 Hello and welcome to the session. In this session we are going to discuss the following question and the question says that Vinny and Mark study in different schools. They both leave together for school from their house. Vinny goes 2 km south and 4 km west and reaches his school. And Mark goes 8 km east and 4 km north and reaches his school. Now using the given figure, find how far apart are the schools. Now we know that Pythagorean theorem states in our right angle triangle, sum of squares of length of x is equal to square of length of hypotenuse that is in our right angle triangle ABC if c represents the length of the hypotenuse and a and b are the length of the legs of the triangle then we have a square plus b square is equal to c square with this key idea let us proceed to the solution. Now in the question we are given that Vinny and Mark study in different schools and they both leave together for school from their house. Vinny goes 2 km south and 4 km west and reaches his school that is if c is the starting point for Vinny and then he goes 2 km south c distance of b and 4 km west c distance a b and reaches his school situated at point a and we can see that coordinates of point a are minus 4 minus 2. Now we are also given that Mark goes 8 km east and 4 km north and reaches his school that is Mark goes 8 km east c distance o c and 4 km north that is distance c d and reaches his school situated at point d and we can see that coordinates of point d are 8, 4 and we need to find the distance between the two schools that is we need to find the distance a d let us suppose a o v x and o d be equal to y that is a o will be equal to x km and o d will be equal to y km then we see that a d is equal to a o plus o d that is x plus y now we find the values of x and y now we can see from the graph that triangle a o b and triangle o d c are right angle triangles and we can apply Pythagoras theorem on both the triangles first we shall find the value of x and here from the graph we see that o b is equal to 2 units and a b is equal to 4 units and let us assume a b as a and o b as b and from the key idea we know that Pythagoras theorem states in a right angle triangle sum of squares of length of legs is equal to square of length of hypotenuse that is if c is the length of the hypotenuse and a and b are the length of the legs of the triangle then a square plus b square is equal to c square and here in triangle a o b x represents the length of the hypotenuse a o and a and b are the length of the legs of the triangle so by Pythagoras theorem we can write x square is equal to a square plus b square and we know that a square is equal to 4 square plus b is equal to 2 so b square will be equal to 2 square so we get x square is equal to 4 square plus 2 square that is x square is equal to 4 square that is 16 plus 2 square that is 4 so we have x square is equal to 16 plus 4 that is 20 now x square is equal to 20 implies that x is equal to square root of 20 now taking the positive square root of x that is positive square root of 20 as distance cannot be negative we get the value of x as 4.4 approximately so we get the value of x as 4.4 now in triangle odc we see that oc is equal to 8 units and cd is equal to 4 units now let us assume oc as a and cd as b that is oc we assume as a and cd as b and now we have to find the value of y since odc is a right angle triangle so we can apply Pythagoras theorem to find the value of y so by Pythagoras theorem we have y square is equal to a square plus b square that is y square is equal to a which is equal to 8 we have 8 square plus b square that is 4 square so we get y square is equal to 8 square that is 64 plus 4 square that is 16 so we get the value of y square as 64 plus 16 that is 80 now y square is equal to 80 implies that y is equal to square root of 80 now taking the positive square root of y we get the value of y as 8.9 approximately that is y is equal to 8.9 now we got the value of x as 4.4 and the value of y as 8.9 so ad will be given by x plus y ad is equal to x plus y that is 4.4 plus 8.9 which is equal to 13.3 approximately that is ad is equal to 13.3 units thus we can say that schools are 13.3 kilometers apart which is the required answer this completes our session hope you enjoyed this session