A square with side length $3$ is inscribed in an isosceles triangle with one side of the square along the base of the triangle. A square with side length $2$ has two vertices on the other square and the other two on sides of the triangle, as shown. What is the area of the triangle?
[asy] //diagram by kante314  draw((0,0)--(8,0)--(4,8)--cycle, linewidth(1.5));  draw((2,0)--(2,4)--(6,4)--(6,0)--cycle, linewidth(1.5));  draw((3,4)--(3,6)--(5,6)--(5,4)--cycle, linewidth(1.5)); [/asy]

$(\textbf{A})\: 19\frac14\qquad(\textbf{B}) \: 20\frac14\qquad(\textbf{C}) \: 21 \frac34\qquad(\textbf{D}) \: 22\frac12\qquad(\textbf{E}) \: 23\frac34$
Let's split the triangle down the middle and label it:

We see that $\bigtriangleup ADG \sim \bigtriangleup BEG \sim \bigtriangleup CFG$ by AA similarity. 
$BE =  \frac{3}{2}$ because $AG$ cuts the side length of the square in half; similarly, $CF = 1$. Let $CG = h$: then by side ratios,
\[\frac{h+2}{h} =  \frac{\frac{3}{2}}{1} \implies 2(h+2) = 3h \implies h = 4\].
Now the height of the triangle is $AG = 4+2+3 =  9$. By side ratios, 
\[\frac{9}{4} = \frac{AD}{1} \implies AD = \frac{9}{4}\]. 
The area of the triangle is $AG\cdot AD = 9 \cdot \frac{9}{4}  = \frac{81}{4} = \boxed{\textbf{(B) }20 \frac{1}{4}}$
~KingRavi