 Hello and welcome to the session. In this session, we discussed the following question which says, find the largest possible area of a right angle to triangle, whose hipop use is 5 centimeters. Let's move on to the solution now. Consider this right triangle ABC, where we have angle ABC is equal to 90 degrees, that is triangle ABC is right angle at V and we take let angle BAC is equal to theta. So we have AB is the base of the right angle triangle ABC, AC is the hipop use and BC is the perpendicular of the right triangle ABC. So now we have AB upon AC is equal to cos theta that is base upon hipop use is cos theta. Now I have given him the question that the hipop use is of measure 5 centimeters. So we have AC is equal to 5 centimeters. So this means that AB is equal to 5 cos theta. Now BC upon AC is equal to sin theta that is the perpendicular upon hipop use is equal to sin theta. Now again AC is equal to 5 centimeters. Therefore BC is equal to 5 sin theta. Now let's find out the area of triangle ABC given by say A. This is equal to half into base that is AB into the height or the perpendicular which is BC. So therefore A is equal to half into AB which is 5 cos theta into BC which is 5 sin theta that is we have A is equal to 25 upon 2 sin theta cos theta or we can say that A is equal to 25 upon 2 into 2 into 2 sin theta into cos theta that is we have multiplied the numerator and denominator by 2. So this becomes A is equal to 25 upon 4 sin 2 theta since we know that sin 2 theta is equal to 2 sin theta into cos theta. Thus we get A equal to 25 upon 4 sin 2 theta. Now we differentiate A with respect to theta that is DA by D theta is equal to 25 upon 4 into 2 into cos 2 theta. Now 2 times is 4 so we have DA by D theta is equal to 25 upon 2 cos 2 theta. Now we find out D to A by D theta 2. So this is equal to 25 upon 2 into 2 into minus sin 2 theta that is we have differentiated DA by D theta with respect to theta. Now this 2 cancels with this 2 so this is equal to minus 25 sin 2 theta this is D to A by D theta 2. Now to get the critical points we take DA by D theta equal to 0 that is 25 upon 2 cos 2 theta is equal to 0 which means that cos 2 theta is equal to 0 or we can say cos 2 theta is equal to cos pi by 2 since we know that cos pi by 2 is equal to 0 this gives us 2 theta is equal to pi by 2 that is we have theta is equal to pi by 4. Now we find out the value for D to A by D theta 2 at theta equal to pi by 4. So this would be equal to minus 25 into sin 2 theta and we put pi by 4 in place of theta so we get sin 2 into pi by 4 or we have D to A by D theta 2 at theta equal to pi by 4 is equal to minus 25 into sin pi by 2 and this is equal to minus 25 since we know that sin pi by 2 is equal to 1 thus we have D to A by D theta 2 at theta equal to pi by 4 is equal to minus 25. Now this minus 25 is less than 0 therefore D to A by D theta 2 at theta equal to pi by 4 is less than 0 therefore we say theta equal to pi by 4 is the point of maxima so the maximum area would be given by putting theta equal to pi by 4 in value for A which is 25 upon 4 sin 2 theta so we have maximum area would be equal to 25 upon 4 into sin 2 into pi by 4 so this would be equal to 25 upon 4 sin pi by 2 which is equal to 25 upon 4 since we know sin pi by 2 is equal to 1 thus we have the maximum area is equal to 25 upon 4 centimeter square or we can say that the largest possible area of a right-angled triangle whose hypotenuse is 5 centimeters is 25 upon 4 centimeter square so 25 upon 4 centimeter square is our final answer. This completes this session hope you have understood the solution for this question.