 Hi, and welcome to the session. Let us discuss the following question. The question says, proof that the ratio of the areas of similar triangles is equal to the ratio of the squares on their corresponding sites. Use the above theorem to solve the following. If triangle ABC is similar to triangle DEF such that area of triangle ABC is 9 centimeters square and the area of triangle DEF is 16 centimeters square and we see is equal to 2.1 centimeters find the length of DEF. It's now beginning with the solution. We are given triangle ABC and triangle DEF such that triangle ABC is similar to triangle DEF We have to prove that area of triangle ABC divided by area of triangle DEF is equal to AB square by DE square is equal to BC square by EF square is equal to CA square by DF square So these are the two triangles given to us and we have to prove that area of triangle ABC divided by area of triangle DEF is equal to AB square by DE square is equal to BC square by EF square is equal to CA square by EF square. Now in construction we will draw AD perpendicular to BC and DG perpendicular to EF. Area of triangle ABC is equal to half base that is BC into height that is AD divided by area of triangle DEF is equal to half DEF into DG. So this is equal to BC by EF into AD by DG. Now in triangle ADB and triangle DGE Angle B is equal to angle E because triangle ABC is similar to triangle DEF also angle ADB is equal to angle DGE because they both are equal to 90 degree. Therefore triangle ADB is similar to triangle BGE. This implies AD by DG is equal to AB by BGE. These two triangles are similar by AA similarity. Triangle ABC is similar to triangle DEF. Therefore AB by DE is equal to BC by EF. Let us name this as one and this as two. Now from one and two we can conclude that AB by DG is equal to BC by EF. Now area of triangle ABC divided of area of triangle DEF is equal to one by two into BC into AD divided by into EF into DG. Now substitute BC by EF in place of AD by DG. So we get BC square by EF square as triangle ABC is similar to triangle DEF. Therefore AB by DE is equal to BC by EF is equal to AC by BF. And hence area of triangle ABC divided by area of triangle PQR is equal to AB square by DE square is equal to BC square by EF square is equal to TF square by AC square. Now we will find the length of EF. We are given that area of triangle ABC is equal to nine centimeter square area of triangle DEF is equal to 16 centimeter square BC is equal to 2.1 centimeters. Now area of triangle ABC divided by area of triangle DEF is equal to BC square by EF square. Area of triangle ABC is 9 centimeters square area of triangle DEF is 60. BC square is 4.41 and we have to find EF square. Now this implies EF square is equal to 4.41 into 16 divided by 9. Now this is equal to 441 into 16 divided by 9 into 100. This is equal to 49 into 16 divided by 100. This is equal to 7 into 7 into 4 into 4 divided by 10 into 10. And this is equal to that is EF is equal to 28 by 10 that is 2.8 centimeters. Hence our required answer is 2.8 centimeters. So this is for the extension. Bye and take care.