Answer:
The answer is K though? Let me elaborate
Step-by-step explanation:
We know that the area of a circle with radius k is A = πk^2.
Let's draw a square inside the circle, such that the corners of the square touch the circle. We can do this by drawing a line from the center of the circle to one of its points, and then drawing a perpendicular line to split that line in half. This creates a right triangle with sides k, k, and hypotenuse √2k.
Now we can use the Pythagorean theorem to find the length of one side of the square:
a^2 + b^2 = c^2
k^2 + k^2 = (√2k)^2
2k^2 = 2k^2
Therefore, the length of one side of the square is a = b = k.
The diagonal of the square can be found using the Pythagorean theorem:
d^2 = a^2 + b^2
d^2 = k^2 + k^2
d^2 = 2k^2
d = √2k
The area of the square is:
A = a^2
A = k^2
So we have a square with a diagonal of √2k and an area of k^2. To show that this square has an area that is half the area of the circle, we can find the area of the circle and divide by 2:
A(circle) = πk^2
A(square) = k^2
A(circle)/A(square) = (πk^2)/(k^2) = π
So the area of the square is π times smaller than the area of the circle. Therefore, there must be a square that has a diagonal with a length between k and 2k and an area that is half the area of the circle.