 Hi and welcome to the session. Let us discuss the following question. Question says, figure 12.26 depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60 meters and they are each 106 meters long. If the track is 10 meters wide, find the distance around the track along its inner edge and second part is find the area of the track. This is the given figure 12.26. First of all, let us understand that area of semicircle of radius r is equal to pi r square upon 2 and circumference of semicircle is equal to 2 pi r upon 2 which is further equal to pi r. Here r is the radius of the given semicircle. Also area of rectangle is equal to length into breadth. L represents the length and B represents the breadth. Now we will use these formulas as our key idea to solve the given question. Let us now start with the solution. First of all, we will join diameters of these two inner semicircles. Now we know distance between the two inner parallel line segments is equal to 60 meters. So this distance is 60 meters and we also know that distance between these two diameters is equal to 106 meters. We are given that length of these two line segments is equal to 106 meters. Now in first part of the question we have to find the distance around the track along its inner edge. So we have to find this distance. Now clearly we can see if we add circumference of these two semicircles and length of these two line segments we get the distance around the track along its inner edge. So first of all let us find out circumference of this semicircle. Now we know diameter of semicircle is equal to distance between the two inner parallel line segments which is equal to 60 meters. Now this implies radius of semicircle is equal to r is equal to 30 meters. We know radius is equal to half of diameter so radius is equal to half multiplied by 60 equal to 30 meters. Now from key idea we know circumference of semicircle is equal to pi r. Now substituting corresponding value of pi and r in this formula we get 22 upon 7 multiplied by 30 meters. Now this is further equal to 660 upon 7 meters. Now clearly we can see diameters of these two semicircles are equal and this implies their radii are also equal. So circumference of this circle is also equal to circumference of this circle. So we can write circumference of two semicircles is equal to 2 multiplied by 660 upon 7 meters. Now this is further equal to 1320 upon 7 meters. Now we will find out distance around the track along its inner edge. Now this is equal to circumference of two semicircles plus length of these two parallel line segments. This is equal to 1320 upon 7 plus 2 multiplied by 106 meters. We know circumference of two semicircles is equal to 1320 upon 7 meters and length of two inner parallel line segments is equal to 106 plus 106 meters. So we can write it as 2 multiplied by 106. Now simplifying we get 1320 plus 1484 upon 7 meters. Now this is further equal to 2804 upon 7 meters. So we get distance around the track along its inner edge is equal to 2804 upon 7 meters. Now this completes the first part of the question. Let us now start with the second part. Now we have to find area of the track. Now clearly we can see area of the track is given by area of this rectangle plus area enclosed between these two semicircles plus area of this rectangle plus area enclosed between these two semicircles. We are given that width of the track is 10 meters. Now let us find out area of this rectangle. We know area of rectangle is equal to length multiplied by breadth. Here L represents length and V represents breadth. Now in this rectangle length is equal to 106 meters and breadth is equal to 10 meters. Now area of given rectangle is equal to 106 multiplied by 10 meters square. Now this is further equal to 1060 meters square. Now clearly we can see length and breadth of this rectangle are equal to length and breadth of this rectangle. So area of this rectangle is equal to area of this rectangle. So we can write area of two given rectangles is equal to 2 multiplied by 1060 meters square. Now this is further equal to 2120 meters square. Now we will find out area enclosed between these two semicircles. Now we know radius of the inner semicircle is equal to 30 meters. So we can write radius of inner semicircle that is r is equal to 30 meters. Now radius of outer semicircle is equal to 30 plus 10. So we can write radius of outer semicircle is equal to r is equal to 30 plus 10 meters. Now this is further equal to 40 meters. Now we can find area of this semicircular region by subtracting area of inner semicircle from area of outer semicircle. So we can write area of semicircular region is equal to pi r square upon 2 minus pi r square upon 2. We know area of the outer semicircle is equal to pi r square upon 2 and area of the inner semicircle is pi r square upon 2 where capital R is the radius of the outer semicircle and small r is the radius of inner semicircle. Now this expression can be further written as pi upon 2 r square minus r square. Now this is equal to pi upon 2 multiplied by square of 40 minus square of 30 meters square. We know value of r is equal to 40 meters and value of small r is equal to 30 meters. So substituting these two values in this expression we get this expression. Now this is further equal to pi upon 2 multiplied by 1600 minus 900 meters square. We know square of 40 is equal to 1600 and square of 30 is 900. Now substituting 22 upon 7 for pi and subtracting these two terms we get 22 upon 7 multiplied by 2 multiplied by 700 meters square. Now we will cancel common factor 14 from numerator and denominator both and we get area of semicircular region is equal to 1100 meters square. Now clearly we can see radius of this semicircle is equal to radius of this semicircle and radius of this semicircle is equal to radius of this semicircle. So area enclosed between these two semicircles is equal to area enclosed between these two semicircles. Now we can write area of two semicircular regions is equal to 2 multiplied by 1100 meters square. We know area of this semicircular region is equal to 1100 meters square. So area of this semicircular region is also equal to 1100 meters square. So we get area of two semicircular regions is equal to 2 multiplied by 1100 meters square. Now this is further equal to 2200 meters square. Now clearly we can see area of the track is equal to area of these two semicircular regions plus area of these two rectangles. Now substituting corresponding values of area of two semicircular regions and area of two rectangles in this expression we get 2200 plus 2120 meters square is equal to area of the track. Now adding these two terms we get 4320 meters square. So we get area of the track is equal to 4320 meters square. Now this completes the second part of the question. This is our required answer. This completes the session. Hope you understood the solution. Take care and keep smiling.