NCERT Solutions Class 6 Mathematics Chapter 14 Practical Geometry

Ans : The required circle can be drawn as follows. 

step 1 

First, open the compasses for the required radius 3.2 cm. 

step 2 

Mark a point 'O' where we want the centre of the circle to be. 

step 3 

Place the pointer of compasses on O. 

Step 4 

Turn the compasses slowly to draw the circle. 


 

Ans : The required circle can be drawn as follows. 

Step 1 

First, open the compasses for the required radius 4 cm. 

step 2 

Mark a point 'O' where we want the centre of the circle to be. 

step 3 

Place the pointer of compasses on O. 

step 4 

Turn the compasses slowly to draw the circle. 

step 5 

Now, open the compasses for 2.5 cm. 

Step 6 

Again put the pointer of the compasses on point 'O' and turn the compasses slowly to draw the circle. 


 

Ans : A circle can be drawn of any convenient radius, also having its centre as O. Let AB and CD be two diameters of this circle. When we join the ends of these diameters, a quadrilateral ABCD is formed. 





As we know that the diameters of a circle are equal in length, therefore, the quadrilateral so formed will have its diagonals of equal length. 
Also, OA = OB = OC = OD radius r and if a quadrilateral has its diagonals of same 
length which are bisecting each other, then it will be a rectangle. 
Let DE and FG be two diameters of this circle such that these are perpendicular to each other. A quadrilateral is formed by joining the ends of these diameters. 





Here, OD = OE = OF = OG = radius r 
In this quadrilateral DEFG, the diagonals are equal and perpendicular to each other. 
Also, since these are bisecting each other, it will be a square. The length of the sides of the quadrilateral so formed can be measured to check our answers. 

Ans : A circle and three required points A, B, C can be drawn as follows.


 

Ans : Let us draw two circles of same radius which are passing through the centres of the other circle. 





Here, point A and B are the centres Of these circles and these circles are intersecting 
each other at point c and O. 

In quadrilateral ADBC, 

AD = AC (Radius of circle centered at A) 

BC = BD (Radius of circle centered at B) 

As radius of both circles are equal, therefore, AD = AC = BC = BD 

Hence, ADBC is a rhombus and i an rhombus, the diagonals bisect each other at \( 90 ^ { \circ } \). 

Hence, \( \overline { \mathrm { AB } } \)  and \( \overline { \mathrm { CD } } \) are at right angles.