Advertisements
Advertisements
प्रश्न
In a regular pentagon ABCDE, draw a diagonal BE and then find the measure of:
(i) ∠BAE
(ii) ∠ABE
(iii) ∠BED
Advertisements
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°
APPEARS IN
संबंधित प्रश्न
Find the number of sides in a regular polygon, if its interior angle is: `1 1/5` of a right angle
The ratio between the exterior angle and the interior angle of a regular polygon is 1 : 4. Find the number of sides in the polygon.
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.
If the difference between the exterior angle of a 'n' sided regular polygon and an (n + 1) sided regular polygon is 12°, find the value of n.
Calculate the number of sides of a regular polygon, if: its interior angle is five times its exterior angle.
Find a number of side in a regular polygon, if it exterior angle is: 30°.
Is it possible to have a regular polygon whose interior angle is: 135°
Is it possible to have a regular polygon whose exterior angle is: 36°
What is the measure of each interior angle of a regular hexagon?
What is the sum of all exterior angles of any regular polygon?
