 Hi and welcome to the session. I am Shashi and I am going to help you with the following question. Question says, find the area of the shaded region in figure 12.19 if PQ is equal to 24 centimeters, PR is equal to 7 centimeters and O is the center of the circle. This is the figure 12.19. First of all let us understand that area of circle is equal to pi r square where r is the radius of circle and area of triangle is equal to half multiplied by base, multiplied by height. This is the key idea to solve the given question. Let us now start with the solution. Now we are given in circle with center rho PQ is equal to 24 centimeters and RP is equal to 7 centimeters. Now clearly we can see RQ is a line segment passing through center of the circle. So it is a diameter of the circle. So we can say RQ is diameter of circle. We know any line passing through the center of the circle having its end points on the circle is the diameter of the circle. Now angle RPQ is equal to 90 degrees. We know angle in a semicircle is a right angle. Now we get triangle RQP is a right triangle. Now we know PQ is equal to 24 centimeters and RP is equal to 7 centimeters. So by applying Pythagoras theorem we can find RQ in right triangle RQP. RQ square is equal to RP square plus PQ square. Now substituting corresponding values of PQ and RP in this expression we get square of RQ is equal to 7 square plus square of 24. Now this implies square of RQ is equal to 49 plus 576. We know square of 7 is 49 and square of 24 is 576. Now this further implies square of RQ is equal to 625. Now taking square root on both the sides we get RQ is equal to 25 centimeters. Now we know RQ is diameter of circle. So diameter that is RQ is equal to 25 centimeters. Now radius of circle that is R is equal to half of diameter which is equal to 25 upon 2 centimeters. From key idea we know area of circle is equal to pi R square. Now substituting corresponding value of pi and R in this formula we get 22 upon 7 multiplied by 25 upon 2 multiplied by 25 upon 2 centimeters square is equal to area of circle. We have to find area of this shaded region. Now if we subtract area of this triangle from area of semicircle then we get area of this shaded region. So first of all we will find out area of this semicircle. So area of semicircle is equal to half area of circle so it is equal to pi R square upon 2 that is multiplying this expression by half we get area of semicircle. So area of semicircle is equal to half multiplied by 22 upon 7 multiplied by 25 upon 2 multiplied by 25 upon 2 centimeters square. Now we will find out area of this triangle. From key idea we know area of triangle is equal to half multiplied by base multiplied by height. Now in this triangle base is equal to rp that is 7 centimeters and height is equal to pq that is 24 centimeters. So area of triangle is equal to half multiplied by 7 multiplied by 24 centimeters square. Now we know area of shaded region is equal to area of semicircle minus area of triangle rqp. Now substituting corresponding values of area of semicircle and area of triangle rqp in this expression we get area of shaded region is equal to half multiplied by 22 upon 7 multiplied by 25 upon 2 multiplied by 25 upon 2 minus half multiplied by 7 multiplied by 24 centimeter square. Now we will simplify this expression we know 2 multiplied by 11 is equal to 22. So we will cancel common factor 2 from numerator and denominator both. Now simplifying we get 625 multiplied by 11 upon 28. We know 25 multiplied by 25 is 625 and 7 multiplied by 4 is equal to 28. Now multiplying these three terms we get 84. So area of the shaded region is equal to 625 multiplied by 11 upon 28 minus 84. Now subtracting these two terms by taking their LCM we get 6875 minus 2352 upon 28 centimeter square. Is equal to area of shaded region. Now this is further equal to 4523 upon 28 centimeter square. So we get area of this shaded region is equal to 4523 upon 28 centimeter square. So this is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.