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
The sum of interior angles of a regular polygon is twice the sum of its exterior angles. Find the number of sides of 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.
Two alternate sides of a regular polygon, when produced, meet at the right angle. Calculate the number of sides in the polygon.
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 interior angle is: 155°
A regular polygon has each exterior angle measuring 40°. How many sides does it have?
If each interior angle of a regular polygon is 144°, what is its corresponding exterior angle?
What is the sum of all exterior angles of any regular polygon?
Which formula correctly represents the sum of interior angles of an n-sided polygon?
