wise problems using the NCERT Solutions Class 6 Maths designed by subject-matter experts, at BYJU’S.
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Exercise 14.1
Access NCERT Solutions for Class 6 Maths Chapter 14: Practical Geometry Exercise 14.1
1. Draw a circle of radius 3.2 cm.
Solutions:
The required circle can be drawn as follows:
Step 1: For the required radius of 3.2 cm, first open the compasses.
Step 2: For the centre of the circle, mark a point ‘O’.
Step 3: Place a pointer of compasses on ‘O’.
Step 4: Now, turn the compasses slowly to draw the required circle.
2. With the same centre O, draw two circles of radii 4 cm and 2.5 cm.
Solutions:
The required circle may be drawn as follows:
Step 1: For the required radius of 4 cm, first open the compasses.
Step 2: For the centre of the circle, mark a point ‘O’.
Step 3: Place a pointer of compasses on ‘O’.
Step 4: Turn the compasses slowly to draw the circle.
Step 5: Next, open the compasses for 2.5 cm.
Step 6: Again, place a pointer of compasses on ‘O’ and turn the compasses slowly to draw the circle.
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?
Solutions:
We will draw a circle having its centre ‘O’, also of any convenient radius. Let AB and CD be the two diameters of this circle. A quadrilateral is formed when we join the ends of these diameters.
We know that the diameter of a circle is equal in length, hence quadrilateral formed will be having its diagonals of equal length.
Also, OA = OB = OC = OD = radius r, and if a quadrilateral has its diagonals of the same length bisecting each other, it will be a rectangle.
Let DE and FG be the diameters of the circle such that both are perpendicular to each other. Now, we can find that a quadrilateral is formed by joining the ends of these diameters.
We can find that OD = OE = OF = OG = radius r.
In this quadrilateral DFEG, diagonals are equal and perpendicular to each other. Also, they are bisecting each other; hence it will be a square.
To check our answers, we can measure the length of the sides of the quadrilateral formed.
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.
Solutions:
We will draw a circle with three required points, A, B, and C, as follows:
5. Let A and 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 
and 
are at right angles.
Solutions:
Let us draw two circles having the same radius, which are passing through the centre of the other circle.
Here, points A and B are the centres of these circles, and these circles intersect each other at points C and D, respectively.
Now, in quadrilateral ADBC, we may observe that:
AD = AC [radius of circle centered at A]
BC = BD [radius of circle centered at B]
The radius of both circles is equal.
Therefore, AD = AC = BC = BD.
Hence quadrilateral ADBC is a rhombus, and in a rhombus, diagonals bisect each other at 900. Hence 
and 
are at right angles.
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