Find the sum of interior angles of a polygon with twelve sides.

1200^{\circ}

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1800^{\circ}

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2400^{\circ}

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3600^{\circ}

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Solution

The correct option is B $1800^{\circ}$
Number of sides, $n = 12$

Sum of interior angles of polygon $= (n - 2) \times 180^{\circ}$

$= (12 - 2) \times 180^{\circ}$

$= (12 - 2) \times 180^{\circ}$

$= 10 \times 180^{\circ}$

$= 1800^{\circ}$

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Find the sum of interior angles of a polygon with twelve sides.

The correct option is B 1800∘Number of sides, n=12 Sum of interior angles of polygon =(n−2)×180∘ =(12−2)×180∘ =(12−2)×180∘ =10×180∘ =1800∘

The correct option is B $1800^{\circ}$
Number of sides, $n = 12$

Sum of interior angles of polygon $= (n - 2) \times 180^{\circ}$

$= (12 - 2) \times 180^{\circ}$

$= (12 - 2) \times 180^{\circ}$

$= 10 \times 180^{\circ}$

$= 1800^{\circ}$