Question

a square is inscribed in a circle of diameter 12√2 millimeters. what is the area of the shaded region? recall that in a 45°−45°−90° triangle, if the legs each measure x units, then the hypotenuse measures x√2 units. ○ (72π − 144) mm² ○ (72π − 72) mm² ○ (288π − 288) mm² ○ (288π − 144) mm²

Explanation:

Step1: Find the radius of the circle

The diameter of the circle \(d = 12\sqrt{2}\text{ mm}\), so the radius \(r=\frac{d}{2}=\frac{12\sqrt{2}}{2}=6\sqrt{2}\text{ mm}\).
The area of the circle \(A_{circle}=\pi r^{2}=\pi(6\sqrt{2})^{2}=72\pi\text{ mm}^2\)

Step2: Find the side - length of the square

Let the side - length of the square be \(a\). Since the diagonal of the square is equal to the diameter of the circle \(d = 12\sqrt{2}\text{ mm}\).
For a square of side - length \(a\), using the Pythagorean theorem \(d^{2}=a^{2}+a^{2}\) (because in a square with diagonal \(d\) and side \(a\), \(d^{2}=2a^{2}\)). Given \(d = 12\sqrt{2}\), then \((12\sqrt{2})^{2}=2a^{2}\), \(288 = 2a^{2}\), \(a^{2}=144\), and the area of the square \(A_{square}=a^{2}=144\text{ mm}^2\)

Step3: Find the area of the shaded region

The area of the shaded region \(A = A_{circle}-A_{square}\)
Substitute the values of \(A_{circle}\) and \(A_{square}\): \(A=(72\pi - 144)\text{ mm}^2\)

Answer:

A. \((72\pi - 144)\text{ mm}^2\)