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)

The correct answer is option (E):

\(\frac{128}{\pi^2} sq.units\)

Let the area of the circle be A and its diameter be d. We are given that A = 4d.
We know that the area of a circle is given by the formula A = πr², where r is the radius.
The diameter d is twice the radius, so d = 2r.
We can substitute the expressions for A and d into the given equation A = 4d.
πr² = 4(2r)
πr² = 8r
Since r cannot be zero (otherwise, the circle wouldn't exist), we can divide both sides by r.
πr = 8
r = 8/π

Now, consider the square inscribed in the circle. The diagonal of the square is the diameter of the circle. Let s be the side length of the square.
By the Pythagorean theorem, the diagonal of the square, which is the diameter of the circle (d), is related to the side length s by d = s√2.
Since d = 2r, we can write 2r = s√2.
We found r = 8/π, so 2(8/π) = s√2
16/π = s√2
s = (16/π) / √2
s = 16 / (π√2)

The area of the square is s².
Area of the square = (16 / (π√2))²
Area of the square = (16²)/(π² * (√2)²)
Area of the square = 256 / (π² * 2)
Area of the square = 128 / π²

Therefore, the area of the square is 128/π² square units.