# Any solution anyone?

Need Help !!

Note by Syed Baqir
2 years, 9 months ago

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

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

- 2 years, 9 months ago

Can you elaborate more how you succeeded in finding the radius of circle,

It will be helpful if you add more lines , Thanks , (Upvoted)

- 2 years, 9 months ago

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:

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

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.

- 2 years, 9 months ago

Thank you very much :D

By the way the Co-ordinate is too complicated to spot !!

- 2 years, 9 months ago

You can write it as:

$$\frac{\pi ab}{2(a^{2}+b^{2})}$$

- 2 years, 9 months ago