20. A square ink pad has sides of length $1 \mathrm{~cm}$. It is covered in black ink and carefully placed in the middle of a piece of white paper. The square pad is then rotated $180^{\circ}$ about one of its corners so that all of the pad remains in contact with the paper throughout the turn. The pad is then removed from the paper.
What area of paper, in $\mathrm{cm}^{2}$, is coloured black?
A $\pi+2$
B $2 \pi-1$
C 4
D $2 \pi-2$
$\mathrm{E} \pi+1$
Solution
E
We let $O$ be the corner of the square about which it rotates. The initial position of the square is shown by solid lines, and its final position by broken lines, so that the corner initially at the position $P$ rotates to the position $Q$.

The area that is in contact with the paper at some time during the rotation is shown shaded. This is the area that gets coloured black as the square rotates. It is made up of the semicircle with diameter $P Q$ and the two shaded half squares outside this semicircle.

The length of the sides of the squares, in $\mathrm{cm}$, is 1 . We let the length of $O P$ be $x \mathrm{~cm}$. Then, by Pythagoras' Theorem, $x^{2}=1^{2}+1^{2}=2$. Therefore, the area of the semicircle with diameter $P Q$ is, in $\mathrm{cm}^{2}, \frac{1}{2} \pi x^{2}=\frac{1}{2} \pi \times 2=\pi$.
The area of the two half squares outside the semicircle is equal to the area of one square with side length $1 \mathrm{~cm}$, that is, $1 \mathrm{~cm}^{2}$.
Therefore, the total area that is coloured black is, in $\mathrm{cm}^{2}, \pi+1$.