What is the internal angle sum of a hexagon?

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What is the internal angle sum of a hexagon?

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Answer 1 · 2018-03-16T09:47:38

$720^{\circ}$ $720^circ$

Explanation:

First, we split the hexagon into 6 equal isoceles triangles, each have the angles ( $60, θ, θ$ $60,theta,theta$ ) ( $\frac{360}{6} = 60$ $360/6=60$ ).

$θ = \frac{180 - 60}{2} = \frac{120}{2} = 60$ $theta=(180-60)/2=120/2=60$

$Sum of internal angles = 6 (120) = 720^{\circ}$ $"Sum of internal angles"=6(120)=720^circ$

Answer 2 · 2018-05-23T20:23:46

$720^{0}$ $720^0$

Explanation:

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The internal sum of four triangles is $4 \times 180^{0}$ $4 times 180^0$

Answer 3 · 2018-06-01T15:17:37

Or, it can be directly calculated using direct formula,

$\to (n - 2) \cdot 180^{\circ}$ $rarr(n-2)*180^@$ where $n$ $n$ is the number of sides of polygon.

In case of hexagon, $n = 6$ $n=6$

So, internal angles sum $= (6 - 2) \cdot 180^{\circ} = 4 \cdot 180^{\circ} = 540^{\circ}$ $=(6-2)*180^@=4*180^@=540^@$

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What is the internal angle sum of a hexagon?

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