 Hello and welcome to the session. In this session, we are going to discuss the following question. And the question says that an equilateral triangle has sites of length 10 centimeters. Find the length of one of its altitudes and hence its area. We know that in an isosceles of an equilateral triangle, the altitudes dissect the base of the triangle. That is, if we have an equilateral triangle ABC and AD is the altitude, then we can say that BD will be equal to DC. We also know that Pythagoras theorem states, in our right-angle triangle, with hypotenuse C, let A and B, we have C squared is equal to A squared plus B squared. That is, we can say that square of length of hypotenuse is equal to sum of squares of length of legs. With this key idea, let us proceed to the solution. In this question, we are given an equilateral triangle with site 10 centimeters. So, here we have the given equilateral triangle ABC with site 10 centimeters. From vertex A, we draw a line perpendicular to BC and it meets BC in D such that AD is the altitude of the triangle. And from the key idea, we know that in an isosceles or an equilateral triangle, the altitudes dissect the base of the triangle. In triangle ABC, which is an equilateral triangle, since altitude deflects the base of an equilateral triangle, therefore we say that CD is equal to BD which is equal to half of BC and BC is 10 centimeters. So, half of BC would be half into 10 centimeters and therefore we get 5 centimeters. So, we say CD is equal to BD is equal to 5 centimeters and we are given AC is equal to 10 centimeters. And we see that triangle ADC is a right angle triangle and we need to find the length of the altitude AD. From the key idea, we know that Pythagoras theorem states in our right angle triangle with hypotenuse C and like A and B, we have C squared is equal to A squared plus B squared. So, using Pythagoras theorem, in triangle ADC, we have square of the hypotenuse AC is equal to sum of squares of length of lengths of the triangle that is AD squared plus DC squared. Let the length of AD be equal to X centimeters. So, we have AC squared that is 10 squared is equal to AD squared that is X squared plus DC squared which is equal to 5 squared which implies that X squared is equal to 10 squared minus 5 squared. And therefore we get X squared is equal to 10 squared that is 100 minus 5 squared that is 25. So, we get the value of X squared as 75 which implies that X is equal to square root of 75. Now, taking the positive square root here as height cannot be negative, we get the value of X as 8.6 approximately which implies that AD is equal to 8.6 centimeters. Thus, we can say that the length of its altitude AD is equal to 8.6 centimeters approximately. We also need to find its area. Now, we shall find area of triangle ABC. We know that area of any triangle say ABC is given by 1 by 2 into base into height that is in triangle ABC we have 1 by 2 into base that is BC into height that is AD. So, we have 1 by 2 into BC that is 10 into AD which is equal to 8.6 which is equal to 5 into 8.6 that is 43 centimeters square approximately. Thus, we say that area of triangle ABC is given by 43 centimeters square approximately which is the required answer. This completes our session. Hope you enjoyed this session.