 Hi and welcome to the session I am Shashi and I am going to help you with the following question. Question says calculate the area of the design region in figure 12.34 common between the two quadrants of circles 8 centimeter each. Now this is the given figure 12.34 let us now start with the solution. Now we are given that these are the two quadrants of the circles of radius 8 centimeter each. Now these radii form a parallelogram we know a quadrilateral in which both pairs of opposite sides are equal is a parallelogram and a parallelogram in which all sides are equal and one angle is right angle is a square. So this quadrilateral is a square. So let us name it as ABCD. So we can write suppose ABCD is a square formed by radii of circles of radius 8 centimeter each. Let us do one construction join diagonal of the square that is BD. Now we can find area of segment BD or we can say area of design on left hand side of BD by subtracting area of triangle ABD from area of quadrant ABD. So first of all let us write area of sector is equal to theta upon 360 multiplied by pi r square where theta is the angle of the sector and r is the radius of the circle. Now we know quadrant is nothing but a sector whose angle is 90 degrees. So we can write area of quadrant ABD is equal to 90 upon 360 multiplied by 22 upon 7 multiplied by square of 8 centimeter square. We know radii of both the circles is equal to 8 centimeters. Now this is further equal to 90 upon 360 multiplied by 22 upon 7 multiplied by 64 centimeter square. Simplifying we get 352 upon 7 centimeter square. Now we will find out area of triangle ABD. We know area of triangle is equal to half multiplied by base multiplied by height. Now in triangle ABD base and height both are equal to 8 centimeters. So we can write area of triangle ABD is equal to half multiplied by 8 multiplied by 8 centimeter square. Now simplifying we get 32 centimeter square. Now we know area of segment BD is equal to area of quadrant ABD minus area of triangle ABD. Now substituting corresponding values of area of quadrant ABD and area of triangle ABD in this expression we get 352 upon 7 minus 32 centimeter square. Now subtracting these two terms by taking their LCM we get 352 minus 224 upon 7 centimeter square. Now this is further equal to 128 upon 7 centimeter square. Now we know diagonal of a square divides it into two triangles of equal areas. So area of triangle ABD is equal to area of triangle CBD. Similarly we also know that area of quadrant ABD is equal to area of quadrant CBD. We know these are the quadrants of the circles of same radii that is 8 centimeters. So this implies if we subtract area of triangle CBD from area of quadrant CBD we get area of this segment BD is equal to 128 upon 7 centimeter square. So we can write area of designed region is equal to 128 upon 7 plus 128 upon 7 centimeter square. We know area of this segment is equal to area of this segment. So area of complete design is equal to 128 upon 7 plus 128 upon 7 centimeter square. Now this is further equal to 256 upon 7 centimeter square. So our required answer is area of designed region common between the two quadrants is equal to 256 upon 7 centimeter square. Now this completes the session. Hope you understood the solution. Take care and keep smiling.