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प्रश्न
The difference between the exterior angles of two regular polygons, having the sides equal to (n – 1) and (n + 1) is 9°. Find the value of n.
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उत्तर
We know that sum of exterior angles of a polynomial is 360°
(i) If sides of a regular polygon = n – 1
Then each angle = `360^circ/("n" - 1)`
and if sides are n + 1, then
each angle = `360^circ/("n" + 1)`
According to the condition,
`360^circ/("n" - 1) - 360^circ/("n" + 1)=9`
`=> 360 [1/("x" - 1) - 1/("x" + 1)] = 9`
`=> 360 [("n" + 1 - "n" + 1)/("n" - 1)("n" + 1)] = 9`
`=> (2 xx 360)/("n"^2 - 1) = 9`
`=> "n"^2 - 1 = (2 xx 360)/9 = 80`
`=> n^2 - 1 = 80`
`=> n^2 = 1 - 80 = 0`
⇒ n2 - 81 = 0
⇒ (n)2 - (9)2 = 0
⇒ (n + 9)(n - 9) = 0
Either n + 9 = 0. then n = -9 which is not possible being negative,
or n - 9 = 0, then n = 9
∴ n = 9
∴ No. of. sides of a regular polygon = 9
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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°____ |
The measure of each interior angle of a regular polygon is five times the measure of its exterior angle. Find :
(i) measure of each interior angle ;
(ii) measure of each exterior angle and
(iii) number of sides in the polygon.
AB, BC and CD are three consecutive sides of a regular polygon. If angle BAC = 20° ; find :
(i) its each interior angle,
(ii) its each exterior angle
(iii) the number of sides in the polygon.
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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Is it possible to have a regular polygon whose exterior angle is: 100°
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