Question

AED is an isosceles triangle with AE=AD.ABCD is a parallelogram and EGF is a line segment.If Angle DCF=65° and Angle EFB=100°,then the number of diagonals of a regular polygon having its each exterior angle equal to the measure of Angle AE FC, is..

Answer 1

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

Number of diagonals in a polygon of 12 sides=54

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Hope this will help you...