Respuesta :
Given
Area of the regular pentagon is 6.9 cm².
Find out the perimeter of a regular pentagon
To proof
Formula
Area of regular pentagon is
[tex]= \frac{1}{4}\sqrt{5(5+2\sqrt{5})}\ a^{2}[/tex]
As given in the question
area of regular pentagon = 6.9 cm²
now equating the area value with the area formula.
[tex]6.9 =\frac{1}{4}\sqrt{5(5+2\sqrt{5})} a^{2}[/tex]
Now put
√5 = 2.24 ( approx)
put in the above equation
[tex]\frac{6.9\times 4}{\sqrt{5(5+2\times2.24)}}=a^{2}\\\frac{27.6}{\sqrt{47.4}}=a^{2}\\\frac{27.6}{6.88}=a^{2}[/tex]
thus
a² = 4.01
a = √ 4.01
a = 2.0 cm ( approx)
As perimeter represented the sum of all sides.
i.e regular pentagon have five sides of equal length.
Thus
perimeter of the regular pentagon = 5 × side length
= 5 ×2.00
therefore the perimeter of the regular pentagon = 10cm
option c is correct
Hence proved
The perimeter of the regular pentagon is 10cm.
Given that
The area of the regular pentagon is 6.9 cm2.
We have to determine
What is the perimeter?
According to the question
The area of the regular pentagon is 6.9 cm2.
The area of a pentagon is the region that is enclosed by all five sides of the pentagon.
In order to find the area of pentagon when only the side length is given, we use the formula:
[tex]\rm Area = \dfrac{1}{4}\sqrt{5(5+2\sqrt{5})} s^2[/tex]
Here, s is the side length.
Substitute all the values in the formula;
[tex]\rm Area = \dfrac{1}{4}\sqrt{5(5+2\sqrt{5})} s^2\\ \\ \rm 6.9= \dfrac{1}{4}\sqrt{5(5+2\sqrt{5})} s^2\\ \\ 6.9 \times 4= \sqrt{5(5+2\sqrt{5})} s^2\\ \\ \dfrac{27.6}{\sqrt{5(5+2\sqrt{5})} }= s^2\\ \\ s^2=\dfrac{27.6}{6.88}\\ \\ s^2=4.01\\ \\ s=\sqrt{4.01}\\ \\ s=2[/tex]
Therefore,
The Pentagon has 5 sides of equal length.
The perimeter of the regular pentagon = 5 × side length
The perimeter of the regular pentagon = 5 × 2
The perimeter of the regular pentagon = 10 cm
Hence, The perimeter of the regular pentagon is 10cm.
To know more about Regular Pentagon click the link given below.
https://brainly.com/question/3556808
