 Hello and welcome to the session I am Deepika here. Let's discuss a question which says a chord of a circle of radius 15 cm subtends an angle of 60 degree at the centre. Find the areas of the corresponding minor and major segments of this circle. Use pi is equal to 3.14 and root 3 is equal to 1.73. In this figure AB is a chord of the circle at the centre of associated region. ABB is called a minor segment of the circle to be which is unshaded is called the major segment of the circle. Now an ABB is equal to major segment AQB is equal to area of circle minor segment APB. So this is a 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. The question it is given that radius of the circle is 15 cm and according to our key idea the area of the minor segment APB is equal to APB angle OAB. So give this as number 1. Now first we will find the area of sector OABB equal to top on 316 into pi R square. So this is equal to now 3 times 60 degree here upon 360 into 3.14 into 15 into 15 cm square. So on cancellation we have this is equal to 1.57 into 75 cm square and this is equal to 117.75 cm square. Let us give this as number 2. Now we will find the area of triangle OAB. Let us draw the triangle separately. Now for finding the area of triangle OAB draw OM perpendicular to AB. Now for finding the area of triangle OAB draw perpendicular to AB. Now triangle OAM and triangle OBM we have is equal to OB, radii of the circle, prime is equal to OM. So by R it is congruency condition. Triangle OAM is congruent to triangle OBM and is the midpoint of AB. OAM equal to angle OBM is equal to half of 60 degree and that is equal to 30 degree. For area of triangle OAB we have the formula this is equal to half into base AB into height OM. Now we will find AB as well as OM. Let OM is equal to x cm. So in triangle OAM we have equal to cos. Now this is equal to x upon OAM is 15 cm. Now cos 30 degree is root 3 upon 2. This implies x is equal to upon 2 cm equal to sin 30 degree. 15 cm is equal to 1 by 2 as sin 30 degree is 1 by 2. Therefore OM is equal to 10 upon 2 cm. AB is equal to which is twice AM because M is the midpoint of AB is equal to 2 into 15 by 2 cm or this is equal to 15 cm. Therefore area of triangle OAB is equal to into AB this is equal to into AB is 15 cm into OM is our x cm. Here x is equal to 15 root 3 by 2 cm. So this is equal to 25 root 3 upon 4 cm square. Now use the value of root 3 which is given in the question 225 into 1.73 upon 4 cm square and this is equal to 389.25 upon 4 cm square or this is equal to 97.3125 cm square. So let us give this as number 3. On substituting the values from 2 and 3 in 1 we have area of minus segment ABB is equal to 117.75 minus 97.3125 cm square and this is equal to 20.4375 cm square. Now let us give this as number 4. Now according to our key idea area of major segment OAB is equal to area of circle minus area of minus segment ABB. Now let us give this as number 5. Now we will find the area of circle because we know the area of minus segment ABB. Now area of circle is equal to IR square and R is our 15 cm. So this is equal to 3.14 into 15 into 15 cm square and this is equal to 706.5 cm square. So let us give this as number 6. Now on substituting the values from 4 we get area of major segment ABB is equal to 706.5 minus 20.4375 cm square and this is equal to 86.0625 cm square. The answer for the ABB question is that area of the minus segment is 20.4375 cm square and area of major segment is 686.0625 cm square. I hope the solution is clear to you. Why I will check here.