Question

b. explain why the area of this circle is more than 18 square units and less than 36 square units.

Explanation:

Step1: Find the area of the inscribed square

The side - length of the inscribed square: Assume the side - length of each small square is \(1\). By the Pythagorean theorem, if the side - length of the inscribed square is \(a\), and we consider the right - triangle formed by two adjacent sides of the small squares. The diagonal of the inscribed square \(d\) (which is equal to the diameter of the circle) can be calculated. If we count the number of small - square sides, we find that the area of the inscribed square \(A_{square}\) can be calculated as composed of \(8\) right - triangles with base and height equal to \(3\) (assuming the side - length of each small square is \(1\)). The area of a right - triangle is \(S=\frac{1}{2}\times base\times height\). So \(A_{square}=8\times\frac{1}{2}\times3\times3 = 18\) square units.

Step2: Find the area of the circumscribed square

The side - length of the circumscribed square \(s = 6\) (if the side - length of each small square is \(1\)). The area of a square is \(A = s^{2}\), so \(A_{circum - square}=6\times6=36\) square units.

Step3: Use the property of the circle's area

For a circle inscribed in a square and circumscribing another square, the area of the circle \(A_{circle}\) satisfies the inequality \(A_{inscribed - square}

Answer:

The area of the inscribed square is \(18\) square units (calculated as the sum of the areas of \(8\) right - triangles with base and height \(3\): \(A = 8\times\frac{1}{2}\times3\times3\)) and the area of the circumscribed square is \(36\) square units (\(A = 6\times6\)). Since the circle contains the inscribed square and is contained within the circumscribed square, its area \(A\) satisfies \(18 < A<36\) square units.