Question

What is the area of the square, if four vertices lie on the circumference of a circle where the area of the circle is four times its diameter in magnitude?

• A. \(\frac{8}{\pi^2} sq.units\)
• B. \(\frac{16}{\pi^2} sq.units\)
• C. \(\frac{32}{\pi^2} sq.units\)
• D. \(\frac{64}{\pi^2} sq.units\)
• E. \(\frac{128}{\pi^2} sq.units\)

Answer 1

Answer: (E)

To solve this problem, let's go through the following steps:

1. First, understand the given conditions. We know that the area of the circle is four times its diameter. The formula for the area of a circle is \(\pi r^2\), and the diameter is \(2r\).
2. According to the problem, \(\pi r^2 = 4 \times 2r\). Simplifying this, we get:
3. So, \(\pi r^2 = 8r\).
4. Divide both sides by \(r\) (assuming \(r \neq 0\)) to find \(r\):
5. \(r = \frac{8}{\pi}\).
6. Since the vertices of the square lie on the circumference of the circle, the diagonal of the square is equal to the diameter of the circle, which is \(2r\). Thus, the diameter of the circle is \(2 \times \frac{8}{\pi} = \frac{16}{\pi}\).
7. For a square with diagonal length \(d\), we use the formula relating the diagonal to the side length \(a\): \(d = a\sqrt{2}\).
8. Thus, \(\frac{16}{\pi} = a\sqrt{2}\).
9. Solve for \(a\):
10. \(a = \frac{16}{\pi\sqrt{2}}\).
11. To find the area of the square, compute \(a^2\):
12. \(a^2 = \left(\frac{16}{\pi\sqrt{2}}\right)^2 = \frac{256}{2\pi^2} = \frac{128}{\pi^2}\) square units.

Thus, the area of the square is \(\frac{128}{\pi^2} \text{ sq.units}\), which matches the correct answer.