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

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