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Question
In a convex hexagon, prove that the sum of all interior angle is equal to the sum of its exterior angles formed by producing the sides in the same order.
In a convex hexagon, prove that the sum of all interior angle is equal to the sum of its exterior angles formed by producing the sides in the same order.
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Solution
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In a convex hexagon, prove that the sum of all interior angle is equal to twice thethe sum of its exterior angles formed by producing the sides in the same order.
InIn a hexagon number of sides = 6.
∴ Sum of the interior angles of a hexagon = (2n - 4) x 90°
= ( 2x 6 - 4) x 90° = 8 x 90° = 720°. -------(1)
Sum of the exterior angles of a hexagon = 360°.
Given that 2 times the sum of the exterior angles of a hexagon i.e 2 x 360°.= 720° -----(2)
from (1) and (2) we get
∴ In a convex hexagon, the sum of all interior angle is equal to twice the sum of its exterior angles.
In a convex hexagon, prove that the sum of all interior angle is equal to twice thethe sum of its exterior angles formed by producing the sides in the same order.
InIn a hexagon number of sides = 6.
∴ Sum of the interior angles of a hexagon = (2n - 4) x 90°
= ( 2x 6 - 4) x 90° = 8 x 90° = 720°. -------(1)
Sum of the exterior angles of a hexagon = 360°.
Given that 2 times the sum of the exterior angles of a hexagon i.e 2 x 360°.= 720° -----(2)
from (1) and (2) we get
∴ In a convex hexagon, the sum of all interior angle is equal to twice the sum of its exterior angles.
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