Find the number of sides in a polygon if the sum of its interior angles is :

(i) $900^{o}$

(ii) $1620^{o}$

(iii) 16 right - angles

(iv) right - angles.

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Solution

(i) Let no. of sides = n

Sum of angles of polygon = $900^{o}$

$∴ (n - 2) \times 180^{o} = 900^{o} n - 2 = \frac{900}{180} n - 2 = 5 n = 5 + 2 n = 7$ .

(ii) Let no. of sides = n

Sum of angles of polygon $= 1620^{o}$

$∴ (n - 2) \times (180^{o}) = 1620^{o} n - 2 = \frac{1620}{180} n - 2 = 9 n = 9 + 2 n = 11$ .

(iii) Let no. of sides = n

Sum of angles of polygon = 16 right angles

$= 16 \times 90 = 1440^{o} ∴ (n - 2) \times 180^{o} = 1440^{o} n - 2 = \frac{1440}{180} n - 2 = 8 n = 8 + 2 n = 10$ .

(iv) Let no. of sides = n

Sum of angles of polygon = 32 right angles

$= 32 \times 90 = 2880^{o} ∴ (n - 2) \times 180^{o} = 2880 n - 2 = \frac{2880}{180} n - 2 = 16 n = 16 + 2 n = 18$ .

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