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, 1 month ago

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@Julian Poon
–
Can you elaborate more how you succeeded in finding the radius of circle,

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

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@Syed Baqir
–
Let the x-coordinate of the point where the circle will touch the side be \(x_{0}\). Now, for a certain point on the side of the rhombus, the coordinates will be \(\left(k,\frac{a}{b}\left(k-\frac{b}{2}\right)+a\right)\). The distance between that point and the origin can thus be found using Pythagoras theorem and is given as:

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.

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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, 1 month ago

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

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

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

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