Points $T, U, V, W, X, Y$ lie on square $P Q R S$, as shown. If $P T=T U=U Q=Q V=V W=W R=X S=S Y$, what fraction of the area of square $P Q R S$ is shaded?
(A) $\frac{5}{18}$
(B) $\frac{1}{3}$
(C) $\frac{2}{9}$
(D) $\frac{1}{4}$
(E) $\frac{1}{6}$

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-022.jpg?height=366&width=374&top_left_y=1427&top_left_x=1358)
Begin by constructing $A V$ perpendicular to $T X$ and $U B$ perpendicular to $Y W$, as shown.

The four segments $T X, U B, A V$, and $Y W$ divide $P Q R S$ into 9 identical squares.

Label the intersections of the perpendicular pairs of these four segments as points $C, D, E$, and $F$.

Segment $U Y$ is a diagonal of square $P U C Y$ and so $U Y$ passes through $E$, the centre of square $P U C Y$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-090.jpg?height=363&width=374&top_left_y=160&top_left_x=1561)

Segment $U E$ is a diagonal of square $T U F E$.

Segment $T W$ is a diagonal of square $T Q W D$ and $T F$ is a diagonal of square $T U F E$.

The diagonals in any square divide the square into 4 identical triangles.

For example, the diagonals $T F$ and $U E$ divide the square $T U F E$ into 4 identical triangles, 3 of which are shaded.

Similarly, we can show that diagonals $F W$ and $V C$ divide square $F V W C$ into 4 identical triangles, 3 of which are shaded.

We may construct the missing diagonals in each of the 9 squares, as shown.

These diagonals divide square $P Q R S$ into $9 \times 4=36$ identical triangles.

Since 10 of these triangles are shaded, then $\frac{10}{36}=\frac{5}{18}$ of square $P Q R S$ is shaded.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-090.jpg?height=371&width=379&top_left_y=885&top_left_x=1556)

ANSWER: (A)

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