Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that)
What can you say about the angle sum of a convex polygon with $n$ number of sides?

(n) 180^{0}

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(n - 1) 180^{0}

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(n - 2) 180^{0}

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(n - 3) 180^{0}

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Solution

The correct option is B $(n - 2) 180^{0}$
$A f t e r a n a l y s i n g t h e g i v e n t a b l e i t i s o b s e r v e d t h a t w h e n (i) n u m b r e o f s i d e s o f t h e p o l y g o n = 3 ⟶ s u m o f t h e i n t e r n a l a n g l e s = 180^{o} = (3 - 2) \times 180^{o} (i i) n u m b r e o f s i d e s o f t h e p o l y g o n = 4 ⟶ s u m o f t h e i n t e r n a l a n g l e s = 360^{o} = (4 - 2) \times 180^{o} (i i i) n u m b r e o f s i d e s o f t h e p o l y g o n = 5 ⟶ s u m o f t h e i n t e r n a l a n g l e s = 540^{o} = (5 - 2) \times 180^{o} (i v) n u m b r e o f s i d e s o f t h e p o l y g o n = 6 ⟶ s u m o f t h e i n t e r n a l a n g l e s = 720^{o} = (6 - 2) \times 180^{o} ∴ B y i n d u c t i o n, i t c a n b e c o n c l u d e d t h a t W h e n t h e n u m b r e o f s i d e s o f t h e p o l y g o n = n ⟶ s u m o f t h e i n t e r n a l a n g l e s = (n - 2) \times 180^{o} A n s - I f t h e n u m b r e o f s i d e s o f a c o n v e x p o l y g o n = n ⟶ s u m o f t h e i n t e r n a l a n g l e s = (n - 2) \times 180^{o}$

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Similar questions

Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.)

Figure

Side

Angle sum

180°

2 × 180°

= (4 − 2) × 180°

3 × 180°

= (5 − 2) × 180°

4 × 180°

= (6 − 2) × 180°

What can you say about the angle sum of a convex polygon with number of sides?

(a) 7

(b) 8

(d) n

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