$ ext{Consider a unit circle with center at the origin. A square is inscribed in the circle, and a triangle is inscribed in the square. Determine the area of the triangle.}$

$\text{First, determine the side length of the square inscribed in the unit circle. The diagonal of the square is equal to the diameter of the circle, which is 2.}$

$\text{Using Pythagoras’ theorem, the side length } a \text{ of the square is given by:}$

$a\sqrt{2} = 2$

$a = \frac{2}{\sqrt{2}} = \sqrt{2}$

$\text{Next, consider an equilateral triangle inscribed in the square. The side length of the triangle is the same as the side length of the square, } a = \sqrt{2}.\text{ The area of an equilateral triangle with side length } a \text{ is given by: }$

$A = \frac{\sqrt{3}}{4} a^2$

$A = \frac{\sqrt{3}}{4} (\sqrt{2})^2 = \frac{\sqrt{3}}{4} \times 2 = \frac{\sqrt{3}}{2}$

$\text{Therefore, the area of the triangle is } \frac{\sqrt{3}}{2}. $

$ ext{First, calculate the side length of the inscribed square. The diagonal of this square is equal to the diameter of the circle, which is 2.}$

$ ext{Thus, the side length } a ext{ of the square is}$

$a sqrt{2} = 2$

$a = frac{2}{sqrt{2}} = sqrt{2}$

$ ext{Next, calculate the area of the triangle inscribed in the square. If it is an equilateral triangle, each side is } a. ext{ The area of an equilateral triangle is given by:}$

$A = frac{sqrt{3}}{4} a^2$

$A = frac{sqrt{3}}{4} (sqrt{2})^2 = frac{sqrt{3}}{4} imes 2 = frac{sqrt{3}}{2}$

$ ext{Thus, the area of the triangle is } frac{sqrt{3}}{2}. $

$ ext{The diagonal of the inscribed square is 2, hence}$

$ a = frac{2}{sqrt{2}} = sqrt{2}$

$ ext{The area of the equilateral triangle inscribed in the square is}$

$ A = frac{sqrt{3}}{4} (sqrt{2})^2 = frac{sqrt{3}}{2}$

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