ABCDEF is a regular hexagon . Find each angle of triangle ABE

Question

Hello sir , im in 8th standard , and im stuck at a question , thats about a triangle in a hexagon

Answer 1

Draw regular hexagon ABCDEF with triangle ABE inside it

 

All sides of hexagon are equal in length

 

Sum of angles of a polygon = (n - 2)180 

       n = number of sides

 

Sum of angles of a hexagon is 720°, since it is a regular 

hexagon each interior angle has measure 120°

 

Quadrilateral ABEF contains half the angle measure of the

hexagon; its interior angle measure = 360°

 

angle AFE = 120° (angle of hexagon ABCDEF)

 

triangle AFE is isosceles; angle FAE = angle FEA = 30°

           sum of angles of triangle = 180°

 

angle FAE + angle EAB = 120°; therefor angle EAB = 90°

 

angle ABE is one half of 120° = 60°

 

angle ABE + angle EAB + angle AEB = 180°

         therfor angle AEB = 30°

Answer 2

Let the center of the hexagon be O.  Since this is a hexagon, m <BOA = 360⁰/6 = 60⁰ .  Then, as OA = OB, m <OBA = m<OAB.  Thus, as the three angles sum to 180^o, each is 60⁰_. 

Since OA = OE = FA = EF, ΔOAE ≅ ΔFAE, and they are isosceles.

Then, m <OAE = m <FAE, and they sum to m <OAF, which is 60 ⁰, so m <OAE = 30^o.

Then, m <BAE = m <BAO + m <OAE = 60^o + 30^o = 90^o.  

Then, for ΔABE, m <A = M<BAE = 90^o. m<B = m <ABE = m<ABO = 60^o.  m <E = 180⁰ - 90⁰ - 60⁰ = 30^o.

 

Answer 3

Each angle of regular hexagon is 120°.

ΔAFE is isosceles.

m∠FAB = 30°

m∠EAB = 90°

BCDE is an isosceles trapezoid

m∠CBE = 60°

m∠EBA = 60°

m∠AEB = 30° 

Answer 4

If you draw segments inside the hexagon in which the segments intersect at the center of the hexagon, you will get 6 equilateral triangles. 

 

What is the measure of one angle in an equilateral triangle?  Use this fact to find the other angles as well as the fact that the sum of the angles in a triangle is 180 degrees.