Circle A is perfectly inscribed in a square, and the square is perfect

Circle B= \(PI* (\frac{a}{2})^2\)
Circle A= \(PI* ( \sqrt{2} \frac{a}{2} )^2\) = \(2*PI* (\frac{a}{2} )^2\)

Area of Circle A is 100% greater than that of area B.

B

We can solve this task by picking numbers. Radius of A = 1 and area of \(A = piR^2 = pi\)

As radius of A it's one half of side square than side of square = 2. Diagonal of square will be equal to diameter of circle B.
Diagonal of square = \(\sqrt{2^2 + 2^2} = \sqrt{8}\)

And radius of B = \(pi\sqrt{8}/2\)
Area of B = pi2

pi2 on 100% more than pi
So answer is B
_________________

Square's side is the diameter for the Circle A and diagonal is the diameter for the Circle B

if side of square be x,
then increase in area

= \(\frac{Area of Circle B - Area of Circle A}{Area of Circle A}\) *100%

= \(\frac{\pi(\frac{\sqrt{2}*x}{2})^2 - \pi(\frac{x}{2})^2}{\pi(\frac{x}{2})^2}\) *100%

= \(\frac{\frac{1}{2} - \frac{1}{4}}{\frac{1}{4}}\) *100%

= 1 * 100%

Answer B

Let Side of Square be a.
Radius of smaller circle A= a/2. Area = pi*a^2/4

Radius of Circle B = Diagonal of Square/2 = a\(\sqrt{2}\)/2. Area = 2pi*a^2/4

(Area B- Area A)/Area A = 1. Therefore, 100%

VERITAS PREP OFFICIAL SOLUTION:

One important note about inscribed shapes is that both squares and circles are perfectly symmetrical, meaning that if you know one length (radius, diameter, circumference, area; or side, diagonal, area, perimeter) you can solve for everything else.

In this case, is we call the radius of the smaller circle r, then the diameter of that circle is 2r and the area is pi*r^2.

If that circle is perfectly inscribed inside a square, then that means that the length of the diameter will perfectly fit within the boundaries of the square, making the side of the square also equal to 2r.

Now, if a circle is perfectly inscribed around that square, then the circle will hit each corner exactly once, making the diameter of the larger circle equal to the diagonal of the square. Using what we know about squares (or using Pythagorean Theorem), we know that the diagonal is equal to the side * ?2, making the diameter of the larger circle equal to 2r*?2, and the radius of the larger circle then equal to half that: r*?2.

The area of the larger circle is the pi*(r*?2)^2, or 2*pi*r^2. Therefore, the area of the larger circle (2*pi*r^2) is twice that of the smaller circle (pi*r^2). But keep in mind that you must answer the right question! The question asks “how much GREATER is the larger circle than the smaller” not “the larger circle is what percent OF the smaller”. To get to twice the size, we only ADD 100%, so the correct answer is B.
_________________

Let the radius of the bigger outer circle be R ; area = \(pi*R^2\) {this radius is called circumradius}
Let the radius of the smaller inner circle be r; area = \(pi*r^2\) {this radius is called inradius}

Now square shares a particular relationships with circumradius and inradius
Area of square in terms of circumradius = \(2*R^2\)
Area of square in terms of inradius = \(4*r^2\)
therefore \(2R^2=4r^2\)
or \(R^2=r^2*2\)
Therefore area of bigger circle is twice that of smaller circle
(2Area-1Area)= 1 times greater Area = 100% greater

ANSWER IS B
_________________

The best thing to do here is make up a value for the side of the square (e.g. 10).

As a result, the area of circle A is 25 x pi

The radius of circle B is 5√2 (because the diagonal is equal to the diameter)

So, the area of circle B is 50 x pi or twice the size of circle A.

2x larger means 100% larger.

B.
_________________

Since squares and circles are symmetric figures the relationship between them can be deduced by using the radius/ diagonal/ diameter. Please find attached image.

_________________