The sum of interior angles a convex polygon having 'n' sides is equal to .

n ×180°

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(n-2) × 180°

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(n-3) × 180°

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(n-4) × 180°

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Solution

The correct option is B (n-2) × 180°

Generalising the above, the sum of interior angles of a convex polygon with 'n' sides is equal to ' $(n - 2) \times 180^{\circ}$ '.

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Similar questions

Sum of interior angles of a polygon of $(n + 1)$ sides $=$

180^{o}

360^{o}

= \frac{360^{\circ}}{n}

= 180^{\circ}

n = \frac{360^{\circ}}{E x t e r i o r a n g l e}

= \frac{(n - 2) \times 180^{\circ}}{n}

(\frac{180^{0}}{n})

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