 Hello and welcome to the session that has discussed the following question. It says prove that the ratio of the area of two similar triangles is equal to the ratio of the squares of their corresponding sites. Using the above, prove that the area of the equilateral triangle, described on the side of a right-angled isosceles triangle, is half the area of the equilateral triangle, described on its high coordinates. So let's now move on to the solution. Let us see what we have to prove. We have to prove that the ratio of the area of two similar triangles is equal to the ratio of the squares of their corresponding sites. So what is given to us, we are given that triangle ABC is similar to triangle PQR. What we have to prove that the ratio of the area of the triangles, that is, area of triangle ABC upon area of triangle PQR, is equal to the ratio of the square of their corresponding sites. That is, AB square upon PQ square is equal to BC square upon QR square is equal to RP square. This is what we have to prove. Let's now do some construction to prove this. We know that the area of the triangle is given by the formula 1 by 2 into base into triangle ABC upon area of triangle PQR is equal to the area of triangle ABC will be 1 by 2 into base of triangle ABC which is BC into the height of the triangle ABC which is AD upon area of triangle PQR which is 1 by 2 into base of triangle PQR which is QR into height of triangle PQR which is PS. So this is equal to BC upon QR into AD upon PS 1 by 2 gets cancelled with 1 by 2 in triangles ADB ASQ angle B is equal to angle Q. This is because triangle ABC is similar to triangle PQR and also angle ADB is equal to angle PSQ because each is 90 degrees. This implies triangle ADB is similar to triangle PSQ by ASimilarity implies that the ratio of their corresponding sites are equal. So we have AB upon PQ AB upon PS AB upon PQ is equal to BC upon QR because triangle ABC is similar to triangle PQR because if two triangles are similar then their ratio of their corresponding sites are equal. AB upon PQ is equal to BC upon QR now AB upon PQ is equal to AD upon PS but AB upon PQ is also equal to BC upon QR. So from 1 and 2 AD upon equal to BC upon QR name this as area of triangle area of triangle PQR is equal to BC upon QR into AD upon we have ABC upon area of triangle QR into BC upon PS. Since AD upon PS is equal to BC upon QR so this is equal to BC square upon QR square angle ABC is similar to triangle PQ as corresponding sites are equal AB upon QR RP. So this implies AB square upon PQ square is equal to BC square upon QR square is equal to CA square upon RP square name this as 4 and this is 5. So from 4 and 5 ABC upon area of triangle BC square upon QR square and RP square angle ABC upon area of triangle PQR is equal to BC square upon QR square and BC square upon QR square is equal to AB square upon PQ square. And it is also equal to CA square upon RP square the ratio of the area of two similar triangles is equal to the ratio of the squares of their corresponding sites. Hence we have through the second part of the question what it says it says that using the above prove that the area of the equilateral triangle described on the site of a right angled isosceles triangle is half the area of the equilateral triangle described on its hypotenuse. Let us understand this through a figure we have given around the base of the triangle will be equal since it is an isosceles triangle and one triangle is described on the ABC and it is an equilateral triangle on the base of the triangle ABC. So BC is equal to VD is equal to CD because this triangle is an equilateral triangle similarly one triangle is described on the hypotenuse of the triangle ABC. Angle BCD is an equilateral triangle and we know that in an equilateral triangle each angle is 60 degrees is similar to triangle BCD because each angle is 60 degrees so by a similarity both triangles are similar. So let us now write what we have just seen right angle BC is similar to triangle AC is equal to angle is equal to 60 degrees similarly angle C is equal to 60 degrees so by a similarity angle BCD is similar to triangle ACE. Also triangle ABC B square AB is equal to BC angle BCD area of triangle BCD upon area of the ratio of the squares of their corresponding sites. So area of triangle BCD upon area of triangle ACE is equal to BC square upon AC square AC square is equal to 2 BC square so we have BC square upon 2 BC square which is equal to 1 by 2. So this implies area of triangle BCD upon area of triangle equal to 1 by 2 and this implies area of triangle BCD equal to half the area of triangle ACE. And this is what we have to prove that the triangle described on the site of the right angled isosceles triangle has area half the area of the triangle described on the hypotenuse of this right angled isosceles triangle hence the result is proved. So this completes the question and the session why for now take care have a good day.