Geometry Again!

If the radius of the large circle is $R$ and that of the small circles is $r$ , we have $\pi R^2=4 \pi r^2$ or $R=2r$ .

So the distance $OR=3r$ .

The line $OR$ is a part of the diagonal and at $45^\circ$ angle with the vertical, so its projection into the vertical, $ER=3r cos(45^\circ)=\frac{3r}{\sqrt{2}}$ .

Add to this the distance $RT=r$ and we’ll have half the side of the square, $a$ .

Therefore $a=2(\frac{3r}{\sqrt{2}}+r)=r(2+3\sqrt{2})$ and the ration $\frac{a}{r}=2+3\sqrt{2}.$

ABCD is a square with side "a" cm.

The diagonal of the square =a root 2.

Let the radii of the larger circle be "R" and smaller circles "r" cm.

Then TT R^2 =4* TT r^2

R^2 = 4 r^2

R=2r -----------------------------------------------------1

Then we can easily find the ratio with the diagonal= a root 2

Just don't forget to include the ends of the diagonal.

You get the answer as 2+3 root 2