Question

Given a square with area A. A circle lies inside the square, such that the circle touches all sides of the square. Another square
with area B lies inside the circle, such that all its vertices lie on the circle.
Find the value of A/B
​

Answer 1

answer:

Let the area of Larger circle be 'r' and the area of smaller circle by 'r1'

In triangle ACR,

CR=r=AR (radius of the circle)

AC=CD+BD+AB

Now, CD=r

DB=r1

To find AB, we need to apply pythagoras theorem in triangle ABQ.

In triangle ABQ,

AQ=BQ=r1 (radius of the circle)

and AB=

(2)

r1

⇒AC=r+r1(1+

(2)

)

Applying pythagoras theorem in triangle ACR,

2r

2

=(r+r1(1+

(2)

))

2

solving, we get r=r1(3+2

(2)

)------(1)

Sum of areas of all small circles = 4π(r1)

2

Area of larger circle = π(r)

2

Ratio of areas =

4π(r1)

2

πr

2

Using equation (1), we get ratio of areas =

4

17+2

(2)