Given,

• A right angle triangle is inscribed inside the circle.

• The radius of the circle is given.

Let us consider,

• ‘r’ is the radius of the circle.

• ‘x’ and ‘y’ be the base and height of the right angle triangle.

• The hypotenuse of the ΔABC = AB² = AC² + BC²

AB = 2r, AC = y and BC = x

Hence,

4r² = x² + y²

y² = 4r² – x²

--- (1)

Now, Area of the ΔABC is

Now substituting (1) in the area of the triangle,

[Squaring both sides]

------ (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (2) with respect to x:

[Since and if u and v are two functions of x, then ]

------ (3)

To find the critical point, we need to equate equation (3) to zero.

[as the base of the triangle cannot be negative.]

Now to check if this critical point will determine the maximum area of the triangle, we need to check with second differential which needs to be negative.

Consider differentiating the equation (3) with x:

----- (4)

[Since ]

Now, consider the value of

As , so the function A is maximum at .

Now substituting in equation (1):

As , the base and height of the triangle are equal, which means that two sides of a right angled triangle are equal,

Hence the given triangle, which is inscribed in a circle, is an isosceles triangle with sides AC and BC equal.