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 ,

we have:

.

Similarly, for ,

we have:

.

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:

≈44.209 cm^2,

or

≈5.791 cm^2,

--Kevin C.