 Hello and welcome to the session I am Deepika here. Let's discuss a question which says a chord of a circle of radius 12 centimeter subtends an angle of 120 degree at the center find the area of the corresponding segment of the circle. Use pi is equal to 3.14 and root 3 is equal to 1.73. Now in this figure AB is a chord of the circle and this region ABB is called a segment of the circle. Now we know that area of the segment ABB is equal to A of the sector OABB minus area of triangle OAB. So this is the key idea behind our question. We will take the help of this key idea to solve the above question. So let's start the solution. Now we are given a chord of a circle of radius 12 centimeter of 120 degree at the center and the area of the corresponding segment that is the area of the segment ABB. Now according to our key idea area of segment ABB is equal to area of sector OABB minus area of triangle OAB. So let us give this as number 1. First we will find the area of sector OABB. Now we know that theta is equal to theta upon 360 into pi R square theta is 120 degree. So this is equal to 120 upon 360 into 3.14 into 12 into 12 centimeter square upon cancellation we have this is equal to 3.14 into 48 centimeter square or this is equal to 150.72 centimeter square. Now let us give this as number 2. Now we will find the area of triangle OABB. Let us draw the triangle separately. Now for finding the area of triangle OAB draw M perpendicular to AB. In triangles BM we have OA is equal to OB, radii of the circle OM is equal to OM common. So by RHS congruence condition triangle OAM is congruent to triangle OBM hence is the midpoint of AB angle AOM is equal to angle BOM is equal to half of 120 degree which is equal to 60 degree. For the area of triangle OAB we have a formula that is equal to 1 by 2 into base AB into height OM. So for finding the area of triangle OAB we want OM as well as AB that OM is equal to x centimeter. So in triangle AM we have M upon OA is equal to now this is 60 degree and this is also 60 degree over equal to 1 by 2. Therefore x is equal to 12 upon 2 which is equal to 6 centimeter. Now also we have OA is equal to sine 16 upon 12 is equal to root 3 by 2 as sine 60 degree is root 3 by 2. Therefore AM is equal to 12 root 3 by 2 or this is equal to 6 root 3 centimeter. Hence AB is equal to which is twice AM because AM is the midpoint of AB is equal to 2 into 6 root 3 centimeter which is equal to 12 root 3 centimeter. Therefore area of triangle OAB is equal to into AB into OM. Now AB is equal to 12 root 3 centimeter into OM which is x centimeter and this is equal to 6 centimeter. Hence area of triangle OAB is equal to 36 root 3 centimeter square. Now the value of root 3 is 1.73. So this is equal to 36 into 1.73 centimeter square and this is equal to 62.28 centimeter square. Let us give this as number 3. Now on substituting the values from in the triangle 1 we have area of segment ABB is equal to 150.72 centimeter square minus 62.28 centimeter square and this is equal to 88.44 centimeter square. Hence the answer for the ABB question is 88.44 centimeter square. I hope the solution is clear to you. Bye and take care.