area of square between tangent(externally) circles Two externally tangent circles,have a square between them,standing on the same base as the two circles.The circles have a radius of 1 unit each.The top two vertices of the square are touching one circle each.How do I find the area of the square?
 A: A picture would help. I hope it will be clear how to draw it. 
Let the two circles have equations $(x+1)^2+(y-1)^2=1$ and $(x-1)^2+(y-1)^2=1$. 
So they are tangent to the $x$-axis, and symmetrical about the $y$-axis.  The centre of the bottom side of the square is then at $(0,0)$. 
Let the square have side $2t$. Then the point $(t,2t)$ is on the right-hand circle. 
It follows that $(t-1)^2+(2t-1)^2=1$. This simplifies to $5t^2-6t+1=0$. The relevant root is $t=\frac{1}{5}$. So the square has area $\frac{4}{25}$.
Remark: We could strip away the coordinatization, and write the solution using only the Pythagorean Theorem. But coordinatization is a (provably) powerful tool in geometry, so one might as well use it.  
If we really want to use Pythagorean Theorem only, let $O$ be the centre of the right-hand circle, and let $P$ be the point where the top right corner of the square meets the circle. Drop a perpendicular from $O$, towards the $X$-axis, and let $Q$ be the point where this perpendicular meets the top edge of the square, extended. 
Then $PQ=1-t$ and $OQ=1-2t$. By the Pythagorean Theorem, $(PQ)^2+(OQ)^2=(OP)^2$, and therefore $(1-t)^2+(2-t)^2=1^2$. Expand, and continue as in the main solution. 
