6. 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 does the perimeter of the parallelogram $A^{\prime} B B^{\prime} P$ measure?
(A) 24
(B) 22
(C) 26
(D) 28
(E) 27

![](https://cdn.mathpix.com/cropped/2024_04_17_414eb95eef0ec41d5707g-1.jpg?height=306&width=388&top_left_y=1111&top_left_x=2413)
Problem 6. The correct answer is (A).

Since in a parallelogram opposite sides are congruent to each other, we have $\overline{P C^{\prime}}=\overline{D^{\prime} D}=\overline{B^{\prime} C}$, $\overline{P D^{\prime}}=\overline{C^{\prime} D}=\overline{A^{\prime} A}, \quad \overline{P A^{\prime}}=\overline{B^{\prime} B}=\overline{D^{\prime} A}, \quad \overline{P B^{\prime}}=\overline{A^{\prime} B}=\overline{C^{\prime} C}$. The perimeter of $A B C D$ is therefore equal to the sum of the perimeters of $A A^{\prime} P D^{\prime}$ and $C C^{\prime} P B^{\prime}$, as well as to the sum of the perimeters of $D D^{\prime} P C^{\prime}$ and $B B^{\prime} P A^{\prime}$. This latter perimeter is thus $31+42-49=24$ meters.