Question -
Answer -
We know that the sum of the interior anglesof a polygon = (n – 2) π
And each angle ofpolygon = sum of interior angles of polygon / number of sides
Now, let uscalculate the magnitude of
(i) Pentagon
Number of sides inpentagon = 5
Sum of interiorangles of pentagon = (5 – 2) π = 3π
∴ Eachangle of pentagon = 3π/5 × 180o/ π =108o
(ii) Octagon
Number of sides inoctagon = 8
Sum of interiorangles of octagon = (8 – 2) π = 6π
∴ Eachangle of octagon = 6π/8 × 180o/ π =135o
(iii) Heptagon
Number of sides inheptagon = 7
Sum of interiorangles of heptagon = (7 – 2) π = 5π
∴ Eachangle of heptagon = 5π/7 × 180o/ π =900o/7 = 128o 34′17”
(iv) Duodecagon
Number of sides induodecagon = 12
Sum of interiorangles of duodecagon = (12 – 2) π = 10π
∴ Eachangle of duodecagon = 10π/12 × 180o/π = 150o