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    Is there a variation of Heron's Formula for pentagons, hexagons, etc? Prof. Loh showed us one for quadrilaterals.

    Day 16
    herons formula
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    • M
      mirthfulcygnet M2★ M3★ M4 M5★ M6
      last edited by debbie

      Module 2 Week 4 Day 16 Bonus Explanation Part 1

      Title says it all!

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      • N
        nastya MOD M0 M1 M2 M3 M4 M5
        last edited by debbie

        Hi @mirthfulcygnet,
        Nice to hear from you! It is great that you are continuing to be interested and ask questions even after having finished the lessons!

        It is true that there're exists a generalizations of Heron's Formula for quadrilaterals and pentagons, but they are much-much more complicated and oriented for higher-level mathematicians. The same relatively easy formulas for the areas of polygons with more than three angles, analogous to Heron's Formula, do not exist.

        To begin with, Heron's formula is a special case of Brahmagupta's formula for the area of a cyclic quadrilateral. Brahmagupta's formula, in its turn, is a special case of Bretschneider's formula for the area of any quadrilateral.
        Heron's formula is also a special case of the formula for the area of a trapezoid based only on its sides. Heron's formula is obtained by setting the smaller parallel side to zero.
        One of the even more high-level formulas is a formula for Robbins pentagons: a cyclic pentagons whose side lengths and area are all rational numbers. If you really want to, you can read about it here (page 15), but I think it would be better for you to learn more in this area first. 🙂

        Try to get used to the Heron's formula first and be ready to use it any time you need to! Then you will be ready to learn more such formulas.

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