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The number of sides of polygon A is 6 more than the number of sides of polygon B. Interior angle of polygon A is 10° more than the interior angle of polygon B. Find the interior angle of polygon whose side is equal to sum of number of sides of polygon A and B.

The number of sides of polygon A is 6 more than the number of sides of polygon B. Interior angle of polygon A is 10° more than the interior angle of polygon B. Find the interior angle of polygon whose side is equal to sum of number of sides of polygon A and B. Correct Answer 168°

Calculation:

Let number of sides of polygon A and polygon B be a and b respectively.

⇒ a = b + 6

Interior angle of polygon A = (a - 2) × 180°/a

Interior angle of polygon B = (b - 2) × 180°/b

But,

⇒ (a - 2) × 180/a = (b - 2) × 180/b + 10

⇒ (b + 4) × 180/(b + 6) = (b - 2) × 180/b + 10

⇒ 180 = 10

⇒ 18 × 12 = b2 + 6b

⇒ b2 + 6b - 216 - 0

⇒ b2 + 18b - 12b - 216 = 0

⇒ b(b + 18) - 12(b + 18) = 0

⇒ (b + 18)(b - 12) = 0

⇒ b = 12

⇒ a = 12 + 6 = 18

Number of sides of new polygon = 12 + 18 = 30

∴ Interior angle of new polygon = (30 - 2) × 180°/30 = 168°

Important Points

Interior angle of n sided polygon = (n - 2) × 180°/n

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