A square is inscribed in a circle of diameter [tex]$12 \sqrt{2}$[/tex] millimeters. What is the area of the shaded region?

Recall that in a [tex]45^{\circ}-45^{\circ}-90^{\circ}[/tex] triangle, if the legs each measure [tex]$x[tex]$[/tex] units, then the hypotenuse measures [tex]$[/tex]x \sqrt{2}$[/tex] units.

A. [tex](72 \pi - 144) \text{ mm}^2[/tex]
B. [tex](72 \pi - 72) \text{ mm}^2[/tex]
C. [tex](288 \pi - 288) \text{ mm}^2[/tex]
D. [tex](288 \pi - 144) \text{ mm}^2[/tex]



Answer :

Let's break down the problem step-by-step.

1. Identify the diameter of the circle and calculate the radius:
The diameter of the circle is given as \(12 \sqrt{2}\) millimeters.

Calculation of the radius:
[tex]\[ \text{Radius} = \frac{\text{Diameter}}{2} = \frac{12 \sqrt{2}}{2} = 6 \sqrt{2} \text{ millimeters} \][/tex]

2. Calculate the area of the circle:
The formula for the area of a circle is \(A = \pi r^2\), where \(r\) is the radius.

Calculation of the area:
[tex]\[ \text{Area of the circle} = \pi (6 \sqrt{2})^2 = \pi \times 72 = 72\pi \text{ square millimeters} \][/tex]

3. Determine the side length of the inscribed square:
In an inscribed square, the diagonal of the square is equal to the diameter of the circle. For a square, \(s \sqrt{2}\) (the diagonal) can be equated to the given diameter.

Calculation of the side length:
[tex]\[ \text{Side length} = \frac{\text{Diagonal}}{\sqrt{2}} = \frac{12 \sqrt{2}}{\sqrt{2}} = 12 \text{ millimeters} \][/tex]

4. Calculate the area of the square:
The area of a square is given by \(A = s^2\), where \(s\) is the side length.

Calculation of the area:
[tex]\[ \text{Area of the square} = 12^2 = 144 \text{ square millimeters} \][/tex]

5. Calculate the area of the shaded region:
The shaded region is the area of the circle minus the area of the inscribed square.

Calculation of the shaded area:
[tex]\[ \text{Area of the shaded region} = \text{Area of the circle} - \text{Area of the square} = 72\pi - 144 \text{ square millimeters} \][/tex]

Therefore, the area of the shaded region is \((72\pi - 144) \text{ mm}^2\). The correct answer choice is:

[tex]\[ \boxed{(72\pi - 144) \text{ mm}^2} \][/tex]