 Hello and welcome to the session. In this session we will discuss some constructions of triangles. First we should learn how to construct a triangle given its base, a base angle and sum of other two sides. In a triangle ABC we are given the base BC then base angle and will be of the triangle ABC and the sum AB plus AC of the other two sides of the triangle ABC. Our first type of construction would be we draw the base BC and at the point B we make an angle equal to the given angle. So as you can see we have drawn the base B and an angle at the point B. Then in the next step we cut a line segment BD which is equal to AB plus AC from the ray BX. This is the line segment BD such that it is equal to the sum of the other two sides of the triangle. In the next step we join DC and make an angle DC by such that it is equal to the angle BDC. So we get this angle BDC is equal to the angle DCY. Let's mark this point as A. So this ABC is the required triangle and this is the required triangle in which we are given the base BC, the base angle at B and the sum of the two sides of the triangle AB and AC. Thus we have ABC is the required triangle. Now we have one more method for the above construction in which the first two steps of the triangle AB and AC are the same. That is we need to perform step one and step two of the above method. Then in the next step we draw a perpendicular bisector of CD to intersect BD. So this PQ is the perpendicular bisector of CD. Let this perpendicular bisector of CD intersect BD at a point A. Now we join AC. So this ABC is the required triangle. We need to note an important point which says that construction of the triangle is not possible if the sum AB plus AC is less than equal to BC. Next we learn to construct a triangle given its base, a base angle and the difference of the other two sides. Given the base BC of the triangle ABC and the base of the base angle say we are given the angle B and the difference of the other two sides that is AB minus AC or we can say AC minus AB then let's see how we can construct a triangle ABC. There would be two cases in this case. Case one would be when we have AB is greater than AC that is we have AB minus AC is given. In the first step we draw the base BC and at point B we make an angle equal to the given angle. So this is the base B and we have drawn angle at the point B which is equal to the given angle. In the next step we cut the line segment BD equal to AB minus AC from the ray BX. This is the line segment BD equal to AB minus AC. Now in the next step we join DC minus AC and at the point B which is equal to AB minus AC and then we draw the perpendicular bisector of DC. This PQ is the perpendicular bisector of DC. Let this perpendicular bisector intersect the ray BX at a point A. Now we join AC then this ABC is the required triangle. Then we have one more case that is when we are given AB is less than AC that is AC minus AB is given to us. Then again a first step is the same that is we draw the base BC and at B we make an angle equal to the given angle. So we have drawn the base BC and angle at the point B. Then in the next step we cut a line segment BD which is equal to AC minus AB from the ray BX extended on opposite side of the line segment BC. This is the line segment BD such that BD is equal to AC minus AB. Now in the next step what we do is we join DC and then we draw the perpendicular bisector of DC. This PQ is the perpendicular bisector of DC such that this PQ intersects the ray BX at a point say A. Then this point A is joined to the point C. Hence this ABC is the required triangle. Next we discuss how we construct a triangle given its perimeter and its two base angles. We need to construct a triangle ABC in which its base angles angle B and angle C are given and also the perimeter of this triangle is given to us that is we are given AB plus BC plus CA. In this the first step of construction is we draw a line segment say XY such that XY is equal to AB plus BC plus CA that is XY is equal to the perimeter of the triangle. So this is the line segment XY such that its length is equal to the perimeter of the given triangle. In the next step we make angle LXY which is equal to the given base angle angle B and angle MYX which is equal to the given base angle angle C. So we have drawn angle LXY which is equal to the given base angle angle B and angle MYX which is equal to the given base angle angle C. Then in the next step we bisect angle LXY and angle MYX and let these bisect angles be the base angle angle C. So we have drawn the bisectors of the angle LXY and angle MYX let these bisectors intersect at a point say A. Then we draw perpendicular bisectors of AX, AY. So this PQ is the perpendicular bisector of AX and RS is the perpendicular bisector of AX and RS is the perpendicular bisector of AY. Now let the perpendicular bisector of AX that is PQ intersect XY at a point B and perpendicular bisector of AY that is RS intersect XY at a point C. Now we join AB and AC. Now this ABC is the requirement of triangular triangle in which the base angle B is given to us and the base angle C is also given to us and the perimeter of the triangle that is AB plus BC plus AC is given to us. So here we have ABC is the required triangle. This completes this session. Hope you have understood the constructions of triangles.