 Hello and welcome to the session. In this session we are going to discuss the following question and the question says that construct a square within a circle of diameter 6 centimeters. Now let us start with its solution. Now in this question we have to construct a square inscribed in a circle with diameter that is 6 centimeters. To draw a square within a circle we follow some steps of construction which we will discuss one by one. First we shall draw a straight line with the help of a ruler and here we have drawn a straight line M. Now in the question it is given that we have to construct a square within a circle of diameter 6 centimeters. The diameter of the circle is given as 6 centimeters so its radius will be equal to half of the diameter that is equal to 6 by 2 centimeters which is equal to 3 centimeters that is we have got the radius of the circle as 3 centimeters so we open the compass to length equal to radius that is 3 centimeters with the help of a ruler. Now we take any point O on the line we place the compass at point O and draw the circle of radius 3 centimeters. The circle will intersect the line at two points. Let us label the two points as A and B that is with O S center and radius O A which is equal to 3 centimeters. We have drawn a circle which intersects the line at two points A and B. AB is the diameter of the circle which is equal to 6 centimeters. We need to construct a square within this circle so points A and B will be the two vertices of the square. Now we shall draw perpendicular bisector of diameter AB. For this we keep the compass at center A and open it to our length more than half of AB. We draw two arcs equidistant from line segment AB. Now again we keep the compass at point B and with the same radius as in previous step we draw two arcs on both sides of the line so that they intersect the first two arcs and let these points of intersection of the two arcs be point P and point P prime and now we join point P to the point P prime. Now we see that the perpendicular bisector intersects the circle at the point C and point D. Now we join A to C, C to B, B to D and D to A thus ACBD is the required square. This completes our session. Hope you enjoyed this session.