7. Consider a parallelogram $A B C D$, the lines parallel to the sides passing through an internal point $P$ intersect the sides at $A^{\prime}, B^{\prime}, C^{\prime}$ and $D^{\prime}$, as shown in the figure. The perimeters (in meters) of the parallelograms $D^{\prime} A A^{\prime} P, C^{\prime} D D^{\prime} P$ and $B^{\prime} C C^{\prime} P$ are those indicated in the figure. How many meters is the perimeter of the parallelogram $A B C D$?

![](https://cdn.mathpix.com/cropped/2024_04_17_18ab8bf5b0aa1ddf2201g-1.jpg?height=304&width=377&top_left_y=1269&top_left_x=2411)
(A) 78
(B) 63
(C) 72
(D) 114
(E) 67
Problem 7. The correct answer is (E).

Since in a parallelogram opposite sides are congruent to each other, we have $\overline{P C^{\prime}}=\overline{D^{\prime} D}, \overline{P D^{\prime}}=\overline{C^{\prime} D}$, $\overline{P A^{\prime}}=\overline{B^{\prime} B}, \overline{P B^{\prime}}=\overline{A^{\prime} B}$. The perimeter of $A B C D$ is therefore equal to the sum of the perimeters of $A A^{\prime} P D^{\prime}$ and $C C^{\prime} P B^{\prime}$, that is $29+38=67$ meters.