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What is the sum of the interior angles of a regular pentagon?

A. 360°
B. 720°
C. 540°
D. 900°


Sagot :

To find the sum of the interior angles of a regular pentagon, we use the formula for the sum of the interior angles of any polygon with [tex]\( n \)[/tex] sides:

[tex]\[ \text{Sum of interior angles} = (n - 2) \times 180^\circ \][/tex]

A pentagon has [tex]\( n = 5 \)[/tex] sides.

Now substituting [tex]\( n = 5 \)[/tex] into the formula:

[tex]\[ \text{Sum of interior angles} = (5 - 2) \times 180^\circ \][/tex]

[tex]\[ \text{Sum of interior angles} = 3 \times 180^\circ \][/tex]

[tex]\[ \text{Sum of interior angles} = 540^\circ \][/tex]

Thus, the sum of the interior angles of a regular pentagon is [tex]\( 540^\circ \)[/tex].

Given the answer choices:
A. 3600
B. 7200
C. 5400
D. 9000

The correct answer is not listed explicitly; it seems there might be a typo or ambiguity in the options. Assuming a conversion error occurs in the options given (as 540 degrees might be multiplied inaccurately to 5400), None of the given options are correct as per standard notation..

Using rigorous standard calculations, the sum of the interior angles of a regular pentagon is indeed:

[tex]\[ 540^\circ \][/tex]
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