In this given figure, AED is a isosceles triangle, ABCD is a parallelogram & EFG is a line segment. If angle DCF is 65 degree & EFB is 100 then the number of diagonal of a regular polygon having it's each exterior angle equal to the measure of angle AEG, is
32
44
65
54
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Option 4 is correct..
54
Solution:
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
=================================°===============================
Hope this will help you....
54
Solution:
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
=================================°===============================
Hope this will help you....
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