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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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5 0
3 years ago
A container of hydrogen at 172 kPa was decreased to 85.0 kPa producing a new volume of 3L. What was the original volume?
Aneli [31]

Answer:

<h2>The answer is 1.48 L</h2>

Explanation:

In order to find the original volume we use the same for Boyle's law which is

P_1V_1 = P_2V_2

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the original volume

V_1 =  \frac{P_2V_2}{P_1}  \\

From the question

P1 = 172 kPa = 172000 Pa

P2 = 85 kPa = 85000 Pa

V2 = 3 L

We have

V_1 =  \frac{85000 \times 3}{172000}  =  \frac{255000}{172000}  =  \frac{255}{172}  \\  = 1.482558139...

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