7. In a square $A B C D$ with side length 2, a segment $M N$ of length 1 is constrained to have its endpoint $M$ on side $A B$ and its endpoint $N$ on side $B C$. This segment divides the square into a triangle $T$ and a pentagon $P$. What is the maximum value that the ratio of the area of $T$ to that of $P$ can assume?
(A) $\frac{1}{4}$
(B) $\frac{2}{5}$
(C) $\frac{1}{7}$
(D) $\frac{2}{15}$
(E) $\frac{1}{15}$.

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7. The answer is $\mathbf{( E )}$. The ratio between the two areas is maximum when the area of T is maximum: indeed, in this case, the area of $\mathrm{P}$ assumes the minimum value. $\mathrm{T}$ is a right triangle whose hypotenuse is 1, and it has the maximum area when it is isosceles. Indeed, such a triangle can be inscribed in a semicircle with diameter 1 and achieves maximum area when the height relative to the hypotenuse is maximum, i.e., $\frac{1}{2}$. In this case, the area of $\mathrm{T}$ is $\frac{1}{4}$ and that of $\mathrm{P}$ is $4-\frac{1}{4}=\frac{15}{4}$, and thus the ratio is $\frac{1}{15}$.