What is the area of the smaller right triangle? 
The two diagonal lines are parallel, and the area of the region between them is $42$

What is $A,$ the area of the smaller right triangle?

So I named the missing base and height of the smaller right triangle x & y respectively and came up with the solution:
$\dfrac{(x + 4)(y + 3)}{2} - \dfrac{xy}{2} = 42$
Solving the equation we get:
$3x + 4y = 72$
I don't know what's next here so I just assumed that $x = 12, y = 9$ which satisfies the equation and resulting to $A = 54$. My question is, am I missing something so that I can solve the equation? 
 A: 
As shown above, consider $\triangle ABC$ and $\triangle ADE$. You have the common right angle of $\angle BAC$. Also, $\angle CBA = \angle EDA$ and $\angle BCA = \angle DEA$ because of the $2$ parallel lines on either side of the blue section, so the corresponding angles of intersection with the transversal are congruent, such as described in item #$3$ in https://en.wikipedia.org/wiki/Parallel_(geometry)#Euclidean_parallelism. Thus, you have that $\triangle ABC$ and $\triangle ADE$ are similar to each other. This means
$$\begin{equation}\begin{aligned}
\frac{|AE|}{|AD|} & = \frac{|AC|}{|AB|} \\
\frac{x}{y} & = \frac{x + 4}{y + 3} \\
x(y + 3) & = y(x + 4) \\
xy + 3x & = xy + 4y \\
3x & = 4y
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Using this in your determined equation gives
$$4y + 4y = 8y = 72 \implies y = 9 \tag{2}\label{eq2A}$$
This also gives from \eqref{eq1A} that
$$3x = 4(9) \implies x = 12 \tag{3}\label{eq3A}$$
These results confirm your guess was correct for the side lengths, so your answer of $A = \frac{1}{2}(9)(12) = 54$ is also correct.
A: 
So I named the missing base and height of the smaller right triangle x & y respectively and came up with the solution:
$\dfrac{(x + 4)(y + 3)}{2} - \dfrac{xy}{2} = 42$

Continuing your equation, we get:
$\dfrac{xy + 3x + 4y + 12 - xy}{2} = 42$
$xy + 3x + 4y + 12 - xy = 2 \cdot 42$
$3x + 4y + 12 = 84$
$\boxed{3x + 4y = 72}$

My question is, am I missing something so that I can solve the equation?

The lines are parallel, thus their slopes will be equal
$\implies \dfrac{x}{y} = \dfrac{x + 4}{y + 3}$
$(y + 3)(x) = (x + 4)(y)$
$xy + 3x = xy  + 4y$
$3x = 4y$
$\color {red}{3x + 4y = 72},3x + (3x)  = 72$
$x = 12, y = 9$
$\boxed{Area = \dfrac{xy}{2} = \dfrac{12 \cdot 9}{2} = \dfrac{108}{2} = 54}$
A: 
$\displaystyle \frac{42}{\frac{1}{2} (3)(4)} = 7$ so seven $3$ by $4$ triangles can fit into the blue area.
This means there are $6$ cuts into the blue region (so that $7$ pieces are formed), so there are $3$ horizontal cuts and $3$ vertical cuts.
Therefore the area of the red triangle is $\frac{1}{2} \times (3 \times 3) \times (3 \times 4) = 54$.
