Three touching circles inscribed in a rectangle 'Three equal circles, with radius r, are inscribed in a rectangle in a way that all three of them touch each other only once. Find the area of rectangle.'  The thing i can't understand is that the arrangement of circles seem limitless to me,  do they need to be touching the sides of the rectangle or not ? 
 A: I think the problem should say the circles are inscribed in a rectangle, not a circle.  Yes, inscribed means they are tangent to the rectangle.  To have the circles touch in pairs, the centers need to be on an equilateral triangle of side $2r$  One side of the rectangle will be tangent to two circles, the other three will be tangent to one circle each.

A: A little thought will reveal that, if the three-circle grouping is to be tangent to the sides of a rectangle, then (at least) one of those circles must be tangent to two adjacent sides of the rectangle. This gives rise to the following diagram:

Let $\bigcirc O$, of radius $r$, be tangent to two sides of the rectangle. Place $\bigcirc A$ and $\bigcirc B$, such that $\overline{OA}$ makes an angle of $\alpha$ with a "horizontal" radius of $\bigcirc O$, and such that $\overline{OB}$ makes an angle of $\beta$ with a "vertical" radius. Of course, $\alpha + \beta = 30^\circ$ (and both $\alpha$ and $\beta$ are non-negative).
We see that the sides of the rectangle must have lengths $2r+2r\cos\alpha$ and $2r+2r\cos\beta$, so that the area is
$$\begin{align}
2r(1+\cos\alpha)\cdot 2r(1+\cos\beta) &= 4r^2(1+\cos\alpha)(1+\cos\beta) \tag{1a}\\
&= 16 r^2 \cos^2\frac{\alpha}{2}\cos^2\frac{\beta}{2} \tag{1b}\\
&= 16 r^2 \left( \cos\frac{\alpha}{2} \cos\frac{\beta}{2}\right)^2 \tag{1c}
\end{align}$$
But then the final factor of $(1c)$ is the square of ...
$$\frac{1}{2}\left( \cos\frac{\alpha+\beta}{2} + \cos\frac{\alpha-\beta}{2} \right) = \frac{1}{2}\left( \cos 15^\circ + \cos\frac{\alpha-\beta}{2} \right) \tag{2}$$
... so we have:

$$\text{area} = 4 r^2 \;\left(\;  \cos 15^\circ + \cos\frac{\alpha-\beta}{2} \;\right)^2 \tag{3}$$

This value depends upon $\alpha$ and $\beta$, so there is no unique area. However, we can see that the area is minimized when $|\alpha-\beta|$ is maximized, and vice-versa. So,

$$\begin{align}
|\alpha - \beta| = 30^\circ &\;\to\; \,\text{min area} = 16 r^2 \;\cos^2 15^\circ = 8 r^2 \left( 1 + \cos 30^\circ \right) = 4 r^2 \left(2 +\sqrt{3}\right) \tag{4a} \\[4pt]
|\alpha - \beta| = \phantom{3}0^\circ &\;\to\; \text{max area} = \phantom{1}4 r^2 \;\left(\; 1 + \cos 15^\circ \;\right)^2 = \text{exercise for reader}\tag{4b}
\end{align}$$


Note: The minimum area occurs when either $\alpha$ or $\beta$ vanishes; that is, when two of the circles are tangent to the same side of the rectangle (equivalently, when two circles are each tangent to two adjacent edges). This is the configuration shown in @RossMillikan's answer.
