perfectly inscribed

[sanju09]

Circle A is perfectly inscribed in a square, and the square is perfectly inscribed within circle B. The area of circle B is what percent greater than the area of circle A?
(A) 50
(B) 100
(C) 150
(D) 200
(E) 250

[rohu27]

Let x be the side of the square.

Radius of circle A will be half of the diagonal of square = x/sqrt2.
radius of circle B will be x/2

%= pi*x^2[1/2-1/4] / pi*x^2*1/4
[spoiler]
200% - option D[/spoiler]

Circle A is perfectly inscribed in a square, and the square is perfectly inscribed within circle B. The area of circle B is what percent greater than the area of circle A?
(A) 50
(B) 100
(C) 150
(D) 200
(E) 250

[BarryLi]

I drew this out on paper, please excuse me for not uploading an image of circle A perfectly inscribed in a square perfectly inscribed in circle B.

Let O represent the centre of circle A (which is also the centre of the square and circle B)

Suppose the radius of circle A is x.
The distance to the corner of the square is sqrt(2)*x, which is also the radius of circle B.

The area of a circle is pi*r^2, therefore:
Area of circle A: pi*(x^2)
Area of circle B: 2*pi*(x^2)

As a result, the area of circle B is 100% greater than the area of circle A.

B

[Night reader]

B cannot be correct;
@rohu, mark the choice brother! percentage of square circle B greater than that of A means that A's value increased by rate (here 2)

radius circle A=a/2, radius circle B=a/sqrt(2), square of circle = pr^2
sq B/sq A= a^2/2 : a^2/4 = 2, 200% increase of value for the square of circle A to get the value of circle square B

answer D

I drew this out on paper, please excuse me for not uploading an image of circle A perfectly inscribed in a square perfectly inscribed in circle B.

Let O represent the centre of circle A (which is also the centre of the square and circle B)

Suppose the radius of circle A is x.
The distance to the corner of the square is sqrt(2)*x, which is also the radius of circle B.

The area of a circle is pi*r^2, therefore:
Area of circle A: pi*(x^2)
Area of circle B: 2*pi*(x^2)

As a result, the area of circle B is 100% greater than the area of circle A.

B

[rohu27]

oops my bad. edited

i do this mistake while practising prblems. instead of writing dwn the answer choice i end up writign the answer (on paper) and dnt realise it until i go bck to verify my answers.

B cannot be correct;
@rohu, mark the choice brother! percentage of square circle B greater than that of A means that A's value increased by rate (here 2)

radius circle A=a/2, radius circle B=a/sqrt(2), square of circle = pr^2
sq B/sq A= a^2/2 : a^2/4 = 2, 200% increase of value for the square of circle A to get the value of circle square B

answer D

I drew this out on paper, please excuse me for not uploading an image of circle A perfectly inscribed in a square perfectly inscribed in circle B.

Let O represent the centre of circle A (which is also the centre of the square and circle B)

Suppose the radius of circle A is x.
The distance to the corner of the square is sqrt(2)*x, which is also the radius of circle B.

The area of a circle is pi*r^2, therefore:
Area of circle A: pi*(x^2)
Area of circle B: 2*pi*(x^2)

As a result, the area of circle B is 100% greater than the area of circle A.

B

[Night reader]

thanks

[GMATGuruNY]

Circle A is perfectly inscribed in a square, and the square is perfectly inscribed within circle B. The area of circle B is what percent greater than the area of circle A?
(A) 50
(B) 100
(C) 150
(D) 200
(E) 250

Circle A:
Plug in r=1.
A = π(1^2) = π.

Square:
s = 2r = 2*1 = 2.
Diagonal = 2√2

Circle B:
R = 1/2(diagonal of square) = 1/2*(2√2) = √2.
A = π(√2)^2 = 2π.

Since Circle B is twice as big as Circle A, the area of Circle B is 100% greater.

The correct answer is B.

To avoid making a careless error, we must understand the following distinction:
The area of Circle B is 200% of the area of Circle A (because the area of Circle B is 2 times the area of Circle A).
The area of Circle B is 100% greater than the area of Circle A (because we have to increase the area of Circle A by 100% to get the area of Circle B).

To be 200% greater (answer choice D), the area of Circle B would need to be 3 times the area of Circle A.
For example, if Circle A = π and Circle B = 3π, the percent increase would be:
Difference/Circle A * 100 = (3π - π)/π * 100 = 2π/π * 100 = 200%.

[Night reader]

thanks Mitch, I guess drawing helps to understand that two circles area can't be 3:1

[rohu27]

thanks Mitch, got to knw how dnagerous it is to visualize nd answer.