Question

ABCD is a square, in which a circle is inscribed touching all the sides of square. In the four corners of square, 4 smaller circles of equal radii are drawn, containing maximum possible area.
What is the ratio of the area of larger circle to that of sum of the areas of four smaller circles?

Answer 1

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 $AB=\sqrt{\left(2\right)}r1$

$\Rightarrow AC=r+r1(1+\sqrt{\left(2\right)})$

Applying pythagoras theorem in triangle ACR,

$2r^2=(r+r1(1+\sqrt{\left(2\right)}){)}^2$

solving, we get $r=r1(3+2\sqrt{\left(2\right)})$------(1)

Sum of areas of all small circles $=$ $4\pi (r1{)}^2$

Area of larger circle $=$ $\pi (r{)}^2$

Ratio of areas $=$ $\frac{\pi r^2}{4\pi (r1{)}^2}$

Using equation (1), we get ratio of areas $=$ $\frac{17+2\sqrt{\left(2\right)}}{4}$