 Hello and welcome to the session. Let's work out the following problem. It says prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides. Use the above result to prove the following. A, B, C is an isosceles triangle right-angled at B. Similar triangles A, C, D and A, B are constructed on sides A, C and A, B. Find the ratio between the areas of triangle A, B, E and A, C, D. So let's now move on to the solution. And let's first proceed on with the first part of the question. Where we have to prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides of the two triangles. So let's first write what is given to us. We are given two similar triangles that is triangle A, B, C is similar to triangle P, Q, R. Let's now write what we have to prove. We have to prove that the ratio of the areas of two similar triangles that is area of triangle A, B, C upon area of triangle P, Q, R is equal to the ratio of the squares of their corresponding sides that is A, B square upon P, Q square is equal to V, C square upon Q, R square is equal to C, A square upon R, P square. And for this let's do some construction. Draw perpendicular A, D on B, C and P, S on Q, R. So we have drawn A, D perpendicular to B, C and P, S perpendicular to Q, R. Let's now start the proof. Now the area of triangle A, B, C upon area of triangle P, Q, R is equal to 1 by 2 base of the triangle A, B which is B, C into height of the triangle A, B, C which is A, D as A, D is perpendicular to B, C upon area of triangle P, Q, R is given by 1 by 2 into base of the triangle P, Q, R which is Q, R into the height of the triangle P, Q, R which is P, S as area of triangle is given by 1 by 2 into base into height or from this we have area of triangle A, B, C upon area of triangle P, Q, R is equal to B, C upon Q, R into A, D upon P, S cancelling 1 by 2. We have this. Now in triangles A, D, B and P, S, Q in triangle A, D, B and P, S, Q we have angle B is equal to angle Q. This angle is equal to this angle because we are given triangle A, B, C is similar to triangle P, Q, R. So, angle B is equal to angle Q and angle A, D, B is equal to angle P, S, Q because each is 90 degrees. Therefore, triangle A, D, B is similar to triangle P, S, Q using angle angle similarity. Now since triangle A, D, B is similar to triangle P, S, Q the ratio of the corresponding sides will be same. So, this implies A, D upon P, S is equal to A, B upon P, Q. Now again since triangle A, B, C is similar to triangle P, Q, R therefore again the ratio of the corresponding sides will be same that is A, B upon P, Q will be equal to BC upon Q, R. So, we have but A, B upon P, Q is equal to BC upon Q, R as triangle A, B, C is similar to triangle P, Q, R. Let us name this as 1, this as 2 and this as 3. Now we have seen that A, D upon P, S is equal to A, B upon P, Q but A, B upon P, Q is equal to BC upon Q, R. So, from 2 we have A, D upon P, S is equal to A, B upon, area upon P, S is equal to BC upon Q, R. This is by using 2 and 3. Now from 1 we have area of triangle A, B, C upon area of triangle P, Q, R is equal to BC upon Q, R into A, D upon P, S but A, D upon P, S is equal to BC upon Q, R. So, we have area of triangle A, BC upon area of triangle P, Q, R is equal to BC upon Q, R into BC upon Q, R. This is by using 1 and 4. So, from this we have area of triangle A, B, C upon area of triangle P, Q, R is equal to BC square upon Q, R square but since triangle A, B, C is similar to triangle P, Q, R. The ratio of the corresponding sides will be same that is A, B upon P, Q is equal to BC upon Q, R is equal to CA upon RP. So, this implies AB square upon P, Q square is equal to BC square upon Q, R square is equal to CA square upon RP square. Let's name this as 5 and this as 6. So, using 5 and 6 we have area of triangle ABC upon area of triangle P, Q, R is equal to AB square upon P, Q square is equal to BC square upon Q, R square is equal to CA square upon RP square using 5 and 6. Hence, we have proved that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. Now, in the next part we are given that ABC is an isosceles triangle right-angled at B. Similar triangles ACD and ABE are constructed on the sides AC and AB. Find the ratio between the areas of the triangle ABE and ACD. Now, since triangle ABE and triangle ACD are similar to triangle ABC and since triangle ABC is an isosceles triangle therefore triangle ABE will also be isosceles triangle and similarly triangle ACD will be an isosceles triangle. So, if ABE is X units and BC is X units therefore AE will be X units. Now, AC is the hypotenuse of the triangle ABC. So, we can find this by using Pythagoras theorem. So, let us now move on to the second part. Now, in right triangle ABC, AC is equal to under the root AB square plus BC square by Pythagoras theorem. Now, let AB is equal to BC is equal to X units since we are given that triangle ABC is an isosceles triangle therefore AC is equal to under the root X square plus X square which is equal to root 2X. So, AC is root 2X. Now, since triangle ACD is similar to triangle ABC and ABC is an isosceles triangle therefore triangle ACD will also be isosceles triangle and it is also a right-angled triangle. So, if this is root 2X, this will also be root 2X. Now, we are given that triangle ABE and triangle ACD similar there are similar triangles triangle ABE is similar to triangle ACD therefore by above theorem the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sites. So, we have area of triangle ABE upon area of triangle ACD is equal to AB square upon AC square that is square of the base of the triangle ABE and square of the base of the triangle ACD. Now, AB is X units and AC is root 2X units and we have to take the square. So, it is X square upon 2X square. So, this is equal to 1 by 2 thus the ratio of the areas of the triangle ABE and ACD is 1 is to 2 equal to 1 is to 2. So, this completes the question and the session. Bye for now. Take care. Have a good day.