$P Q R S$ is a square with side length 60 and centre $C$. Point $W$ lies on $P S$ so that $W S=53$. Point $X$ lies on $S R$ so that $X R=40$. The midpoint of $Q R$ is $Y$. Point $Z$ lies on $P Q$. What is the length of $Z Q$ so that the total area of the shaded regions is equal to the total area of the non-shaded regions?
(A) 21
(B) 15
(C) 23
(D) 19
(E) 17

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-021.jpg?height=388&width=401&top_left_y=318&top_left_x=1317)

## Part C: Each correct answer is worth 8.
Solution 1

Square $P Q R S$ has side length 60 , and so its area is $60^{2}=3600$.

We require the total area of the shaded regions to equal the total area of the non-shaded regions, and so the total area of the shaded regions is to be half the area of the square, or 1800 .

Since $C$ is the centre of the square, its distance to each of the sides of the square is $\frac{1}{2}(60)=30$.

To determine the area of each of the shaded regions, we begin by labelling each of the given lengths (as well as those that we can determine), and constructing diagonal $S Q$, as shown.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-098.jpg?height=390&width=421&top_left_y=1862&top_left_x=1535)

Quadrilateral $W C X S$ is made up of two triangles: $\triangle S C X$ and $\triangle S C W$.

In $\triangle S C X$, the base $S X=20$ and the perpendicular height from $C$ to $S R$ is 30 (since $C$ is the centre of the square $P Q R S)$.

Thus, the area of $\triangle S C X$ is $\frac{1}{2}(20)(30)=300$.

In $\triangle S C W$, the base $S W=53$ and the perpendicular height from $C$ to $S W$ is 30 .

The area of $\triangle S C W$ is $\frac{1}{2}(53)(30)=795$.

Therefore, the area of quadrilateral $W C X S$ is $300+795=1095$.

Quadrilateral $Z Q Y C$ is made up of two triangles: $\triangle C Q Y$ and $\triangle C Q Z$.

The midpoint of $Q R$ is $Y$ and $C$ is the centre of the square, so $C Y$ is perpendicular to $Q Y$.

In $\triangle C Q Y$, the base $C Y=30$, the perpendicular height $Y Q=30$, and so the area of $\triangle C Q Y$ is $\frac{1}{2}(30)(30)=450$.

In $\triangle C Q Z$, if the base is $Z Q$, then the perpendicular height from $C$ to $P Q$ is 30 .

The area of $\triangle C Q Z$ is $\frac{1}{2}(Z Q)(30)=15(Z Q)$.

Therefore, the area of quadrilateral $Z Q Y C$ is $15 \times Z Q+450$.

Finally, adding the areas of the shaded regions, we get $1095+15 \times Z Q+450=1800$ or $15 \times Z Q=1800-1095-450=255$, and so $Z Q=\frac{255}{15}=17$.

## Solution 2

As in Solution 1, we require the total area of the shaded regions to equal 1800 .

We begin by labelling $T$, the midpoint of side $P S$.

Since $Y$ is the midpoint of $Q R$, then $T Y$ passes through $C$ and is perpendicular to each of $P S$ and $Q R$.

Since $C$ is the centre of the square, its distance to each of the sides of the square is $\frac{1}{2}(60)=30$.

We label each of the given lengths (as well as those that we can determine), as shown.

Quadrilateral $W C X S$ is made up of $\triangle W C T$ and trapezoid $T C X S$ ( $T C$ is parallel $S X$ and thus $T C X S$ is a trapezoid).

In $\triangle W C T$, base $W T=W S-T S=53-30=23$ and the perpendicular height $C T$ is 30 . Thus, the area of $\triangle W C T$ is $\frac{1}{2}(23)(30)=345$.

In trapezoid $T C X S$, the parallel sides have lengths $T C=30$ and $S X=20$, and the perpendicular height is $T S=30$.

The area of trapezoid $T C X S$ is $\frac{1}{2}(30)(30+20)=750$.

Therefore, the area of quadrilateral $W C X S$ is $345+750=1095$.

Quadrilateral $Z Q Y C$ is also a trapezoid (since $Z Q$ is parallel to $C Y$ ).

The area of trapezoid $Z Q Y C$ is $\frac{1}{2}(Q Y)(Z Q+30)=15(Z Q+30)$.

Adding the areas of the shaded regions, we get $1095+15(Z Q+30)=1800$ or $15(Z Q+30)=1800-1095=705$ or $Z Q+30=\frac{705}{15}=47$, and so $Z Q=17$.

ANswer: (E)