 Hello and welcome to the session, let's discuss the following question. It says two equal sides of an isosceles triangle with fixed base B are decreasing at the rate of 3 centimeter per second. How fast is the area decreasing when the two sides equal sides are base? Now we are given that the sides of an isosceles triangle are decreasing at the rate of 3 centimeter per second and we have to find the corresponding degrees in the area when the two equal sides becomes equal to the base. So let's now move on to the solution. Equal sides equal to x centimeter that is AB is equal to AC is equal to x centimeter and BC be the base which is equal to B and let AL be the altitude. So AL would be minus B by 2 whole square that is B square by 4. Here we have used the Pythagoras theorem to the triangle ALC since ALC is the right angle triangle. So AL will be equal to the square root of the hypotenuse minus LC and the length of LC will be B by 2 since the whole length is B so half of it would be B by 2. Now area of triangle ABC will be given by 1 by 2 into base into height. Now base is B and height is AL which is square root of x square minus B square by 4 area. Let this be denoted by A B equal to 1 by 2 into B into the square root of x square minus B square by 4. Now we have to find the corresponding decrease in the area when the side two equal sides are increasing at the rate of 3 centimeter per second. We have to find the corresponding decrease when the two equal sides becomes equal to the base. So we have to find DA by DT so differentiating both sides we have 1 by 2 into B into 1 by 2 into x square minus B square by 4 whole to the power minus 1 by 2 into derivative of x square minus B square by 4. The derivative of x square would be 2x and derivative of B square by 4 would be 0 because it's a constant. So this is equal to since we are differentiating this with respect to T so we also need to differentiate x with respect to we have to multiply this with DX by DT. This is equal to 1 by 2 into B into 1 upon square root of x square minus B square by 4 into x into DX by DT. Now we are given that the sides are increasing at the rate of 3 centimeter per second so DX by DT is equal to 3. So here we substitute 3 in place of DX by DT so we have 1 by 2 into B into 1 upon square root of B square by 4 into x into DX by DT that is 3. Now we have to find DA by DT when x is equal to B so when x is equal to B DA by DT becomes 1 by 2 into B into 1 upon square root of B square minus B square by 4 into 3 into B. Again this is equal to 1 by 2 into B into 1 upon square root of 4 B square minus B square upon 2 square root of 4 is 2 into 3B. Again this is equal to 3B square here 2 gets cancelled with 2 3B square upon square root of 3B square this is further equal to 3B square upon under root 3 into B so this is equal to under the root 3B. DA by DT is equal to root 3B that is the area is decreasing at the rate of root 3B centimeter square per second so the answer is root 3 centimeter square per second. So this completes the question and the session by for now take care have a good day.