A problem on conditional geometric probability The point (x, y) is chosen randomly in the unit square. What is the conditional probability of $x^2+y^2 \leq 1/4$ given that $xy \leq 1/16$

I started solving this and while calculating I got some very unpleasant numbers but the problem is not marked as difficult in this book where I found it. So I start suspecting that I am having some conceptual mistake.

The two curves intersect at $x=\sqrt{2+\sqrt {3}} /4$ and at $x=\sqrt{2-\sqrt {3}} /4$   

Now I have to find a few areas via definite integrals. Is that what this problem is about?! 

The integral of $\sqrt{1/4-x^2} $ is very unpleasant, it seems. So?!

Is there some smarter / nicer approach?
$xy \leq \frac{1}{16}$ iff $2xy \leq \frac{1}{8}$.

Conditional on this, 

$x^2 + y^2 \leq \frac{1}{4}$ iff $x^2 + 2xy + y^2 = (x+y)^2 \leq \frac{3}{8}$ iff $x+y \leq \sqrt{\frac{3}{8}}$.  

Now do you know how to calculate the probability that the sum of two random numbers is less than a given number?
Suppose that $xy\le \frac{1}{16}.$ Then $x^2+y^2\le \frac14\iff (x+y)^2\le \frac38$  This should lead to an easier way to approach it.