What is the sum, in degrees, of the measures of the interior angles of a pentagon?

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What is the sum, in degrees, of the measures of the interior angles of a pentagon?

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Answer 1 · 2017-01-03T00:44:24

$540^o$ or $5xx108^o$

Explanation:

In any $n$ -gon, the sum of the interior angles is $(n-2)*180^o$ .
This can be proven by taking a regular n-gon (why not?), and drawing a (isosceles) triangle from the centre to one of the sides.

The centre angle will be $360^o/n$ and the other angles of that triangle will be $1/2(180-360/n)$ .
The inner angle of the n-gon will be just twice that, or $180-360/n$

The sum of all inner angles will then be:
$n*(180-360/n)=n*(180xxn/n-360/n)=$

$canceln*((180n-360)/canceln)=(n-2)*180^o$

In case of a regular n-gon, each angle will be $(n-2)/n*180^o$

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What is the sum, in degrees, of the measures of the interior angles of a pentagon?

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