 Hi and welcome to the session. Let us discuss the following question. Question says, in figure 12.30, OACB is a quadrant of circle with center O and radius 3.5 cm. If OD is equal to 2 cm, find the area of the quadrant OACB and we have to find the area of the shaded region. This is the given figure 12.30. First of all let us understand that area of sector of a circle is equal to theta upon 316 multiplied by pi r square where theta is the angle of the sector in degrees and r is the radius of the circle. Now we will use this formula as our key idea to solve the given question. Let us now start with the solution. Now we are given that OACB is a quadrant of a circle with center O and it has radius equal to 3.5 cm. Now we are also given that OD is equal to 2 cm. OD is 2 cm. Now we have to find area of quadrant OACB. Now clearly we can see OACB is a sector of the circle and this arc AB subtends 90 degrees angle at the center. You know quadrant of a circle makes 90 degree angle at the center of the circle. So clearly we can see in sector OACB angle made by the sector at the center of the circle is 90 degrees and radius of the circle is 3.5 cm. And we know that area of sector is equal to theta upon 316 multiplied by pi r square where theta is the angle of the sector in degrees and r is the radius of the circle. Now substituting corresponding values of theta and r in this formula we can find area of sector OACB. We get area of sector OACB is equal to 90 upon 360 multiplied by pi multiplied by square of 3.5 cm square. Now substituting pi is equal to 22 upon 7 in this expression we get 90 upon 360 multiplied by 22 upon 7 multiplied by 3.5 multiplied by 3.5 cm square. Now simplifying this expression we get 1 upon 4 multiplied by 22 multiplied by 0.5 multiplied by 3.5 cm square. Now this expression can be further written as 1 upon 4 multiplied by 22 multiplied by 5 upon 10 multiplied by 35 upon 10 cm square. We know 2 multiplied by 2 is equal to 4 and 2 multiplied by 11 is equal to 22. Now 5 multiplied by 1 is 5 and 5 multiplied by 2 is equal to 10. Similarly 5 multiplied by 7 is equal to 35 and 5 multiplied by 2 is equal to 10. Now we get 77 upon 8 cm square is equal to area of sector OACB. Now this completes the first part of the question. Now we have to find area of this shaded region. We know area of this shaded region is equal to area of quadrant OACB minus area of triangle ODB. So we will find area of triangle ODB we know area of triangle is equal to half into base into height. Here base is OB that is 3.5 cm and height is OD that is 2 cm so area of triangle ODB is equal to half multiplied by 3.5 multiplied by 2 cm square. Now 2 and 2 will cancel each other and we get area of triangle ODB is equal to 3.5 cm square. We know area of shaded region is equal to area of quadrant OACB minus area of triangle ODB. Now substituting corresponding values of area of quadrant OACB and area of triangle ODB in this expression we get 77 upon 8 minus 3.5 cm square. Now subtracting these 2 terms by taking their LCM we get 77 minus 28 upon 8 cm square. Now simplifying further we get 49 upon 8 cm square is equal to area of shaded region. So we get area of shaded region is equal to 49 upon 8 cm square. Now this completes the second part of the given question. So the required answer is area of quadrant OACB is equal to 77 upon 8 cm square and area of shaded region is equal to 49 upon 8 cm square. So this completes the session. Hope you understood the solution. Take care and have a nice day.