# Answer: 5

## Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_bdfe3271f0fede0d79bbg-1.jpg?height=1016&width=1034&top_left_y=977&top_left_x=274)

Notice that triangles $\triangle A B C$ and $\triangle A B D_{1}$ are congruent, hence their areas are equal. Triangles $\triangle A B D_{1}$ and $\triangle A_{1} B D_{1}$ share the same height $D_{1} H$ and have equal bases, thus they are equal in area. Therefore, $S\left(\triangle A A_{1} D_{1}\right)=1$. It can be shown that $S\left(\triangle B B_{1} A_{1}\right)=S\left(\triangle C C_{1} B_{1}\right)=S\left(\triangle D D_{1} C_{1}\right)=1$. Adding these up, we get $S\left(A_{1} B_{1} C_{1} D_{1}\right)=5$.
1. Given a parallelogram $A B C D$ and points $A_{1}, B_{1}, C_{1}$, and $D_{1}$ are chosen such that point $A$ is the midpoint of segment $D D_{1}$, point $B$ is the midpoint of $A A_{1}$, point $C$ is the midpoint of $B B_{1}$, and point $D$ is the midpoint of $C C_{1}$. Find the area of $A_{1} B_{1} C_{1} D_{1}$, given that $S(A B C D)=1$.