Rhombus $P Q R S$ is inscribed in rectangle $J K L M$, as shown. (A rhombus is a quadrilateral with four equal side lengths.) Segments $P Z$ and $X R$ are parallel to $J M$. Segments $Q W$ and $Y S$ are parallel to $J K$. If $J P=39, J S=52$, and $K Q=25$, what is the perimeter of rectangle $W X Y Z$ ?
(A) 48
(B) 58
(C) 84
(D) 96
(E) 108

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-083.jpg?height=420&width=501&top_left_y=264&top_left_x=1300)
Since $J K L M$ is a rectangle, then the angles at $J$ and $K$ are each $90^{\circ}$, so each of $\triangle S J P$ and $\triangle Q K P$ is right-angled.

By the Pythagorean Theorem in $\triangle S J P$, we have

$$
S P^{2}=J S^{2}+J P^{2}=52^{2}+39^{2}=2704+1521=4225
$$

Since $S P>0$, then $S P=\sqrt{4225}=65$.

Since $P Q R S$ is a rhombus, then $P Q=P S=65$.

By the Pythagorean Theorem in $\triangle Q K P$, we have

$$
K P^{2}=P Q^{2}-K Q^{2}=65^{2}-25^{2}=4225-625=3600
$$

Since $K P>0$, then $K P=\sqrt{3600}=60$.

(Instead of using the Pythagorean Theorem, we could note instead that $\triangle S J P$ is a scaled-up version of a 3-4-5 right-angled triangle and that $\triangle Q K P$ is a scaled-up version of a 5-12-13 right-angled triangle. This would allow us to use the known ratios of side lengths to calculate the missing side length.)

Since $K Q$ and $P Z$ are parallel and $P K$ and $W Q$ are parallel, then $P K Q W$ is a rectangle, and
so $P W=K Q=25$.

Similarly, $J P Z S$ is a rectangle and so $P Z=J S=52$.

Thus, $W Z=P Z-P W=52-25=27$.

Also, $S Y R M$ is a rectangle. Since $J M$ and $K L$ are parallel ( $J K L M$ is a rectangle), $J K$ and $M L$ are parallel, and $P Q$ and $S R$ are parallel ( $P Q R S$ is a rhombus), then $\angle M S R=\angle K Q P$ and $\angle S R M=\angle Q P K$.

Since $\triangle S M R$ and $\triangle Q K P$ have two equal angles, then their third angles must be equal too. Thus, the triangles have the same proportions. Since the hypotenuses of the triangles are equal, then the triangles must in fact be exactly the same size; that is, the lengths of the corresponding sides must be equal. (We say that $\triangle S M R$ is congruent to $\triangle Q K P$ by "angle-side-angle".) In particular, $M R=K P=60$.

Thus, $Z Y=S Y-S Z=M R-J P=60-39=21$.

Therefore, the perimeter of rectangle $W X Y Z$ is $2(21)+2(27)=96$.

ANSWER: (D)