 Hi and welcome to our session. Let us discuss the following questions. The question says show that rectangle on maximum area that can be inscribed in a circle of radius r is a square of side root to r. Let's now begin with the solution. Now let a, b, c, d be a rectangle right in a circle of radius r. We have to show that rectangle of maximum area that can be inscribed in a circle of radius r is a square of side root to r. Now in right triangle a, b, c, theta is equal to a, b, pi, a, c. a, c is equal to 2 r. So this implies a, b is equal to 2 r to b, c, pi, c is equal to 2 r is equal to length of the rectangle that is a, b which is equal to 2 r cos theta into back of the rectangle that is b, c that is 2 r sin theta. And this is equal to 4 r square cos theta sin theta is equal to 2 r square sin 2 theta. Let us name this as equation up one. Now we will differentiate one with respect to theta. a by d theta is equal to a square cos 2 theta into 2 and this is equal to i d2 a by r to minus r square cos 0 to 0 is a square of side.