Find the probability that the sum of two randomly chosen positive numbers.... followed by some conditions. Find the probability that the sum of two randomly chosen positive numbers (both $\le 1$) will not exceed $1$ and that their product will be $\le \frac{2}{9}$
My input
I just solved a problem in which we choose two points on a unit length line segment and a condition was given that three segments made by these two points must be greater than $\frac{1}{4}$
Similarly I solved this problem as: 
Considering $AB$ as a line segment of length $1$ unit.
Same situation is here as we have to choose two numbers(As I took points on line). Also distance cant be negative. Therefor I chose two points such that there sum doesn't succeed $1$ and product is $\le \frac{2}{9}$. Two points are chose between $AB$ as $XY$ now length of $AX=x$ ,$XY=y$, And then I just applied condtions as follows: 
$x+y<1$ , $XY\le \frac{2}9$.

I have doubts.
Is this approach is correct?
I don't have its answer behind my book and that's why I didn't calculate shaded region area.  I would love to learn any other approach to solve this problem. Sorry if problem seems confusing because of my bad English. 
 A: Comment: You have a path to the answer; this is just a way to confirm
you are on the right track without actually working the problem.
I assume you have independent random variables $X$ and $Y,$ each distributed $\mathsf{Unif}(0, 1).$ And you seek $P(X+Y \le 1, XY \le 2/9),$ where as usual the comma means intersection.  
Here is a simulation in R statistical software of 100,000 independent $X$'s and $Y$'s that illustrates the problem and your approach. [This method of getting
the answer by simulation is sometimes called 'Monte Carlo integration'. It is
mainly used for problems that are too messy to handle with ordinary Riemann integration (often in more than 2 dimensions).
You should use ordinary integration for your answer.]
set.seed(731);  m = 10^5;  x = runif(m);  y = runif(m)
A = (x + y <= 1);  B = (x*y <= 2/9) # A and B are logical vectors of TRUEs and FALSEs.
mean(A);  mean(B);  mean(A & B)     # mean of a logical vector is its proportion of TRUEs.
[1] 0.50086  # aprx P(A) = 1/2
[1] 0.55795  # aprx P(B)
[1] 0.48821  # aprx answer P(A & B)

Because of the order in which I have plotted the 100,000 points below, it is the percentage
of yellow ones that you want, as in your diagram. (About 0.488.)
plot(x, y, pch=".")
  points(x[B], y[B], pch=".", col="red")        # Read [ ] as "such that"
  points(x[A], y[A], pch=".", col="green2")
  points(x[A&B], y[A&B], pch=".", col="yellow")


For an exact solution it may be easiest to subtract the area of the green
'nibble' from 1/2. (As @copperhat has suggested.)
A: One way is to compute the area of the triangle ($={1 \over 2}$) and subtract the area between
the red and blue lines.
The red and blue lines are $y= {2 \over 9x}$ and $y = 1-x$. Equating and
solving the resulting quadratic gives the bounds of integration
${1 \over 2} (1 \pm {1 \over 3})$. Hence the little area is
$\int_{{1 \over 2} (1 - {1 \over 3})}^{{1 \over 2} (1 + {1 \over 3})} (1-x-{2 \over 9x}) dx = {1 \over 6}-{2 \over 9} \ln 2 $.
The probability is then ${1 \over 3}+{2 \over 9} \ln 2 \approx 0.487$.
