Question

a square is inscribed in a circle of diameter 12\sqrt{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\sqrt{2} units. (72\pi - 144) mm² (72\pi - 72) mm² (288\pi - 288) mm² (288\pi - 144) mm²

Explanation:

Step1: Find the radius of the circle

The diameter $d = 12\sqrt{2}$ mm, so the radius $r=\frac{d}{2}=\frac{12\sqrt{2}}{2}=6\sqrt{2}$ mm.

Step2: Calculate the area of the circle

The area formula of a circle is $A_{circle}=\pi r^{2}$. Substituting $r = 6\sqrt{2}$ mm, we get $A_{circle}=\pi\times(6\sqrt{2})^{2}=\pi\times72 = 72\pi$ $mm^{2}$.

Step3: Find the side - length of the square

The diameter of the circle is the diagonal of the square. In a square, if the side - length is $a$, and the diagonal $D$, for a square (which can be divided into two 45 - 45 - 90 triangles), $D = a\sqrt{2}$. Given $D=12\sqrt{2}$ mm, then $a\sqrt{2}=12\sqrt{2}$, so $a = 12$ mm.

Step4: Calculate the area of the square

The area formula of a square is $A_{square}=a^{2}$. Substituting $a = 12$ mm, we get $A_{square}=12^{2}=144$ $mm^{2}$.

Step5: Calculate the area of the shaded region

The area of the shaded region $A = A_{circle}-A_{square}=72\pi - 144$ $mm^{2}$.

Answer:

$(72\pi - 144)$ $mm^{2}$