A large square is divided into a small square surrounded by four congruent rectangles as shown. The perimter of each of the congruent rectangles is $14$. What is the area of the large square?

[asy]pathpen = black+linewidth(0.7); D((0,0)--(7,0)--(7,7)--(0,7)--cycle); D((1,0)--(1,6)); D((0,6)--(6,6)); D((1,1)--(7,1)); D((6,7)--(6,1)); [/asy]
$\mathrm{(A) \ }49 \qquad \mathrm{(B) \ }64 \qquad \mathrm{(C) \ }100 \qquad \mathrm{(D) \ }121 \qquad \mathrm{(E) \ }196$
Let the length of the longer side be $x$, and the length of the shorter side be $y$.  We are given that $2x+2y=14\implies x+y=7$.  However, note that $x+y$ is also the length of a side of the larger square.  Thus the area of the larger square is $(x+y)^2=7^2=\boxed{49\text{ (A)}}$.