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TopNewestIm getting the answer as

\[\frac{\pi ab}{4\left(a^2+b^2\right)}:0.5\]

So, first, the area of the rhombus is \(\frac{1}{2}ab\)

Now, to find the area of the circle, we can first find the radius of the circle.

Let the center of the circle be at coordinate \((0,0)\). and one of the rhombus's side be described with the equation \(y=\frac{a}{b}\left(x-\frac{b}{2}\right)+a\)

The radius will be the shortest distance between the center of the circle and the side of the rhombus. Finding that distance, we get \[r^{2}=\frac{a^2b^2}{4\left(a^2+b^2\right)}\] The area of the circle is therefore \[\ \pi \times \frac{a^2b^2}{4\left(a^2+b^2\right)}\]

Hence, the ratio is \[\boxed{\frac{\pi ab}{4\left(a^2+b^2\right)}:0.5}\] – Julian Poon · 1 year, 10 months ago

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It will be helpful if you add more lines , Thanks , (Upvoted) – Syed Baqir · 1 year, 10 months ago

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\[\sqrt{k^2+\left(\frac{a}{b}\left(k-\frac{b}{2}\right)+a\right)^2}\]

When \(k=x_{0}\), the above equation would be minimised. Note that \(\sqrt{x_{0}^2+\left(\frac{a}{b}\left(x_{0}-\frac{b}{2}\right)+a\right)^2}=r\), where r is the radius of the circle.

In order to find the minimum, we can find the minimum of \(k^2+\left(\frac{a}{b}\left(k-\frac{b}{2}\right)+a\right)^2\), where \(x_{0}^2+\left(\frac{a}{b}\left(x_{0}-\frac{b}{2}\right)+a\right)^2=r^{2}\)

All we have to do to find \(r^{2}\) is to simplify the quadratic and find the minimum.

This might help. – Julian Poon · 1 year, 10 months ago

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By the way the Co-ordinate is too complicated to spot !! – Syed Baqir · 1 year, 10 months ago

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\( \frac{\pi ab}{2(a^{2}+b^{2})} \) – Syed Baqir · 1 year, 10 months ago

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