## Excel in math and science

### Master concepts by solving fun, challenging problems.

## It's hard to learn from lectures and videos

### Learn more effectively through short, conceptual quizzes.

## Our wiki is made for math and science

###
Master advanced concepts through explanations,

examples, and problems from the community.

## Used and loved by 4 million people

###
Learn from a vibrant community of students and enthusiasts,

including olympiad champions, researchers, and professionals.

## Comments

Sort by:

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

Log in to reply

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

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

Log in to reply

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.

Log in to reply

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

Log in to reply

You can write it as:

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

Log in to reply