Question

in a regular polygon ABCDE draw a diagonal BE and then find the measure of angle BAE angleABE and angleBED

Answer 1

here is your answer OK dude......


Given: ABCDE is a regular pentagon

⇒ Sum of angles of ABCDE = 180° × (5 – 2) = 540°

⇒ Each angle of ABCDE =

⇒ ∠BAE = 108°

Now in ΔABE

AB = AE

⇒ ∠ABE = ∠AEB

and ∠BAE + ∠ABE + ∠AEB = 180°

⇒ 108° + ∠ABE + ∠ABE = 180°

⇒ 2∠ABE = 180° – 108°

angle ABC = 72 upon 2 = 36 degree

and ∠AED = ∠AEB + ∠BED

⇒ ∠BED = ∠AED – ∠AEB

= 108° – 36°

= 72°

I hope I help you

Answer 2

ABCDE is a Regular pentagon.

Sum of interior angles of a regular polygon: (N-2)×180°

here N= 5;

So,

= (5-2)×180°

=3×180

=540°.

As all 5 angle of a regular pentagon are equal measure, each angle is;

540/5 =108°.

So ∠ BAE = 108°.

In ΔABE,AB=AE

∠BAE+∠AEB+∠ABE=180°

108 +2∠ABE=180

∠ABE=36°.

now ∠BED =180-108

=72°