 Hello and welcome to this session. In this session we discuss the following question that says construct a quadrilateral abcd given ab equal to 2.6 cm, bc equal to 4 cm, cd equal to 3.2 cm, ab equal to 2 cm and diagonal bd equal to 3.6 cm, marker point p on diagonal ac equidistant from b and c. Before we move on to the solution I will discuss one result which says that the locus of a point p distant to two given points is the right pi sector in the two points. This is the key idea that we use for this question. Let's proceed with the solution now. Quadrilateral abcd in which we have given ab equal to 2.6 cm, bc equal to 4 cm, cd equal to 3.2 cm, ab equal to 2 cm and the diagonal bd equal to 3.6 cm. For this we would follow some steps of construction. We draw a line segment equal to 2.6 cm. This is the line segment ab of measure 2.6 cm. Now in the next step we have drawn this arc taking a as the centre and radius equal to 2 cm. This is given that the diagonal bd is of measure 3.6 cm. So to locate the point d in the next step we have with b as the centre 3.6 cm the previous arc. This is the arc drawn taking b as the centre and radius equal to 3.6 cm and this arc in the 6th previous arc. Let this point of interception of the two arcs be point b, ab and bd triangle abd of the quadrilateral abcd where we have ab is of measure 2.6 cm then ab is of measure 2 cm and bd is of measure 3.6 cm. Next we are supposed to locate the point c of quadrilateral abcd. So for that our next step would be with the centre 2 cm as the radius. So we have drawn this arc taking b as the centre and radius equal to 4 cm. Next with b as the 3.2 cm we draw another arc so intersect to each other and let this point of interception of the two arcs be point c. In the next step we draw b and bc. So we get this quadrilateral abcd in which ab is of measure 2.6 cm, ab of 2 cm, bd, parallel bd of measures 3.6 cm, bc is of measure 3 cm and bc is of measure 3.2 cm. Abcd is the required quadrilateral. Mark a point p on the diagonal ac of the quadrilateral abcd such that this point p would be equal distance from the points b and c. Now first of all we draw an ac. So we have this diagonal ac and equal distance from two given points is the right line sector of this straight line which joins the two points. So now to mark a point p on the diagonal ac such that that point p is equal distance from the points b and c we have to draw the right line sector of the side bc. Such that it needs to be so in the next step we have draw the perpendicular bi-sector bc of quadrilateral perpendicular bi-sector of the side bc. Now this perpendicular bi-sector intersects the diagonal ac. So it is equal distance from the points b and c. That is the point on the diagonal ac on the points b and c. So required point. Hope you understood the solution of this question.