A rectangle ABCD is inscribed in a circle of radius 'a'. Find the 'area of the rectangle' function and state the domain.

To find the area of the rectangle, we first need to determine the dimensions of the rectangle.

Let's consider the diagonal of the rectangle, BD. This diagonal is also the diameter of the circle, which we know is equal to 2a.

The diagonal BD of the rectangle divides it into two congruent right triangles, ABD and BCD. Each right triangle has legs of length a and a hypotenuse of length 2a.

Using the Pythagorean theorem, we can find the length of either of the legs of the right triangle:

a^2 + a^2 = (2a)^2
2a^2 = 4a^2
2a^2 - 4a^2 = 0
-2a^2 = 0
a^2 = 0

Taking the square root of both sides, we find that a = 0.

Since the length of a side of the rectangle cannot be zero, it means that the given information is not valid. Hence, a rectangle cannot be inscribed in a circle of radius 'a'.

As for the domain, typically the domain for the area of a rectangle is all real numbers, but in this case, since a rectangle cannot be inscribed in a circle of radius 'a', the domain is empty or null.

To find the area of the rectangle, we first need to determine the dimensions of the rectangle. Let's consider the given information.

We know that the rectangle ABCD is inscribed in a circle of radius 'a'. In an inscribed rectangle, the diagonals are equal and bisect each other at right angles. Since the rectangle is inscribed in a circle, the diagonals of the rectangle are the diameters of the circle. Therefore, the length of each diagonal is 2a.

Let's assume the length of the rectangle is 'l' and the width is 'w'. As the diagonals of the rectangle are equal, we can apply the Pythagorean theorem to find the relationship between 'a', 'l', and 'w'.

Using the Pythagorean theorem, we can write:
(l/2)^2 + (w/2)^2 = a^2

Simplifying this equation gives us:
(l^2 + w^2) / 4 = a^2

Now, we can solve this equation for 'l^2 + w^2':
l^2 + w^2 = 4a^2

The rectangle's area, A, can be calculated by multiplying its length and width:
A = l * w

Substituting the value of 'l^2 + w^2' from the previous equation, we get:
A = 4a^2

The area of the rectangle is given by the function A = 4a^2, where 'a' represents the radius of the inscribed circle.

Now, let's consider the domain of the function. Since the radius of a circle cannot be negative, the domain of the function is a ≥ 0, or all non-negative values of 'a'.

Therefore, the area of the rectangle function is A = 4a^2, where 'a' belongs to the domain a ≥ 0.

clearly the domain is [0,2a] and the range is [0,2a^2]

The two sides AB and BC obey

AB^2 + BC^2 = 4a^2