 Hi and welcome to the session. Let us discuss the following question. Question says, in figure 12.27, A, B and C, D are two diameters of a circle. Perpendicular to each other and O, D is the diameter of the smaller circle. If O, A is equal to 7 cm, find the area of the shaded region. This is the given figure 12.27. Let us now start with the solution. Now we are given that A, B and C, D are two diameters of a circle with center row and these diameters are perpendicular to each other and O, A is equal to 7 cm. That is radius of this circle with center row is equal to 7 cm. We are also given that O, D is the diameter of this smaller circle and we know O, D is equal to O, A is equal to 7 cm. We know O, A and O, D are radii of this circle. So if O, A is equal to 7 cm, O, D is also equal to 7 cm. Now we have to find area of this shaded region. Now first of all we will find out area of this smaller circle. We are given that O, A is equal to 7 cm. Now this implies O, D is equal to O, A is equal to 7 cm. We know radii of same circle are equal. Now diameter of smaller circle is equal to O, D is equal to 7 cm. So radius of smaller circle that is R is equal to 7 upon 2 cm. We know radius is equal to half of diameter. Now we know that area of circle is equal to pi R square where R is the radius of the circle. Now substituting corresponding value of R and pi in this formula we get area of smaller circle. It is equal to 22 upon 7 multiplied by square of 7 upon 2 cm square. Now it is further equal to 22 upon 7 multiplied by 7 upon 2 multiplied by 7 upon 2 cm square. Now 7 will get cancelled by 7 and we know 2 multiplied by 11 is equal to 22. So we get 77 upon 2 cm square is equal to area of smaller circle. Now we have to find area of these two shaded regions. This is a segment BC and this is a segment AC. Now clearly we can see this is a semi-circle ACB. If we subtract area of triangle ABC from area of semi-circle ACB we get area of these two segments. So first of all we will find area of triangle ABC. We know area of triangle is equal to half multiplied by base multiplied by height. Now we can write area of triangle ABC is equal to half multiplied by AB multiplied by OC. Here AB is the base and OC is the height. We know OC is the radius of this circle and radius of this circle is equal to 7 cm and AB is the diameter of this circle. So it is equal to 14 cm. We know diameter is equal to twice of the radius. So we will substitute 14 for AB and 7 for OC in this expression and we get half multiplied by 14 multiplied by 7 cm square. Now we will cancel common factor 2 from numerator and denominator both and we get area of triangle ABC is equal to 49 cm square. Now let us find out area of semi-circle ACB. Now we know area of semi-circle is equal to pi r square upon 2. Now we can write area of semi-circle ACB is equal to pi multiplied by square of 7 upon 2. We know radius of this semi-circle is equal to 7 cm. Now substituting 22 upon 7 for pi we get area of semi-circle ACB is equal to 22 upon 7 multiplied by 7 multiplied by 7 upon 2. 7 will get cancelled by 7 and here we know 2 multiplied by 11 is equal to 22. Now this is further equal to 77 cm square. Now we will find out area of minor segment BC and area of minor segment AC. So we can write area of minor segments BC and AC is equal to area of semi-circle ACB minus area of triangle ABC. Now substituting corresponding value of area of semi-circle ACB and area of triangle ABC in this expression we get area of minor segments BC and AC is equal to 77 minus 49 cm square. Now this is further equal to 28 cm square. Now we have to find area of the shaded region. Now we know area of the shaded region is equal to area of this smaller circle plus area of these two minor segments. So we can write area of shaded region is equal to area of smaller circle plus area of minor segments BC and AC. Now we know area of smaller circle is equal to 77 upon 2 cm square and area of minor segments BC and AC is equal to 28 cm square. So substituting corresponding values of area of smaller circle and area of minor segments BC and AC in this expression we get 77 upon 2 plus 28 cm square. Now adding these two terms by taking their LCM we get 77 plus 56 upon 2 cm square. Now this is further equal to 133 upon 2 cm square. Now converting this term into decimals we get 66.5 cm square. So we get area of shaded region is equal to 66.5 cm square. So our required answer is 66.5 cm square. This completes the session. Hope you understood the solution. Take care and keep smiling.