so.. notice the picture below
and thus, we also know, that A = 51, thus
[tex]\bf A=\cfrac{1}{2}bh\implies A=\cfrac{(h+11)h}{2}\implies A=\cfrac{h^2+11h}{2}\qquad A=51
\\\\\\
51=\cfrac{h^2+11h}{2}\implies 102=h^2+11h\implies 0=h^2+11h-102
\\\\\\
0=(h+17)(h-6)\implies
\begin{cases}
0=h+17\implies &-17=h\\
0=h-6\implies &6=h
\end{cases}[/tex]
now, the height is just a value, it can't be negative, thus we can rule out -17, thus h = 6
and surely you'd know what the base is
