 Hi and welcome to the session. Let us discuss the following question. The question says the triangular walls of a flyover have been used for advertisements. The sides of the walls are 122 meters, 22 meters and 120 meters. C figure 12.9. This is the figure 12.9. This is the triangular side walls of a flyover. The sides of these walls are 122 meters, 22 meters and 120 meters. The advertisements yielded an earning of Rs 5000 per meter square per year. A company hired one of its walls for three months. How much rent did it pay? Let us now begin with the solution. In this question, we will first find area of triangular wall by using Hyrol's formula. C equal to 122 meters, D is equal to 22 meters and C is equal to 120 meters. Let us first calculate the semi-perimeter. We know that semi-perimeter is equal to sum of lengths of all sides divided by 2. Now substitute the values of A, B and C. A is equal to 122 meters, D is equal to 22 meters and C is equal to 120 meters divided by 2. This is equal to C4 meters by 2 and this is equal to 132 meters. If S minus A, S is equal to 132 meters, A is equal to 122 meters, so this is equal to 10 meters. S minus B is equal to 132 meters minus 22 meters and this is equal to 110 meters. Z equal to 132 meters minus 120 meters and this is equal to 12 meters. Now area of S minus A into S minus B into S minus C. Substituting the values of S, S minus A, S minus B and S minus C, we get into 10 into 12. 110 is equal to 11 into 10 into 12 meters square and this is equal to 12 into 11 meters square and this is equal to 1320 meters square. The question it is given that advertisement charges are rupees from a company for 3 years. This is rupees 5000 for whole year therefore for 3 months it will charge rupees 5000 into 3 by 12 for 20 meters square. Rupees 16,50,000. This completes the session. Bye and take care.