Solution
B

Let $V$ be the foot of the perpendicular dropped from $S$ to $T R$. Angle $S T V=$ angle $T P U$ since both are equal to $90^{\circ}-$ angle $P T U$. So triangles $S T V$ and $T P U$ are congruent as both contain a right angle and $P T=S T$. Hence $T U=S V$.
The area of the shaded triangle is $\frac{1}{2} \times T U \times S V=\frac{1}{2} \times T U \times T U=1$, so $T U=\sqrt{2}$.
The area of the smaller square is 16 so $P U=4$. Applying Pythagoras' Theorem to triangle $P T U$ gives $P T^{2}=P U^{2}+T U^{2}=16+2=18$. Hence the area of the larger square is 18 .
21. The smaller square in the picture has area 16 and the grey triangle has area 1 .
What is the area of the larger square?
A 17
B 18
C 19
D 20
E 21