 Hello friends, let's discuss the following question. It says, a path in the shape of a quadrilateral ABCD has angle C equal to 90 degrees. AB is 9 meter, BC is 12 meter, CD is 5 meter and AD is 8 meter. How much area does it occupy? So we have a path in the shape of quadrilateral ABCD, angle C is 90 degrees. We have to find the area of this path. To find the area of this path, we will divide the quadrilateral into triangles and then find the area of triangle ABD and area of triangle BCD and we will add the two areas to get the area of quadrilateral ABCD. And for the area of triangle, we will be using your own formula, which says that if we have a triangle, say ABC, its size AB and C, then S is given by A plus B plus C upon 2 and the area of triangle ABC is given by under the root of S into S minus A into S minus B into S minus C. So this knowledge will work as the idea for the question. Let us now proceed on with the solution. We first find the area of triangle BCD. So in triangle BCD, BC is 12 meter, CD is 5 meter and angle C is 90 degrees. To find the area of the triangle BCD, we need to know the length of BD, but it is the hypotenuse of triangle BCD. So the BD is equal to under the root of C square plus CD square which is equal to 12 square plus 5 square which is equal to 144 plus 25 which is equal to root of 169 and this is equal to 13. And the unit of length is given to be in meter. Now we can find the area of triangle BCD using Hiram's formula where S is A plus B plus C by 2 where A is 12, B is 5, C is 13 upon 2 which is equal to 15. Now we find area of triangle BCD which is given by under the root of 15 into 15 minus A which is 12, 15 minus B, 15 minus C which is 13. Now this is equal to 15 into 3 into 10 into 2 which is equal to 13 and unit of area is meter square. Now we are going to find area of triangle ABD so in triangle ABD, AB is 9 meter, BD is 13 meter and AD is 8 meter. Now we find area for that we first need to find S which is A plus B plus C by 2 is 9 plus 13 plus 8 upon 2 which is equal to 15. Now we find area of triangle ABD given by under the root of S which is 15 into S minus A into S minus B into S minus C which is equal to 15 into 6 into 2 into 7 and on finding the square root this will come out to be equal to 35.5 meter square approximately. Hence area quadrilateral ABCD is equal to area of triangle BCD plus area of triangle ABD. Now area of triangle BCD is 30 meter square and area of triangle ABD is 35.5 meter square and the sum is 65.5 meter square approximately. Hence the area of the park is 65.5 meter square approximately. So this completes the question hope you enjoyed this session goodbye and take care.