Third or Sixth Root Unity?

Geometry Level 2

A B C D E F ABCDEF is a regular hexagon with each side of 2 cm 2 \text{ cm} . Then find the area of the \triangle A B G ABG to four decimal places.


The answer is 3.4641.

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7 solutions

Chew-Seong Cheong
Dec 23, 2014

Area of A B G = 1 2 × A B × A E = 1 2 × 2 × 2 × A F × cos 3 0 \triangle ABG = \dfrac {1}{2} \times AB \times AE = \dfrac {1}{2} \times 2 \times 2 \times AF \times \cos{30^\circ}

= 1 2 × 2 × 2 × 2 × 3 2 = 2 3 = 3.464 \quad \quad \quad \quad \quad \quad \space = \dfrac {1}{2} \times 2 \times 2 \times 2 \times \dfrac {\sqrt{3}}{2} = 2\sqrt{3} = \boxed {3.464}

I just calculated the apothem and used the traditional triangle area formula.

Swapnil Das - 6 years, 2 months ago

Look at the figure below: In the figure is the hexagon A B C D E F ABCDEF and in the second figure it is also divided into 6 equilateral triangles.The height of the equilateral triangles divides them into 2 right triangles with base 1 cm and hypotenuse 2 cm.Therefore by the Pythagorean theorem we get: ( H e i g h t ) 2 + 1 2 = 2 2 ( H e i g h t ) 2 = 4 1 = 3 H e i g h t = 3 (Height)^2+1^2=2^2\\(Height)^2=4-1=3\\Height=\sqrt{3}

The height of the equilateral triangles is half the height of A B G ABG .So the height of A B G ABG is 2 × 3 = 2 3 2\times\sqrt{3}=2\sqrt{3} .The base of A B G ABG is 2cm so: A r e a = 1 2 × b r e a d t h × h e i g h t = 1 2 × 2 × 2 3 = 2 3 = 3.464 T o 4 d e c i m a l p l a c e s Area=\frac{1}{2}\times breadth \times height\\=\frac{1}{2}\times2\times2\sqrt{3}\\=2\sqrt{3}=3.464\;To\;4\;decimal\;places

Isn't the upload image feature great

Mehul Chaturvedi - 6 years, 5 months ago

Very easy simply adding the points A,E and B,D then we get ∆ ABG=(1/2)* rectangle ABDE . Again it is very simple that the height of the triangle is a√3 so the area of the triangle is (1/2) a a√3 sq unit so ar(∆ ABG)=3.4641 sq cm. Can't be it done like it!!!!!!!!!!

anna anant - 6 years, 5 months ago
Rishabh Tripathi
Mar 10, 2015

In B C D \bigtriangleup BCD -

By sine rule,

B D sin 120 = B C sin 30 \frac{BD}{\sin120} = \frac{BC}{\sin30}

B D = 4 × sin 120 BD = 4 \times \sin120

Area of Shaded = 1 2 × D E × B D \bigtriangleup = \frac{1}{2} \times DE \times BD

= 1 2 × 2 × 4 × sin 120 = \frac{1}{2} \times 2 \times 4 \times \sin120

= 4 × 0.866 = 4 \times 0.866

= 3.464 = 3.464

Did the same thing.

Sam Maltia - 5 years, 7 months ago
Aareyan Manzoor
Dec 31, 2014

C D = A E = 1 , A B = 2 A D = 2 A C 2 A D 2 = 2 3 CD=AE=1,AB=2AD=2\sqrt{AC^2-AD^2}=2\sqrt{3} A E F = 1 2 × A B × B F = 1 2 × 2 3 × 2 = 2 3 = 3.464 ∆AEF=\dfrac{1}{2}\times AB\times BF=\dfrac{1}{2}\times 2\sqrt{3}\times 2=2\sqrt{3}=\boxed{3.464}

The apothem is calculated with the formula sqrt(side^2 - (side/2)^2)

sqrt(2^2 - (2/2)^2) = sqrt(3)

The height of the triangule is 2 * apothem => 2 sqrt(3)

The area of a triangle is calculated with the formula (base * height)/2

=> 2*2sqrt(3)/2 = 2 sqrt(3)

Ans: 2 sqrt(3)

Anna Anant
Jan 6, 2015

Area of ABG = 1/2(AB)(h) join E and A and h=EA such that angle F meets EA at O, F is divided in two angles F= EFO+AFO=120 (Hexagon's angle) EFO=30 and AFO=90 EO=FE Cos30 , OA= FA Cos30 EO=2(0.866), OA=2(0.866) EO=1.732, OA=1.732 EA=EO+OA EA=3.464 A=1/2(h)(AB) A=1/2(3.464)(2) A=3.464 sq cm

Amerigo Jumalon
Jan 6, 2015

A=1/2 bh; a=1cot 30=sqrt3 and h=2a Substituting, A=1/2 x 2 x 2sqrt3 =2x sqrt3=3.464sqcm.

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