If \(\ell\) is the length and \(w\) the width, then \(\ell=2w-2\).
Since the area of the rectangle is
\(\ell w=40\)
you have
\((2w-2)w=40\implies 2w^2-2w-40=0\implies w^2-w-20=0\)
Factoring the left hand side gives
\((w-5)(w+4)=0\implies w=5,w=-4\)
Obviously the width can't be negative, so \(w=5\). This means D is the answer, since it's the only option with this measure.
