11. Two congruent pentagons are each formed by removing a rightangled isosceles triangle from a square of side-length 1 . The two pentagons are then fitted together as shown.
What is the length of the perimeter of the octagon formed?
A 4
B $4+2 \sqrt{2}$
C 5
D $6-2 \sqrt{2}$
E 6
SolUTION
$\mathbf{E}$
Let $P, Q, R, S, T, U, V$ and $W$ be the vertices of the octagon, as shown in the diagram, and let $X$ be the vertex as shown.

The sides $P X, X U, U V, Q R, R S$ and $S T$ all have length 1 .
Let $P W$ have length $s$. Then, because $P W V$ is a right-angled isosceles triangle, $V W$ also has length $s$.

Because the pentagons $P X U V W$ and $Q R S T X$ are congruent, $Q X$ and $T X$ also have length $s$.

It follows that $P Q$ and $T U$ each have length $1-s$.
We can now deduce the the length of the perimeter of the octagon $P Q R S T U V W$ is
$$
(1-s)+1+1+1+(1-s)+1+s+s=6 \text {. }
$$

Commentary
Note that in order to work out the length of the perimeter we did not need to know the value of $s$. It is, however, not difficult to find the value of $s$. You are asked to do this in Problem 11.1.

FOR INVESTIGATION
11.1 Find the length of $Q X$.
11.2 Find the area of the octagon $P Q R S T U V W$.