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marta [7]
3 years ago
7

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
Chemistry
1 answer:
Degger [83]3 years ago
6 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
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3 years ago
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  • Particles could refer to atoms, molecules, formula units.
  • <em>Knowing that every one mole of a substance contains Avogadro's no. of molecules (NA = 6.022 x 10²³).</em>

<em><u>Using cross multiplication:  </u></em>

1.0 mole → 6.022 x 10²³ molecules.

??? mole → 8.95 x 10²³ molecules.

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