 Hello and welcome to the session. I am Asha and I am going to help you with the following construction which says, construct a triangle PQR in which QR is 6 centimeters, angle Q is 60 degrees and PR minus PQ is equal to 2 centimeters. Now to construct this triangle we shall work according to the construction 11.5 of your book which says to construct a triangle given its base, a base angle and the difference of other two sides. Suppose we have to construct triangle ABC, suppose this is triangle ABC, here we are given the base BC, a base angle and the difference of two sides. Now here we have two cases, case 1 is that site AB is greater than AC and the second case is that AB is less than AC. And here we have to construct a triangle PQR, suppose this is triangle PQR such that PR minus PQ is equal to 2 centimeters. So this implies that PQ is less than PR. So we will work according to the construction of case 2 of 11.5. Now let us draw a rough figure as to how we will proceed on with the construction. First we shall draw a ray Qx which will mark a point R on it such that QR is equal to 6 centimeters. Then we shall draw angle YQR into 30 degrees with the help of a compass. Then we shall extend it on the opposite side of Qy also. Then we will mark on the ray Qy extended on the opposite side such that AQ is equal to 2 centimeters. Then we will join A to R and in the next step we shall find the perpendicular bisector of AR, suppose this is the perpendicular bisector of AR, name it as PQ and the point where PQ intersects with Qy, let us name it as P. Then we shall join P with AR. Let the bisector be named as P dash Q dash. Since this point is P, we have to construct triangle PQR. So this is the required triangle PQR. If we are given that side PQ is less than PQR then we have followed case 2 of construction 11.5. Now let us write down the steps of construction and side by side we shall draw the triangle with the help of geometrical instruments and mark a point QR is equal to 6 centimeters. This we have drawn a ray Q and now let us mark a point R on it such that QR is equal to 6 centimeters. The second step is to construct angle YQR is equal to 30 degrees and that we will construct with the help of a compass the opposite of line segment QR. So first we will construct this angle and then we shall bisect it to get the 30 degrees angle. So this is angle 30 degrees and we have to extend it on the opposite side of QR. So let us name this ray as QZ and the next step we shall mark a point QZ AQ is equal to 2 centimeters. The next step we shall join this point is A we have taken as so we have to join A with R the next step then we have to draw the perpendicular bisector AR the ray QY at a point the perpendicular bisector of line AQ let us name it as MN such that it intersects the ray QY at a point. Now in the next step we shall join P to R so we have joined this PR is the required triangle. So hope you have understood it try to make it out with proper measurements take care and bye for now.