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प्रश्न
In a regular pentagon ABCDE, draw a diagonal BE and then find the measure of:
(i) ∠BAE
(ii) ∠ABE
(iii) ∠BED
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उत्तर
(i) Since number of sides in the pentagon = 5
Each exterior angle = `360/5 = 72^circ`
∠BAE = 180° – 72°= 108°
(ii) In Δ ABE, AB = AE
∴ ∠ABE = ∠AEB
But ∠BAE + ∠ABE + ∠ AEB = 180°
∴ 108° + 2 ∠ABE = 180° – 108° = 72°
⇒ ∠ABE = 36°
(iii) Since ∠AED = 108° ...[∵ each interior angle = 108°]
⇒ ∠AEB = 36°
⇒ ∠BED = 108° – 36° = 72°
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संबंधित प्रश्न
Fill in the blanks :
In case of regular polygon, with :
| No.of.sides | Each exterior angle | Each interior angle |
| (i) ___8___ | _______ | ______ |
| (ii) ___12____ | _______ | ______ |
| (iii) _________ | _____72°_____ | ______ |
| (iv) _________ | _____45°_____ | ______ |
| (v) _________ | __________ | _____150°_____ |
| (vi) ________ | __________ | ______140°____ |
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Calculate the number of sides of a regular polygon, if: its exterior angle exceeds its interior angle by 60°.
Is it possible to have a regular polygon whose interior angle is: 135°
