Advertisements
Advertisements
प्रश्न
KL, LM and MN are three consecutive sides of a regular polygon. If ∠LKM = 20°, find the interior angle of the polygon and the number of sides of the polygon.
उत्तर
In ΔLMK, LM = LK ...[Sides of a regular polygon]
∴ ∠LMK = ∠LKM = 20° ...[Angles opp to equal sides are equal]
∴ ∠LKM + ∠LMK + ∠KLM = 180°
⇒ 20° + 20° + ∠KLM = 180°
⇒ ∠KLM = 140°
∴ Each interior angle of the polygon is 140°.
∴ Each interior angle of a regular polygon
= `(("n" - 2) xx 180°)/"n"`
⇒ `(("n" - 2) xx 180°)/"n"` = 140°
⇒ 180°(n - 2) = 140°n
⇒ 40°n = 360°
∴ n = 9
∴ Number of sides of the polygon = 9.
APPEARS IN
संबंधित प्रश्न
One angle of a six-sided polygon is 140o and the other angles are equal.
Find the measure of each equal angle.
Three angles of a seven-sided polygon are 132o each and the remaining four angles are equal. Find the value of each equal angle.
In a pentagon ABCDE, AB is parallel to DC and ∠A: ∠E : ∠D = 3: 4: 5. Find angle E.
In a parallelogram ABCD, AB = 20 cm and AD = 12 cm. The bisector of angle A meets DC at E and BC produced at F.
Find the length of CF.
Find the number of sides in a regular polygon, when each interior angle is: 120°
Is it possible to have a polygon whose each interior angle is 124°?
Is it possible to have a polygon whose each interior angle is 105°?
The number of sides of two regular polygons are in the ratio 2 : 3 and their interior angles are in the ratio 9 : 10. Find the number of sides of each polygon.
The sum of the interior angles of a polygon is 6.5 times the sum of its exterior angles. Find the number of sides of the polygon.
In a pentagon PQRST, ∠P = 100°, ∠Q = 120° and ∠S = ∠T. The sides PQ and SR, when produced meet at right angle. Find ∠QRS and ∠PTS.