AkkuJi
Here is my explanation

Let the square sheet PQRS have all sides as \(s\)
Area of circle Y with radius \(r\) insribed in it would be \(\frac{πs^2}{4}\)

Remember - Equilateral parallelogram is a Rhombus

Let the rhombus sheet ABCD have all sides as \(a\)
(Refer to the figure below)

In △AXD;
∠DAX = 60°
∠ADX = 30°
∠AXD = 90° (since, diagonals of a rhombus are perpendicular)

Using the 30°-60°-90° property;
AX (opposite to 30°) = \(\frac{a}{2}\)
DX (opposite to 60°) = \(\frac{\sqrt{3}a}{2}\)

In order to find the radius of circle X inscribed in rhombus ABCD, we can equate the area in △AXD
i.e. \(\frac{1}{2}(\frac{a}{2})(\frac{\sqrt{3}a}{2}) = \frac{1}{2}(a)(r)\)
\(\frac{\sqrt{3}a^2}{4} = ar\)
\(r = \frac{\sqrt{3}a}{4}\)

We have been given - the areas of the two circles will be equal
So, \(π(\frac{\sqrt{3}a}{4})^2 = \frac{πs^2}{4}\)
\(\frac{3a^2}{16} = \frac{s^2}{4}\)
\(\frac{3a^2}{4} = s^2\)
\(s = \frac{\sqrt{3}a}{2}\)


Now, Area of rhombus ABCD = \(\frac{1}{2}\)x diagonal AC x diagonal BD = \(\frac{1}{2}(a)(\sqrt{3}a) = \frac{\sqrt{3}a^2}{2}\), and
Area of square PQRS = \(s^2 = (\frac{\sqrt{3}a}{2})^2 = \frac{3a^2}{4}\)

Required ratio = \((\frac{\sqrt{3}}{2})(\frac{4}{3}) = \frac{2}{\sqrt{3}}\)

Thanks for the explanation KarunMendiratta

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