 Hello and welcome to the session. In this session we discussed the following question which says, prove that the area of an equilateral triangle described on the hypotenuse of the right-angled isosceles triangle is twice the area of an equilateral triangle described on one of its sides. Before we move on to the solution, let's recall some results. First we have the AAA that is angle, angle, angle, similarity, criterion. According to this we have that if in two triangles corresponding angles are equal then their corresponding sides are in the same ratio and hence the two triangles are similar and the other result that we should know is that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. This is the key idea that we use for this question. Now let's proceed with the solution. Consider this figure, here we have triangle PQR is an isosceles right-angled triangle where angle PQR is equal to 90 degrees and PQ is equal to QR. Then we also have the triangle PRS is an equilateral triangle which is described on the hypotenuse of triangle PQR and its hypotenuse is PR and this triangle PRS is described on the hypotenuse of this isosceles right-angled triangle. Then we have this equilateral triangle PTQ that is triangle PTQ is an equilateral triangle which is described on the side PQ of the right-angled isosceles triangle PQR and we are supposed to prove that the area of the triangle PRS is equal to 2 times the area of the triangle PTQ. Now let's start with the proof. First of all we take let PQ equal to QR be equal to X units then in right triangle PQR PRS is equal to PQS plus QRS this is by the Pythagoras theorem and so from here we get PRS is equal to X square plus X square that is PRS is equal to 2X square or we can say that PR is equal to root 2 into X units that is the hypotenuse of the right-angled isosceles triangle PQR is equal to root 2X units. Now since triangle PRS is an equilateral triangle therefore PR is equal to RS is equal to SP is equal to root 2X units also the triangle PTQ is an equilateral triangle therefore PT is equal to TQ is equal to QP equal to X units as we have the PQ is equal to X units. Now as each of the triangles PRS PTQ are equilateral therefore each angle of both the triangles is 60 degrees. Now as in triangles PTQ and PRS corresponding angles are equal therefore we can say that triangle PRS is similar to the triangle PTQ by the AAA similarity criterion. As we know that the ratio of the areas of two similar triangles is equal to the square of the ratio of the corresponding sides so as both these triangles are similar therefore the ratio of their areas that is area of triangle PRS upon the area of the triangle PTQ is equal to the square of the ratio of the corresponding sides that is PR upon PT whole square or you can say that area of the triangle PRS upon the area of the triangle PTQ is equal to PR square upon PT square. Now we know that PR is equal to root 2X so this is equal to root 2X square upon PT which is X so this would be X square and so further we get area of the triangle PRS upon the area of the triangle PTQ is equal to 2X square upon X square X square X square cancels and so we get area of the triangle PRS is equal to 2X the area of the triangle PTQ so hence we have proved that the area of the triangle PRS is equal to 2X the area of the triangle PTQ so we were supposed to prove this and hence proved this completes the session hope you have understood the solution of this question.