Ask Question
13 June, 17:31

What is the approximate area of a regular pentagon with a side lenghtt of 6 feet and a distance from the center to a vertex of 6 feet

+4
Answers (1)
  1. 13 June, 17:42
    0
    A regular pentagon may be divided into 5 congruent triangles.

    If the side of the pentagon is 6 feet, each base of the triangles will be 6 feet.

    If the distance from the center to a vertex is 6 feet, all the sides of the triangles will 6 feet.

    So there are 5 congruent, equilateral triangles each with side equal to 6 feet.

    The area one of those triangles is calculateb by:

    base * height / 2

    height ^2 = (side length) ^2 - (side length / 2) ^2 = 6^2 - 3^2 = 36 - 9 = 27

    => height = √ (27) ≈ 5.2 feet

    And, using area = base * height / 2,

    area of a triangle = 6 feet * 5.2 feet / 2 = 15.6 feet^2

    And the area of the entire pentagon equals 5 triangles ≈ 5 * 15.6 feet^2 = 78 feet^2

    Answer: 78 feet^2
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “What is the approximate area of a regular pentagon with a side lenghtt of 6 feet and a distance from the center to a vertex of 6 feet ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers