If a parallelogram and a rectangle are on the same base and between the same parallels, then the perimeter of a parallelogram is __ than the perimeter of a rectangle.

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Given that both a parallelogram and a rectangle are based on the same line segment and limited by the same parallel lines, one might be curious about how their perimeters compare.

Here is a concise explanation:

  • Parallelogram BDFE shares the same base, DF, with the rectangle ACFE, and both are enclosed by the same parallels.
  • Observing segment lengths in the diagram, it's shown that $AE < BE$ and $FC < DF$. This indicates that the sides of the parallelogram (BE and DF) opposite the base are typically longer than those of the rectangle (AE and FC).

This results in the parallelogram having a greater perimeter than the rectangle, given that the cumulative length of its sides surpasses that of the rectangular sides encompassing the same area.

To formalize:

  • Perimeter of rectangle = 2(AE + FC)
  • Perimeter of parallelogram = 2(BE + DF)

Since $AE < BE$ and $FC < DF$, it follows that the perimeter of the parallelogram is greater than that of the rectangle.

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