Answer is 54 but how
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Given :
∠DCF = 65° ,∠ EFB= 100°
∠EFB+∠ EFC= 180°. ( Linear pair)
100° +∠ EFC = 180°
∠EFC = 180°-100°= 80°
In ∆ EFC
∠EFC + ∠DCF +∠FEC= 180°
[ Angle sum PROPERTY]
80° +65°+∠FEC= 180°
∠FEC = 180°-145°= 35°
∠FEC = 35°
∠ADE = ∠DCF = 65° (Corresponding angles)
AED = 65° (AE=AD)
∠AEG = ∠AED - ∠FEC
∠AEG = 65° - 35° = 30°
∠AEG = 30°
Number of sides in a regular polygon= 360°/ exterior angle
Number of sides in a regular polygon with exterior angle 30°= 360°/ 30°= 12
Number of sides in a regular polygon with exterior angle 30°=12
Number of diagonals in a polygon= n(n-3)/2
Number of diagonals in a polygon of 12 sides= 12(12-3)/2= (12 ×9)/2= 6×9 = 54
The number of diagonals in a polygon of 12 sides=54.
∠DCF = 65° ,∠ EFB= 100°
∠EFB+∠ EFC= 180°. ( Linear pair)
100° +∠ EFC = 180°
∠EFC = 180°-100°= 80°
In ∆ EFC
∠EFC + ∠DCF +∠FEC= 180°
[ Angle sum PROPERTY]
80° +65°+∠FEC= 180°
∠FEC = 180°-145°= 35°
∠FEC = 35°
∠ADE = ∠DCF = 65° (Corresponding angles)
AED = 65° (AE=AD)
∠AEG = ∠AED - ∠FEC
∠AEG = 65° - 35° = 30°
∠AEG = 30°
Number of sides in a regular polygon= 360°/ exterior angle
Number of sides in a regular polygon with exterior angle 30°= 360°/ 30°= 12
Number of sides in a regular polygon with exterior angle 30°=12
Number of diagonals in a polygon= n(n-3)/2
Number of diagonals in a polygon of 12 sides= 12(12-3)/2= (12 ×9)/2= 6×9 = 54
The number of diagonals in a polygon of 12 sides=54.
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