Question

In a convex hexagon, prove that the sum of all interior angle is equal to twice the sum of its exterior angles formed by producing the sides in the same order.

Answer 1

$$\text{ For a convex hexagon, interior angle }=(\frac{2n-4}{n}\times 90°)$$
$$\text{ For a hexagon,}n=6$$
$$\therefore \text{ Interior angle }=(\frac{12-4}{6}\times 90°)$$
$$=(\frac{8}{6}\times 90°)$$
$$=120°$$
$$\text{ So, the sum of all the interior angles }=120°+120°+120°+120°+120°+120°=720°$$
$$\therefore \text{ Exterior angle }={\left(\frac{360}{n}\right)}^{°}={\left(\frac{360}{6}\right)}^{°}={60}^{°}$$
$$\text{ So, sum of all the exterior angles }={60}^{°}+{60}^{°}+{60}^{°}+{60}^{°}+{60}^{°}+{60}^{°}={360}^{°}$$
$$\text{ Now, sum of all interior angles }=720°$$
$$=2\left(360°\right)$$
$$=\text{ twice the exterior angles }$$
$$\text{ Hence proved }.$$