Actually, there are two solutions.
Let w be the width of rectangle c and h its height.
Then the area given for triangle a tells us:
and the area given for triangle b gives:
Multiplying both sides of both equations by 2, we have:
Solving the second equation for w, .
Substituting this into the first equation,
Multiplying both sides by (6-h) and doing a little simplifying, we get
Multiplying and rearranging, we get:
Dividing both sides by 3, this becomes:
This is a quadratic equation, and will have two roots:
Both of these numbers are in the range 0<h<6.
Now, what is the width for each of these? We know:
so for ,
Similarly, for ,
Both of these are in the allowed range of 0<w<15; so both solutions are geometrically valid.
The area of rectangle c is , so we see the possible areas are: