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

Hint: Assume the area of the larger circle be ‘${R_1}$ ‘, and the area of the smaller circle be ‘${R_2}$’. And then find the area of circle using the standard formula of area \[A = \pi {r^2}\]

Complete step-by-step answer:
Let the area of the larger circle be ‘${R_1}$ ‘, and the area of the smaller circle be ‘${R_2}$’.

Now, see in triangle ACR,
CR = r = AR (radius of the larger circle)
Now, we can also write,
AC = CD + BD + AB
From the figure, we can say CD = r, DB = ${r_1}$
To find AB, we need to apply Pythagora's theorem in triangle ABQ.
Therefore, in triangle ABQ,
AQ = BQ = r1 (radius of the smaller circles)
Therefore, AB = $\sqrt 2 {r_1}$
Therefore, AB = r + ${r_1}(1 + \sqrt 2 )$
Applying Pythagoras theorem in triangle ACR,
$2{r^2} = {(r + {r_1}(1 + \sqrt 2 ))^2}$
$r = {r_1}(3 + 2\sqrt 2 )$
Now, sum of areas of 4 smaller circles = $4\pi {r_1}^2$
And, the area of the larger circle = $\pi {r^2}$
Therefore, the ratio of areas = $\dfrac{{\pi {r^2}}}{{4\pi {r_1}^2}}$
Using equation (1), we get the ratio of areas = $\dfrac{{17 + 2\sqrt 2 }}{4}$
Hence, the answer is = $\dfrac{{17 + 2\sqrt 2 }}{4}$

Note: Whenever these types of questions appear, assume the radius of larger circle as r and smaller circles as $r_1$. Then, by using the figure, apply Pythagora's theorem to find the relation between the radius and then find the areas of the smaller circles and larger circle. At last find their ratio.